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AP 9th Class Maths Chapter 7 Triangles Important Questions

September 16, 2024September 16, 2024 by Srinivas

AP 9th Class Maths Important Questions Chapter 7 Triangles

These AP 9th Class Maths Important Questions 7th Lesson Triangles will help students prepare well for the exams.

AP Board Class 9 Maths 7th Lesson Triangles Important Questions

9th Class Maths Triangles 2 Marks Important Questions

Question 1.
State SAS congruency rule.
Solution:
Two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of the other triangle.

Question 2.
State SSS congruency rule.
Solution:
If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

Question 3.
State RHS congruency rule.
Solution:
If in two right triangles, the hypotenuse and one side of one triangle are equal to the hypotenuse and corresponding side of the other triangle, then the two triangles are congruent.

Question 4.
Write the properties of a triangle relating to its sides.
Solution:

The sum of any two sides of a triangle is greater than the third side.
The difference of any two sides of a triangle is less than the third side.

Question 5.
What are called congruent figures ?
Solution:
Figures which are identical, i.e., having same shape and size are called congruent figures.

Question 6.
Triangle ABC is an isosceles right angled triangle in which ∠A = 90°. Find ∠B.
(OR)
ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Solution:

∵ ABC is an isosceles right angled triangle in which ∠A = 90°.
∴ ∠A = 90° ……………. (1)
AB = AC ………………. (2)
∴ ∠ABC = ∠ACB …………….. (3)
(Angles opposite to equal sides of a triangle are equal)
Again, in Δ ABC,
∠BAC + ∠ABC + ∠ACB = 180°
(Sum of the angles of a triangle is 180°)
90° + ∠ABC + ∠ABC = 180° [From (1) and (3)]
⇒ 2 ∠ABC = 90°
⇒ ∠ABC = 45°
∴ ∠B = 45°
From (3)
∠B = ∠C = 45°
∴ ∠C = 45°

Question 7.
In the given figure, if AB = AC, then prove that ∠ABD = ∠ACE.

Solution:
In ΔABC,
AB = AC (Given)
∴ ∠ACB = ∠ABC (∵ Angles opposite to equal sides of a triangle are equal)
⇒ ∠ABC = ∠ACB …………… (1)
Again,
∠ABC + ∠ABD = 180° ………….. (2)
(Linear Pair Axiom)
and ∠ACB + ∠ACE = 180° ………… (3)
(Linear Pair Axiom)
From (2) and (3)
∠ABC + ∠ABD = ∠ACB + ∠ACE …………. (4)
Subtracting (1) from (4), we get
∠ABD = ∠ACE

Question 8.
In the given figure, if l || m, ∠ABC = ∠ABD = 40° and ∠BAC = ∠BAD = 90°, then prove that ΔBCD is an isosceles triangle.

In ΔBAC and ΔBAD,
∠ABC = ∠ABD (Each = 40°)
∠BAC = ∠BAD (Each = 90°)
AB = AB (Common)
∴ Δ BAC ≅ ΔBAD (ASA Rule)
∴ BC = BD (CPCT)

Question 9.
In the given figure, ABCD is a square and P is the midpoint of AD. BP and CP are joined. Prove that ∠PCB = ∠PBC.

Solution:
In ΔBAP and ΔCDP,
∠BAP = ∠CDP (Each = 90°) (∵ ABCD is a square)
BA = CD (∵ ABCD is a square)
⇒ AP = DP (∵ P is the midpoint of AD)
∴ ΔBAP = ΔCDP (SAS Rule)
∴ ∠ABP = ∠DCP ………… (1) (CPCT)
Again, ∠ABC = ∠DCB (Each = 90°)
⇒ ∠ABP + ∠PBC
= ∠DCP + ∠PCB …………….. (2)
Subtracting (1) from (2), we get
∠PBC = ∠PCB
∴ ∠PCB = ∠PBC

Question 10.
In the given figure ΔABC and ΔDBC are two triangles. How do you show that ΔABC ≅ ΔDBC ?
Solution:

$\overline{\mathrm{AB}}$ = $\overline{\mathrm{BD}}$ (Given)
$\overline{\mathrm{AC}}$ = $\overline{\mathrm{CD}}$ (Given)
BC = CB (Common side)
∴ ΔABC ≅ ADBC. (By SSS congruency rule)

9th Class Maths Triangles 4 Marks Important Questions

Question 1.
Observe the given figure and show that ΔEBL ≅ ΔECD.

Solution:
Consider ΔEBL and ΔECD,
∠BEL = ∠CED
(vertically opposite angles)
BE = EC
∠EBL = ∠ECD
(DC // AL, DL is the transversal
∴ ∠EBL = ∠ECD alternate interior angles)
∴ ΔEBL ≅ ΔECD (by ASA congruency)

Question 2.
In ΔABC, the lines are drawn parallel to BC, CA and AB respectively through A, B, C intersecting at P, Q and R. Find the ratio of perimeter of ΔPQR and ΔABC.
Solution:

From figure, AB // QP and BC // RQ.
So, in ABCQ parallelogram AB // CQ
Similarly, in BCAR, ABPC parallelograms
BC = AQ and BC = RA.
Mid point of QR is ‘A’.
Similarly mid point of PR and PQ is B and C.
∴ AB = $\frac{1}{2}$PQ,
BC = $\frac{1}{2}$ QR and CA = $\frac{1}{2}$ PR.
∴ Perimeter of ΔPQR = PQ + QR + PR
= 2AB + 2BC + 2CA
= 2 (AB + BC + CA)
= 2 (Perimeter of ΔABC)
∴ Ratio of perimeters of ΔPQR and ΔABC is 2 : 1.

Question 3.
In figure, AP and BQ are perpendiculars to the line segment AB and AP = BQ. Prove that O is the mid-point of the line segments AB and PQ.

Solution:
Given : AP and BQ are perpendiculars to the line segment AB and AP = BQ.
To Prove : O is the mid-point of the line segments AB and PQ.
Proof : In Δ OAP and ΔQBQ,
AP = BQ (Given)
∠OAP = ∠OBQ (Each = 90°)
∠AOP = ∠BOQ (Vertically opposite angles)
∴ ΔOAP = ΔOBQ (AAS Rule)
∴ OA = OB (CPCT)
and OP = OQ (CPCT)
⇒ O is the mid-point of the line segments AB and PQ.

Question 4.
In figure, OA = OB, OC = OD and ∠AOB = ∠COD. Prove that AC = BD.

Solution:
Given : OA = OB, OC = OD and ∠AOB = ∠COD
To Prove : AC = BD
Proof : In ΔAOC and ΔBOD,
OA = OB …………. (1) (Given)
OC = OD …………… (2) (Given)
∠AOB = ∠COD (Given)
⇒ ∠AOB – ∠COB = ∠COD – ∠COB
(Subtracting ∠COB from both sides)
⇒ ∠AOC = ∠BOD ……………. (3)
In view of (1), (2) and (3),
ΔAOC ≅ ΔBOD (SAS Rule)
∴ AC = BD (CPCT)

Question 5.
In figure, AB = BC, AD = EC. Prove that ΔABE ≅ ΔCBD.

Solution:
Given : AB = BC, AD = EC.
To Prove : ΔABE ≅ ΔCBD
Proof : In ΔABC,
AB = BC ………. (1) (Given)
∴ ∠BCA = ∠BAC
Angles opposite to equal sides of a triangle are equal
⇒ ∠BCD = ∠BAE
⇒ ∠BAE = ∠BCD ………….. (2)
AD = EC (Given)
⇒ AD + DE = EC + DE (Adding DE to both sides)
⇒ AE = CD ……………. (3)
Now, in Δ ABE and ∠CBD,
AB = CB [From (1)]
∠BAE = ∠BCD [From (2)]
AE = CD [From (3)]
∴ Δ ABE ≅ Δ CBD (SAS Rule)

Question 6.
In figure, if PS = PR, ∠TPS = ∠QPR prove that PT = PQ.

Solution:
Given : PS = PR, ∠TPS = ∠QPR.
To Prove : PT = PQ.
Proof: In ΔPSR,
PS = PR (Given)
∴ ∠PRS = ∠PSR (Angles opposite to equal sides of a triangle are equal)
⇒ ∠PSR = ∠PRS
⇒ ∠PTS + ∠TPS = ∠PQR + ∠QPR (An exterior angle of a triangle is equal to the sum of its two interior opposite angles)
⇒ ∠PTS = ∠PQR ∵ ∠TPS = ∠QPR (given)
⇒ ∠PTQ = ∠PQT
∴ PQ = PT (Sides opposite to equal angles of a triangle are equal)
∴ PT = PQ

Question 7.
In ΔABC, D is the mid-point of BC. The perpendiculars from D to AB and AC are equal. Prove that ΔABC is isosceles.
Solution:
Given : In ΔABC, D is the mid-point of BC. The perpendiculars from D to AB and AC are equal.
To Prove : ΔABC is isosceles.

Proof : In right triangles DEC and DFB, Hyp. DC = Hyp. DB (∵ D is the mid-point BC)
Side DE = Side DF (given)
∴ ΔDEC ≅ ΔDFB (RHS Rule)
∴ ∠DCE = ∠DBF (CPCT)
⇒ ∠CBA = ∠BCA
⇒ ∠B = ∠C
∴ AC = AB (Sides opposite to equal angles of a triangle are equal)
⇒ AB = AC
∴ Δ ABC is isosceles.

Question 8.
In figure, D is any point on the base BC produced of an isosceles triangle ABC. Prove that AD > AB.

Solution:
Given : D is any point on the base BC produced of an isosceles triangle ABC.
To Prove : AD > AB
Proof : ∵ ABC is an isosceles triangle.
∴ AB = AC
∴ ∠ACB = ∠ABC ………. (1)
(Angle opposite to equal sides of a triangle are equal)
In Δ ACD,
Ext. ∠ACB > ∠CDA (An exterior angle of a triangle is greater than each of its two interior opposite angles) ‘
⇒ ∠ABC > ∠CDA [From (1)]
⇒ ∠ABD > ∠BDA
∴ AD > AB (Side opposite to greater angle is longer)

Question 9.
The following facts are given about triangles ABC, DEF PQR, LMN and XYZ.
AB = ED = PQ, BC = EF = QR,
PQ = LM = XY, AC = DF, LN = XZ
∠E – ∠Q = ∠M, ∠P = ∠L and
∠N = ∠Z = 90°
Based on the above information, answer the following questions:

i) Then by what criteria is ΔABC congruent to ΔDEF
A) SSS
B) SAS
C) ASA
D) RHS
Answer:
A) SSS

ii) Then by what criteria is ΔDEF congruent to ΔPQR ?
A) SSS
B) SAS
C) ASA
D) RHS
Answer:
B) SAS

iii) Then by what criteria is ΔPQR congruent to ΔLMN ?
A) SSS
B) SAS
C) ASA
D) RHS
Answer:
C) ASA

iv) Then by what criteria is ΔLMN congruent to ΔXYZ ?
A) SSS
B) SAS
C) ASA
D) RHS
Answer:
D) RHS

v) Which of the following is ture ?
A) Any two isosceles triangles with same perimeter are congruent
B) Any two right angles are congruent
C) Any two equilateral triangles with same perimeter are always congruent
D) None of the above
Answer:
C) Any two equilateral triangles with same perimeter are always congruent

9th Class Maths Triangles 8 Marks Important Questions

Question 1.
In ΔPQR, PS ⊥ QR and ΔPQS ≅ ΔPRS.

PQ = 2x + 3, PR = 3y + 1, QS = x,
SR = y + 1. Find the area of ΔPQR.
Solution:

In ΔPQR, PS ⊥ QR, so ΔPRS ≅ ΔPQS
From CPCT,
PQ = PR and QS = SR and 2x + 3 = 3y + 1 and x = y + 1
2x – 3y = -2 ……………… (1) and
x – y = 1 …………. (2)
Substitute x = y + 1 in equation (1)
2(y + 1) – 3y = -2
2y + 2 – 3y = -2
-y = -4
⇒ y = 4
by y = 4 in x = y+ 1 ⇒ x = 5
∴ QR = x + y + 1 = 5 + 4 + 1 = 10 units
PQ = PR = 2(5) + 3 = 13 units
QS = x = 5 units.
∴ ΔPQS is a right triangle.
From pythagoras theorem,
PQ² = QS² + PS²
PS² = PQ² – QS²
PS² = (13)² – (5)² = 144
∴ PS = 12 units
∴ Area of ΔPQR = $\frac{1}{2}$ × QR × PS
= $\frac{1}{2}$ × 10 × 12 = 60 sq.units.

Question 2.
In the given figure $\overline{\mathbf{A B}}$ || $\overline{\mathbf{C D}}$ || $\overline{\mathbf{E F}}$ at equal distances equal distances and AF is a transversal $\overline{\mathbf{G H}}$ is perpendicular to $\overline{\mathbf{A B}}$. If AB = = 4.5 cm, GH = 4 cm and FB = 8 cm, find the area of ΔGDF.

Solution:
In the given figure $\overline{\mathbf{A B}}$ || $\overline{\mathbf{C D}}$ || $\overline{\mathbf{E F}}$ at equal distances.
$\overline{\mathbf{A F}}$ is a transversal and $\overline{\mathbf{G H}}$ is perpendicular to $\overline{\mathbf{A B}}$.
So, GH = 4 cm, AB = 4.5 cm, FB = 8 cm To find area of ΔGDF,
From ΔABF, D is mid point of $\overline{\mathbf{B F}}$
Similarly ‘G’ is mid point of $\overline{\mathbf{A F}}$.
So BD ⊥ DF, AG = GF
Median divides a triangle into two equal triangles.
∴ Area of ΔABG = Area of ΔBGF
Similarly $\overline{\mathbf{G D}}$ is median of ΔBGF.
Area of ΔABG = 2 × area of ΔDGF
Area of ΔDFG = $\frac{1}{2}$ area of ΔABG
Area of ΔABG = $\frac{1}{2}$ × 4.5 × 4 = 9 sq.cm
Area of ΔDGE = $\frac{1}{2}$ × 9 = 4.5 sq.cm

Question 3.
In figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Solution:
Given : In figure, PR > PQ and PS bisects ∠QPR.
To Prove : ∠PSR > ∠PSQ
Proof : In Δ PQR
PR > PQ (Given)
∴ ∠PQR > ∠PRQ ………….. (1)
(Angle opposite to longer side is greater)
∵ PS is the bisector of ∠QPR
∴ ∠QPS = ∠RPS ………… (2)
In Δ PQS,
∠PQR + ∠QPS + ∠PSQ = 180°………… (3)
(∵ The sum of the three angles of a Δ is 180°)
In Δ PRS,
∠PRS + ∠SPR + ∠PSR = 180° ……….. (4)
[∵ The sum of the three angles of a Δ is 180°]
From (3) and (4),
∠PQR + ∠QPS + ∠PSQ = ∠PRS + ∠SPR + ∠PSR
⇒ ∠PQR + ∠PSQ = ∠PRS + ∠PSR
⇒ ∠PRS + ∠PSR = ∠PQR + ∠PSQ
⇒ ∠PRS + ∠PSR > ∠PRQ + ∠PSQ [From (1)]
⇒ ∠PRQ + ∠PSR > ∠PRS + ∠PSQ (∵ ∠PRQ = ∠PRS)
⇒ ∠PSR > ∠PSQ

Question 4.
Prove that the medians of an equilateral triangle are equal.
Solution
Given : ABC is an equilateral triangle whose medians are AD, BE and CF.
To Prove : AD = BE = CF.

Proof :
In Δ ADC and Δ BEC,
AC = BC ………….. (1)
(∵ Δ ABC is equilateral ∴ AB = BC = CA)
∠ACD = ∠BCE ……….. (2) (Common angle)
∵ AD is a median.
∴ DC = DB = $\frac{1}{2}$ BC
∵ BE is a median.
∴ EA = EC = $\frac{1}{2}$AC
∵ AC = BC
∴ DC = EC ……………. (3)
In view of (1), (2) and (3),
Δ ADC ≅ Δ BEC (SAS Rule)
∴ AD = BE ………. (4) (CPCT)
Similarly, we can prove that
BE = CF …………. (5)
and CF = AD …………(6)
From (4), (5) and (6), we get
AD = BE = CF

Question 5.
In the given figure, AB = AC. D is a point on AC and E on AB such that AD = ED = EC = BC. Prove that ∠A : ∠B = 1 : 3.

Solution:
Let ∠A = x° …………… (1)
In Δ AED,
∵ AD = ED (given)
∴ ∠A = ∠DEA (Angles opposite to equal sides of a triangle are equal)
⇒ x° = ∠DEA
⇒ ∠DEA = x° …………… (2)
In Δ EDC,
exterior ∠EDC = ∠DEA + ∠A (An exterior angle of a triangle is equal to the sum of its two interior opposite angles)
= x° + x° [From (1) and (2)]
= 2x° …………….. (3)
In Δ CED,
∵ EC = ED (Given)
∴ ∠ECD = ∠EDC (Angles opposite to equal sides of a triangle are equal)
⇒ ∠ECD = 2x° ………(4) [From (3)]
In Δ AEC,
exterior ∠BEC = ∠ECD + ∠EAC (An exterior angle of a triangle is equal to the sum of its two interior opposite angles)
∠BEC = 2x° + x° [From (4) and (1)]
⇒ ∠BEC = 3xc ………….. (5)
In Δ BCE,
∵ BC = CE (Given)
∴ ∠CBE = ∠BEC (Angles opposite to equal sides of a triangle are equal)
⇒ ∠B = 3x° ……………(6)[From (5)]
⇒ ∠B = 3∠A . [From (1)]
⇒ ∠A : ∠B = 1 : 3

Question 6.
In figure, AB = AC, CH = CB and HK || BC. If ∠CAX = 137°, then find ∠CHK.

Solution:
Given : AB = AC, CH = CB and HK || BC.
∠CAX = 137°.
To Determine : ∠CHK
Determination : ∵ BX is a line
∴ ∠BAC + ∠CAX = 180° (Linear Pair Axiom)
⇒ ∠BAC + 137° = 180°
∠BAC = 43° ……………. (1)
In Δ ABC,
∠BAC + ∠ABC + ∠BCA = 180° (The sum of the angles of a triangle is 180°)
⇒ 43° + ∠ABC + ∠BCA = 180° [From (1)]
⇒ ∠ABC + ∠BCA = 137° ………….. (2)
In Δ ABC,
∴ AB = AC (Given)
∠ACB = ∠ABC ……….. (3)
(Angles opposite to equal sides of a triangle are equal)
From (2) and (3),
∠ABC = ∠BCA = $\frac{137^{\circ}}{2}$ = 68.5° …………… (4)
In Δ HBC,
CH = CB (Given)
∴ ∠CBH = ∠CHB (Angles opposite to equal sides of a triangle are equal)
⇒ ∠CHB = ∠CBH
= ∠ABC
= 68.5° …………. (5) [From (4)]
Again, HK || BC (Given)
∴ ∠ABC + ∠BHK = 180° (The sum of the consecutive interior angles on the same side of a transversal is 180°)
⇒ 68.5° + (∠CBH + ∠CHK) = 180° [From (4)]
⇒ 68.5° + 68.5° + ∠CHK = 180° [From (5)]
⇒ 137° + ∠CHK = 180°
⇒ ∠CHK = 180° – 137° = 43°

Question 7.
In figure, ∠ACB is a right angle and AC = CD and CDEF is a parallelogram. If ∠FEC = 10°, then calculate ∠BDE.

Solution:
Given: ∠ACB is a right angle afid AC = CD and CDEF is a parallelogram. ∠FEC = 10°.
To Calculate : ∠BDE
Calculation : AC = CD (Given)
∴ ∠ADC = ∠DAC ………… (1)
(Angles opposite to equal sides of a triangle are equal)
∵ DC || EF (∵ CDEF is a parallelogram and opposite sides of a parallelogram are parallel)
and a transversal EC intersects them
∴ ∠FEC = ∠ECD (Alternate Interior Angles)
⇒ 10° = ∠ECD[∵ ∠FEC = 10° (given)]
∠ECD = 10° …………… (2)
⇒ ∠ACB = 90°(∵ ∠ACB is a right angle)
⇒ ∠ACD + ∠ECD = 90°
⇒ ∠ACD + 10° = 90° [From (2)]
⇒ ∠ACD = 80° ………….. (3)
In Δ ACD,
∠ACD + ∠ADC + ∠DAC = 180°
(∵ The sum of the angles of a triangle is 180°)
⇒ 80° + ∠ADC + ∠DAC = 180° From (3)
⇒ ∠ADC + ∠DAC – 100° ………… (4)
From (1) and (4),
∠ADC + ∠DAC = $\frac{100^{\circ}}{2}$ = 50° ………….. (5)
∵ CDEF is a parallelogram.
∴ DE || CF (Opposite sides of a parallelogram are parallel)
⇒ DE || AF
and a transversal BA intersects them
∴ ∠BDE = ∠BAC (Corresponding angles)
⇒ ∠BDE = 50° [From (5)]

Question 8.
In triangle, prove that sum of two sides of a triangle is greater than the third side,
Solution:
Given : Δ ABC
To Prove:
AB + AC > BC
AB + BC > AC
BC + AC > AB

Construction : Produce side BA to D such that
AD = AC
Proof: In ∠ACD,
AC = AD (By construction)
∴ ∠ADC = ∠ACD (Angles opposite to equal sides of a triangle are equal)
⇒ ∠ACD = ∠ADC
⇒ ∠BCA + ∠ACD > ∠ADC
⇒ ∠BCD > ∠ADC
⇒ ∠BCD > ∠BDC ⇒ BD > BC (Side opposite to greater angle is longer)
⇒ BA + AD > BC
⇒ BA + AC > BC (By construction AD = AC)
⇒ AB + AC > BC
Similarly, we can prove that
AB + BC > AC
and BC + AC > AB.

Question 9.
There is a triangular park ABC whose two corner angles A and B are 50° and 60° respectively. Three friends Rashmi, Sita and Geeta go daily for a morning walk and walk along these three sides AB, BC and AC respectively.
Who walks maximum distance along these three ?
Who walks least ?
Why morning walk is necessary for us?
Solution:
∠A = 50°, ∠B = 60°
∠A + ∠B + ∠C = 180° (The sum of the three angles of a triangle is 180°)
⇒ 50° + 60° + ∠C = 180°
⇒ ∠C = 70°
∵ ∠C > ∠A
∴ AB > BC …………. (1)
(Side opposite to greater angle of a triangle is longer)
∵ ∠C > ∠B
∴ AB > CA ……….. (2)
(Side opposite to greater angle of a triangle is longer)
∴ ∠B > ∠A
AC > BC ……… (3)
(Side opposite to greater angle of a triangle is longer)
From (1), (2) and (3),
AB > AC > BC
⇒ AB is the longest side and BC is the smallest side.
∴ Rashmi walks maximum distance and Sita walks least distance.
(∵ Rashmi walks along AB and Sita walks along BC)
Morning walk is necessary for us to maintain our physical health. We get oxygen from the trees to the greatest amount in the morning. Also, we enjoy calm and quiet atmosphere in the morning. Moreover, we feel fresh in the morning which enables us to complete all the mental activities upto the end of the day quite smoothly and rationally.

Question 10.
In the given figure, ABCD is a square and EF is parallel to the diagonal BD. If EM = FM, then prove that
i) DF = BE
ii) AM bisects ∠BAD.

Solution:
Given : ABCD is a square and EF is par-allel to the diagonal BD.
EM = FM
To Prove : i) DF = BE
ii) AM bisects ∠BAD.

Construction : Join AE and AF.
Proof: i) In Δ CBD,
CB = CD ………… (1)
[∵ ABCD is a square]
∵ ∠CDB = ∠CBD ……………. (2)
[Angles opposite to equal sides of a triangle are equal]
EF || BD (Given)
∵ ∠CBD = ∠CEF ………… (3)
(Corresponding angles) and ∠CDB = ∠CFE ……….. (4)
(Corresponding angles)
From (2), (3) and (4),
∠CEF = ∠CFE
∴ CF = CE
(Sides opposite to equal angles of a triangle are equal)
⇒ CE = CF ………… (5)
Subtracting (5) from (1),
CB – CE = CD – CF
⇒ BE = DF
⇒ DF = BE …………. (6)

ii) In right triangle ADF,
AF² = AD² + DF² (By Pythagoras Theorem)
= AB² + BE² [∵ ABCD is a square and from (6)]
= AE² (By Pythagoras Theorem)
⇒ AF = AE …………… (7)
In Δ AMF and Δ AME,
MF = ME (Given)
AM = AM (Common)
AF = AE [From (7)]
∴ Δ AMF ≅ Δ AME (SSS Rule)
∴ ∠MAF = ∠MAE …………. (8) (CPCT)
Now, in right Δ ADF and right Δ ABE,
AF = AE [From (7)]
AD = AB (∵ ABCD is a square)
∴ Δ ADF ≅ Δ ABE (RHS Rule)
∴ ∠FAD = ∠EAB ……………. (9) (CPCT)
Adding (8) and (9), we get
∠MAF + ∠FAD = ∠MAE + ∠EAB
⇒ ZMAD = ZMAB
⇒ AM bisects ∠BAD.

AP 9th Class Maths Important Questions Chapter 7 Triangles

Question 1.
The interior angles of a triangle are (3x – 10)°, (3x + 10)°, (3x)°. Find the angles.
Solution:
Interior angles of a triangle are (3x- 10)°, (3x + 10)° and (3x)°
Sum of angles = (3x – 10)° + (3x + 10)° + (3x)° = 180°
9x = 180°
x = 20°
∴ Angles = 3 × 20 – 10 = 60 – 10 = 50°
= 3 × 20 + 10 = 60 + 10 = 70°
= 3 × 20° ± 60°
∴ Angles are 50°, 70° and 60°.

Question 2.
If ΔABC ≅ ΔPQR, express equal sides of triangles.
Solution:
If Δ ABC ≅ ΔPQR
⇒ AB = PQ, BC = QR, AC = PR

Question 3.
In ΔABC, the lines are drawn parallel to BC, CA and AB respectively through A, B, C intersecting at P, Q and R. Find the ratio of perimeter of ΔPQR and ΔABC
Solution:

From figure, AB // QP and BC // RQ.
So, in ABCQ parallelogram AB // CQ
Similarly, in BCAR, ABPC parallelograms BC = AQ and BC = RA.
Mid point of QR is ‘A’.
Similarly mid point of PR and PQ is B and C.
∴ AB = $\frac{1}{2}$ PQ,
BC = $\frac{1}{2}$ QR and CA = $\frac{1}{2}$ PR.
∴ Perimeter of ΔPQR = PQ + QR + PR
= 2AB + 2BC + 2CA
= 2 (AB + BC + CA)
= 2(Perimeter of ΔABC)
∴ Ratio of perimeters of ΔPQR and ΔABC is 2 : 1.

Question 4.
In the adjacent figure, prove that ΔPQX ≅ ΔPRY.

Solution:
In ΔPQX, ΔPRY .
Sides QX = RY {Given}
Sides PX = PY {Given}
∠PXY = ∠PYX
180 – ∠PXY = 180° – ∠PYX
⇒ ∠PXQ = ∠PYR {Angle}
By S – A – S congruence criterion.
ΔPQX ≅ ΔPRY

Question 5.
In the given figure, AL is parallel to DC and E Is mid point of BC. Show that triangle EBL Is congruent to triangle ECD.

Solution:
From the adjacent figure AL // CD and E is the midpoint of BC.
i.e., BE = EC
from ΔBEL & ΔCED.
∠B = ∠C [ ∵ Alternate interior angles]
∠DEC = ∠BEL [ ∵ vertically opposite, angles]
∴ By A.S.A rule
Δ EBL ≅ Δ ECD.

Question 6.
ABC is a right angled triangle and M is the midpoint of the hypotenuse AB. A line segment MD, parallel to BC, passes through M and meets AC at D. Prove that (i) D is midpoint of AC, (ii) MD is perpendicular to AC.

Solution:
Given that ΔABC is a right angled triangle in which, M is the midpoint of AB. Also $\overline{\mathrm{MD}}$ / / $\overline{\mathrm{BC}}$.
∠C = 90° ⇒ BC⊥AC then MD JL AC [ ∵ BC // MD]
If a line is drawn parallel to one side and (divides the 2^nd side into two equal parts) passes through the midpoint of second side, it will also passes through the midpoint of 3^rd side, i.e., MD bisects AC.
i.e., D is the midpoint of AC.

Question 7.
In ΔPQR, PS⊥QR and ΔPQS ≅ ΔPRS.

PQ = 2x + 3, PR = 3y + 1, QS = x,
SR = y + 1. Find the area of ΔPQR.
Solution:

In ΔPQR, PS⊥QR, so ΔPRS ≅ ΔPQS From CPCT,
PQ = PR and QS = SR and 2x + 3 = 3y + 1 and x = y + 1
2x – 3y = -2 ………………… (1) and
x – y = 1 ………….. (2)
Substitute x = y + 1 in equation (1)
2(y + 1) – 3y = -2
2y + 2 – 3y = -2
-y = -4
⇒ y = 4
by y = 4 in x = y+ 1 ⇒ x = 5
∴ QR = x + y+ l= 5 + 4+ l = 10units
PQ = PR = 2(5) + 3 = 13 units
QS = x = 5 units.
∴ ΔPQS is a right triangle.
From pythagoras theorem,
PQ² = QS² + PS²
PS² = PQ² – QS²
PS² = (13)³ – (5)² = 144
∴ PS = 12 units
∴ Area of ΔPQR = $\frac{1}{2}$ × QR × PS
= $\frac{1}{2}$ × 10 × 12 = 60,sq.units.

Question 8.
In the given figure $\overline{\mathrm{AB}}$ || $\overline{\mathrm{CD}}$ || $\overline{\mathrm{EF}}$ at equal distances and AF is a transversal. $\overline{\mathrm{GH}}$ is perpendicular to $\overline{\mathrm{AB}}$. If AB = 4.5 cm, GH = 4 cm and FB = 8 cm, find the area of ΔGDF.

Solution:
In the given figure $\overline{\mathrm{AB}}$ || $\overline{\mathrm{CD}}$ || $\overline{\mathrm{EF}}$ at equal distances.
$\overline{\mathrm{AF}}$ is a transversal and $\overline{\mathrm{GH}}$ is perpendicular to $\overline{\mathrm{AB}}$.
So, GH = 4 cm, AB = 4.5 cm, FB = 8 cm To find area of ΔGDF,
From ΔABF, D is mid point of $\overline{\mathrm{BF}}$
Similarly ‘G’ is mid point of $\overline{\mathrm{AF}}$.
So BD ⊥ DF, AG = GF
Median divides a triangle into two equal triangles.
∴ Area of ΔABG = Area of ΔBGF
Similarly $\overline{\mathrm{GD}}$ is median of ΔBGF.
Area of ΔABG = 2 × area of ΔDGF
Area of ΔDFG = $\frac{1}{2}$ area of ΔABG
Area of ΔABG = $\frac{1}{2}$ × 4.5 × 4 = 9 sq.cm
Area of ΔDGE = $\frac{1}{2}$ × 9 = 4.5 sq.cm

Question 9.
In ABC, E and F are mid points of sides AB and AC respectively then prove that i) EF // BC and ii) EF = $\frac{1}{2}$BC.
Solution:
Given : B and F are mid points of AB and AC.

R.T.P. : i) EF // BC, ii) EF = $\frac{1}{2}$ BC
Construction: ..
GC // AB, extend EF upto G.
Proof:
ΔAEF ACGF
∠AFE = ∠CFG (Vertically opposite angles)
AF = FC
∠EAF = ∠GCF (Alternate angles)
∴ ΔAEF = ΔCGF
∴ CG = BE and CG // BF (Construction)
∴ EBCG is a parallelogram.

Question 10.
In triangle ABC, D is a point on BC such that ΔABD and ΔACD are congruent. Find the values of x and y, if AB = 2x + 3, AC = 3y + 1, BD = x and
DC = y + 1.
Solution:
From the adjacent figure ΔABC, D is a point on BC.

Such that ΔABD ≅ ΔACD

AB = 2x + 3
AC = 3y + 1

BD = x
DC = Y + 1

Δ ABD = Δ ACD
⇒ AB = AC
⇒ 2x + 3 = 3y + 1
⇒ 2x – 3y + 2 = 0 …………….(1)
BD = CD ⇒ x = y + 1
⇒ x – y – 1 = 0 ………………….(2)
⇒ from (1) & (2)

y = 4
from (2) ⇒ x – 4 = 1
⇒ x = 1 + 4 = 5
∴ (x, y) = (5, 4)

Question 11.
In triangle ΔABC, D, E and F are mid points of sides AB’, BC and AC respectively. If we join points D, E and F, then show that triangle ABC is divided into four congruent triangles.

Solution:
Given : In ABC, D,E,F are the midpoints of sides AB, BC and AC respectively. .
R.T.P : ΔADF ≅ ΔDBE ≅ ΔDEF ≅ ΔCEF
Proof : D, E are the midpoints of $\overline{\mathrm{AB}}$ and $\overline{\mathrm{BC}}$ of ΔABC respectively.
So by mid-point theorem, we know that DE // AC
Similarly DF//BC and EF // AB.
∴ ADEF, BEFD, CFDE are all parallelograms.
In a parallelogram ADEF, DF is the diagonal.
So, ΔADF ≅ ΔDEF …………….. (1)
[∵ Diagonal divides the parallelogram into two congruent triangles]
Similarly ΔBDE ≅ ΔDEF ………………… (2)
and ΔCEF ≅ ΔDEF …………….. (3)
From (1), (2) & (3) ⇒ ΔADF ≅ ΔDEF ≅ ΔBDE ≅ ΔCEF all the triangles are congruent to each other,
i.e., ΔABC divides into four congruent triangles by joining the mid points of the sides.

AP 9th Class Telugu Important Questions and Answers 2024-2025

September 6, 2024August 31, 2024 by AP Board

AP Board 9th Class Telugu Important Questions and Answers 2021-2022

Andhra Pradesh SCERT AP State Board Syllabus 9th Class Telugu Chapter Wise Important Questions and Answers are part of AP Board Solutions Class 9.

Students can also read AP 9th Class Telugu Gudie for exam preparation.

AP Board 9th Class Telugu Lessons Important Questions and Answers 2024-2025

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AP 9th Class Telugu Important Questions (Old Syllabus)

AP 10th Class Maths Chapter 6 Important Questions Triangles

August 23, 2024August 22, 2024 by Mahesh

AP 10th Class Maths Important Questions Chapter 6 Progressions

These AP 10th Class Maths Chapter Wise Important Questions 6th Lesson Triangles will help students prepare well for the exams.

6th Lesson Triangles Class 10 Important Questions with Solutions

10th Class Maths Triangles 1 Mark Important Questions

Question 1.
Write the properties of similar triangles.
Answer:
Two triangles are said to be similar, if their

corresponding angles are equal.
corresponding sides are proportional.

Question 2.
Write the examples of similar figures.
Answer:
All circles, all squares, all equilateral triangles, etc…

Question 3.
State the Thales theorem.
Answer:
If a line drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

Question 4.
State the converse of basic proportionality theorem.
Answer:
If a line divides any two sides of a tri¬angle in the same ratio, then the line must be parallel to the third side.

Question 5.
State the pythagoras theorem.
Answer:
In a right angled triangle square of the hypotenuse is equal to the sum of the squares of the other two sides.

That is in ΔABC, ∠B = 90° then AC² = AB² + BC².

Question 6.
In ΔABC and ΔDEF, $\frac{\mathrm{AB}}{\mathrm{DE}}$ = $\frac{\mathrm{BC}}{\mathrm{FD}}$.Which of the following makes the two triangles similar ?
A) ∠A = ∠D
B) ∠B = ∠D
C) ∠B = ∠E
D) ∠A = ∠F
Solution:
B) ∠B = ∠D

$\frac{\mathrm{AB}}{\mathrm{DE}}$ = $\frac{\mathrm{BC}}{\mathrm{FD}}$
∴ ∠B = ∠D
ΔABC ~ ΔEDF

Question 7.
In the given fig. DE || BC. Find BC.

Solution:
$\frac{\mathrm{AD}}{\mathrm{AB}}$ = $\frac{\mathrm{DE}}{\mathrm{BC}}$
$\frac{2}{2+3}$ = $\frac{4}{\mathrm{BC}}$
$\frac{2}{5}$ = $\frac{4}{\mathrm{BC}}$
2BC = 20
BC = 10 cm.

Question 8.
In ΔABC, PQ || BC. If PB = 6 cm, AP = 4 cm, AQ = 8 cm, find the length of AC.

Solution:
PQ || BC
By BPT

$\frac{4}{6}$ = $\frac{8}{\mathrm{QC}}$
$\frac{2}{3}$ = $\frac{8}{\mathrm{QC}}$
2QC = 24
QC = 12
AC = 8 + 12 = 20 cm

Question 9.

In the given figure, ΔABC ~ ΔQPR, If AC = 6 cm, BC = 5 cm, QR = 3 cm and PR = x; then the value of x is :
Solution:
ΔABC ~ ΔQPR

$\frac{6}{3}$ = $\frac{5}{2}$
2 = $\frac{5}{x}$
x = $\frac{5}{x}$ = 2.5 cm

Question 10.
If in ΔABC and ΔPQR, $\frac{\mathrm{AB}}{\mathrm{QR}}$ = $\frac{\mathrm{BC}}{\mathrm{PR}}$ = $\frac{\mathrm{CA}}{\mathrm{PQ}}$ then :
A) ΔPQR ~ ΔCAB
B) ΔPQR ~ ΔABC
A) ΔCBA ~ ΔPQR
B) ΔBCA ~ ΔPQR
Solution:
A) ΔPQR ~ ΔCAB

ΔPQR ~ ΔCAB

Question 11.
Which of the following is not a similarity criterion of triangle ?
A) AA
B) SAS
C) AAA
D) RHS
Answer:
D) RHS

Question 12.
If ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm, then ∠B = …………… .
Solution:

AB² = (6√3)² = 36 × 3 = 108
AC² = 12² – 144
BC² = 6² = 36
AC² = AB² + BC²
144 = 108 + 36
144 = 144
∠B = 90°

Question 13.
In the given fig. DE || BC. Which of the following is true ?

A) x = $\frac{a+b}{a y}$
B) y = $\frac{a x}{a+b}$
C) x = $\frac{a y}{a+b}$
D) $\frac{x}{y}$ = $\frac{a}{b}$
Solution:
C) x = $\frac{a y}{a+b}$
DE || BC, ΔADE ~ ΔABC
$\frac{a}{x}$ = $\frac{a+b}{y}$
x(a + b) = ay
x = $\frac{a y}{a+b}$

Question 14.
In ΔABC and ΔDEF, ∠F = ∠C, ∠B = ∠E and AB = $\frac{1}{2}$ DE then the two triangles are
A) congruent, but not similar not congruent
B) similar, but not congruent
C) neither congruent nor similar
D) congruent as well as similar
Solution:
B) similar, but not congruent

AB = $\frac{1}{2}$ DE
2AB = DE
ΔABC ~ ΔDEF but not congruent.

Question 15.
In the given fig. If DE || BC. Find EC.

DE || BC
$\frac{\mathrm{AD}}{\mathrm{DB}}$ = $\frac{\mathrm{AE}}{\mathrm{EC}}$
$\frac{1.5}{3}$ = $\frac{1}{\mathrm{EC}}$
EC = 2 cm

Question 16.
In the given figure, AD = 2 cm, DB = 3 cm, DE = 2.5 cm and DE || BC. Find the value of x.

Solution:
ΔADE ~ ΔABC
$\frac{\mathrm{AD}}{\mathrm{DB}}$ = $\frac{D E}{\mathbf{x}}$
$\frac{2}{5}$ = $\frac{2.5}{x}$
2x = 2.5 × 5
x = 2.5 × $\frac{5}{2}$
x = 2.5 × 2.5
x = 6.25 cm

Question 17.
In two triangles ΔPQR and ΔABC, it is given that $\frac{\mathrm{AB}}{\mathrm{BC}}$ = $\frac{\mathrm{PQ}}{\mathrm{PR}}$ For these two triangles to be similar, which of the following should be true ?
A) ∠A = ∠P
B) ∠B = ∠Q
C) ∠B = ∠P
D) ∠A = ∠R
Solution:
C) ∠B = ∠P

$\frac{\mathrm{AB}}{\mathrm{BC}}$ = $\frac{\mathrm{PQ}}{\mathrm{PR}}$
$\frac{\mathrm{AB}}{\mathrm{PQ}}$ = $\frac{\mathrm{BC}}{\mathrm{PR}}$
ΔPQR ~ ΔBAC
∠P = ∠B.

Question 18.
In ΔABC, DE || BC, AD = 4 cm, BD = 6 cm and AE = 5 cm, find the length of EC.
Solution:

DE || BC
By BPT
$\frac{4}{6}$ = $\frac{5}{\mathrm{EC}}$
$\frac{2}{3}$ = $\frac{5}{\mathrm{EC}}$
EC = $\frac{15}{2}$ = 7.5 cm

Question 19.
In ΔABC, DE || BC, AD = 2 cm, BD = 3 cm, DE: BC is equal to ……………… .

Solution:
DE || BC, ΔADE, ΔBC
∠ A = ∠A (Common angle)
∠D = ∠B (Corresponding angle) by AA Similarity
ΔADE ~ ΔABC
$\frac{\mathrm{AD}}{\mathrm{AB}}$ = $\frac{\mathrm{AE}}{\mathrm{AC}}$ = $\frac{\mathrm{DE}}{\mathrm{BC}}$
$\frac{2}{2 + 3}$ = $\frac{\mathrm{DE}}{\mathrm{BC}}$
$\frac{2}{5}$ = $\frac{\mathrm{DE}}{\mathrm{BC}}$
DE : BC = 2 : 5

Question 20.
In the given fig. DE || BC. If AD = 3 cm, AB = 7 cm and EC = 3 cm, then find the length of AE.

DE || BC, BD = 7 – 3 = 4 cm
By BPT $\frac{3}{4}$ = $\frac{\mathrm{AE}}{3}$
4AE = 9
AE = $\frac{9}{4}$
AE = 2.25 cm.

Question 21.
The perimeters of two similar triangles are 24 cm and 18 cm respectively. If one side of the first triangle is 8 cm then what is the corresponding side of second triangle ?
Answer:
6

Question 22.
In ΔABC, ∠B = 90°, BD ⊥ AC.
If AD = 8 cm and BD = 4 cm, then what is the length of CD ?
Answer:
2 cm.

Question 23.
Among “Circles”, “Squares” and “Triangles”, which are always similar ?
Answer:
Circles and squares are always similar.

Question 24.
In ΔABC and ΔDEF,
if ∠B = ∠E, ∠C = ∠F, then which of the following is a true statement ?
A) $\frac{\mathrm{AB}}{\mathrm{DE}}$ = $\frac{\mathrm{CA}}{\mathrm{EF}}$
B) $\frac{\mathrm{BC}}{\mathrm{EF}}$ = $\frac{\mathrm{AB}}{\mathrm{FD}}$
C) $\frac{\mathrm{AB}}{\mathrm{DE}}$ = $\frac{\mathrm{BC}}{\mathrm{EF}}$
D) $\frac{\mathrm{CA}}{\mathrm{FD}}$ = $\frac{\mathrm{AB}}{\mathrm{EF}}$
Answer:
C) $\frac{\mathrm{AB}}{\mathrm{DE}}$ = $\frac{\mathrm{BC}}{\mathrm{EF}}$

Question 25.
Given DE || BC, AD = 4.5 cm, BD = 9 cm, and EC = 8 cm. What is the length of AE?

Solution:
AD = 4.5 cm, BD = 9 cm, EC = 8 cm
$\frac{\mathrm{AD}}{\mathrm{DB}}$ = $\frac{\mathrm{AE}}{\mathrm{EC}}$ ⇒ $\frac{4.5}{9}$ = $\frac{\mathrm{AE}}{8}$ ⇒ 9AE = 36
∴ AE = 4 cm

Question 26.
The sides of two equilateral triangles ABC and DEF are 4 cm. and 5 cm. respeclively. Find $\frac{{ar}(\triangle D E F)}{{ar}(\triangle A B C)}$.
Solution:
Ratio of areas of Δ = (Ratio of corresponding sides)²
$\frac{(\triangle \mathrm{DEF})}{(\triangle \mathrm{ABC})}$ = $\frac{5^2}{4^2}$ = $\frac{1}{2}$

Question 27.
ΔABC ~ ΔDEF and ∠A = ∠D = 90°, find the measure of ∠B + ∠F.
Solution:
ΔABC ~ ΔDEF
⇒ ∠A = ∠D = 90°
∠B = ∠E; ∠C = ∠F
Now in ΔABC, ∠A + ∠B + ∠C = 180°
Now putting ∠A = ∠D = 90°
⇒ 90° + ∠B + ∠C = 180°
⇒ ∠B + ∠C = 90°
Now putting ∠C = ∠F
⇒ ∠B + ∠F = 90°

Question 28.
Given AB || CD, identify the pair of similar triangles in this image and justify.

Solution:
∠AEB = ∠CED (Vertically opp. angle)
∴ ∠BEA and ∠CDE are similar by AAA – Similarity rule
ΔBEA ~ ΔCDE

Question 29.
ABC is a triangle in which
p) If AB² + BC² = AC² then
q) BC² + AC² = AB² then
r) CA² + AB² = BC² then

i) ∠A = 90°
ii) ∠B = 90°
iii) ∠C = 90°

A) p – i, q – ii, r – iii
B) p – ii, q – i, r – iii
C) p – iii, q – i, r – ii
D) p – ii, q – iii, r – i
Answer:
D) p – ii, q – iii, r – i

Question 30.
Statement-I: The lengths 3 cm, 4 cm, 5 cm form a right angled triangle. Statement-II: If ‘a’ is the side of an equilateral triangle, then its height is √3 a. Now, choose the correct answer.
(A) Statement-I and Statement-II both are True.
(B) Statement-I and Statement-II both are False.
(C) Statement-I is True. Statement-II is False.
(D) Statement-I is False. Statement-II is True.
Answer:
C) Statement-I is True. Statement-II is False.

Question 31.
Give two different examples of pair of similar figures.
Answer:
(i) Two equilateral triangles.
(ii) Two circles.

Question 32.
“In a right triangle, the square of hypotenuse is equal to the sum of the squares of the other two sides.” The above theorem is related to which Mathematician ?
A) Baudhayan
B) Aryabhatta
C) Euclid
D) Bhaskaracfyarya
Answer:
A) Baudhayan

Question 33.
Assertion (A) : The sides of two similar triangles are in the ratio 2 : 5, then the areas of these triangles are in the ratio 4 : 25
Reason (R) : The ratio of the areas of two similar triangles is equal to the square of the ratio of their sides.
A) Both assertion and reason are true and reason is the correct explanation of assertion.
B) Both assertion and reason are true but reason is not the correct explanation of assertion.
C) Assertion is true but reason is false.
D) Assertion is false but reason is true.
Answer:
A) Both assertion and reason are true and reason is the correct explanation of assertion.

10th Class Maths Triangles 2 Marks Important Questions

Question 1.
In the figure, if RS // ID then find AR.

Given,
In ΔADI, RS // ID
By Thales theorem,
$\frac{\mathrm{AS}}{\mathrm{DS}}$ = $\frac{\mathrm{AR}}{\mathrm{IR}}$
$\frac{3}{2}$ = $\frac{\mathrm{AR}}{3}$
AR = $\frac{3 \times 3}{2}$ = $\frac{9}{2}$ = 4.5 cm
Therefore AR = 4.5 cm.

Question 2.
In ΔABC, D and E are point on the sides of AB and AC respectively. Such that DE // BC. If $\frac{\mathrm{AD}}{\mathrm{DB}}$ = $\frac{2}{3}$ and AC = 18 cm, find AE.
Solution:

Given in ΔABC, DE // BC, $\frac{\mathrm{AD}}{\mathrm{DB}}$ = $\frac{2}{3}$,
AC = 18 cm
Let AE = x then EC = 18 – x.
In ΔABC, DE // BC
By Thales theorem
$\frac{\mathrm{AD}}{\mathrm{DB}}$ = $\frac{\mathrm{AE}}{\mathrm{EC}}$
$\frac{2}{3}$ = $\frac{\mathrm{x}}{18-\mathrm{x}}$
3x = 2(18 – x)
3x = 36 – 2x
3x + 2x = 36
5x = 36
x = $\frac{36}{5}$ = 7.2 cm
Therefore, AE = 7.2 cm.

Question 3.
In the figure, if AB ⊥ BC and DE ⊥ AC. Prove that ΔABC ~ ΔAED.

Solution:
Given in ΔABC, ∠B = 90°, ∠E = 90°
that is AB ⊥ BC and DE ⊥ AC.
In ΔABC, ΔAED
∠ABC = ∠AED = 90° (Angle)
∠BAC = ∠DAE (Common angle)
Therefore by Angle-Angle criterian of similarity
ΔABC ~ ΔAED

Question 4.
A vertical stick 10 cm long casts a shadow 8 cm long. At the same time a tower casts a shadow 30 m long. Determine the height of the tower.
Solution:
Given length of stick AB =10 cm
length of its shadow BC = 8 cm
Height of tower PQ = x cm.

length of the shadow of tower at the same time QR = 30 cm.
By Angle-Angle similarity
ΔABC ~ ΔPQR
then the corresponding sides are in proportion.
$\frac{\mathrm{AB}}{\mathrm{PQ}}$ = $\frac{\mathrm{BC}}{\mathrm{QR}}$
$\frac{10}{x}$ = $\frac{8}{30}$
8x = 30 × 10
x = $\frac{30 \times 10}{8}$ = $\frac{75}{2}$ = 37.5 cm
Therefore, height of tower = 37.5 cm.

Question 5.
A vertical stick of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts the shadow 28 m long. Find the height of the tower.
Solution:

Let length of stick AD = 6 m
Length of its shadow DI = 4 m
Let height of tower SR = x m
and its shadow at the same time BE = 28 m.
By Angle-Angle similarity
ΔADI ~ ΔSRE
then the corresponding sides are in proportion

$\frac{\mathrm{AD}}{\mathrm{SR}}$ = $\frac{\mathrm{DI}}{\mathrm{RE}}$
$\frac{6}{4}$ = $\frac{x}{28}$
x = $\frac{6}{4}$ × 28 = 42 m
Therefore, height of tower = 42 m.

Question 6.
The perimeter of two similar triangles ABC and PQR are respectively 36 cm and 24 cm. If PQ = 10 cm find AB.
Solution:
Given ΔABC ~ ΔPQR
So, ratio of the corresponding sides = ratio of their perimeters.

Therefore, AB = 15 cm

Question 7.
Is a square similar to a rectangle? Justify your answer.
Solution:
In a square and rectangle, the corresponding angles are equal. But the corresponding sides are not proportional.

∴ A square and a rectangle are not similar.

Question 8.
In ΔABC, LM//BC and $\frac{\mathrm{AL}}{\mathrm{LB}}$ = $\frac{2}{3}$, AM = 5 cm, find AC.
Solution:
In ΔABC, LM//BC
⇒ $\frac{\mathrm{AL}}{\mathrm{LB}}$ = $\frac{\mathrm{AM}}{\mathrm{MC}}$ (By B.P.T)

But = $\frac{\mathrm{AL}}{\mathrm{LB}}$ = $\frac{2}{3}$, so $\frac{\mathrm{AM}}{\mathrm{MC}}$ = $\frac{2}{3}$
Given AM = 5 cm and AM : MC = 2 : 3
$\frac{\mathrm{AM}}{\mathrm{MC}}$ = $\frac{2}{3}$ ⇒ $\frac{5}{\mathrm{MC}}$ = $\frac{2}{3}$
⇒ MC = $\frac{5 \times 3}{2}$ = $\frac{15}{2}$ = 7.5
∴ AC = AM + MC = 5 + 7.5 = 12.5 cm

Question 9.

In the figure, ∠BAC = ∠CED, then verify whether the value of ‘x’ is 3 or not.
Solution:
Given : In ΔABC and ΔECD
∠A = ∠E
∠ACB = ∠ECD [∵ Vertically opposite]
∴ ∠B = ∠D [∵ Angle sum property]
∴ ΔABC – ΔEDC
⇒ $\frac{\mathrm{AB}}{\mathrm{ED}}$ = $\frac{\mathrm{BC}}{\mathrm{DC}}$ = $\frac{\mathrm{AC}}{\mathrm{EC}}$
⇒ $\frac{36}{12}$ = $\frac{9}{x}$
⇒ 36x = 108
⇒ x = $\frac{108}{36}$ = 3

Question 10.
In ΔABC, DE || BC and AC = 5.6 cm, AE = 2.1 cm, then find AD : DB.
Solution:

In ΔABC DE || BC,
AE = 2.1 cm
EC = AC – AE = 5.6 – 2.1 = 3.5 cm
As per B.P.T, AD : DB = AE : EC = 2.1 : 3.5 = 3 : 5

Question 11.
In the given figure, $\overline{\mathbf{A B}}$||$\overline{\mathbf{Q R}}$ and PA = 2 cm, AQ = 3 cm, then find the ratio of areas of ΔPQR and ΔPAB.

Solution:
Given AB || QR; PA = 2, AQ = 3
then, ΔPQR ~ ΔPAB
∴ PQ = 2 + 3 = 5
∴ Ratio of area of similar triangles is equal to square of ratio of their corresponding sides.

Question 12.
In ΔPQR and ΔXYZ, it is given that ΔPQR ~ ΔXYZ, ∠Y + ∠Z = 90° and XY : XZ = 3 : 4. Then find the ratio of sides in ΔPQR.
Solution:
Given that ΔPQR – ΔXYZ means ∠P = ∠X, ∠Q = ∠Y and ∠R = ∠Z (∵ Corresponding angles are equal)
Similarly $\frac{\mathrm{PQ}}{\mathrm{XY}}$ = $\frac{\mathrm{QR}}{\mathrm{YZ}}$ = $\frac{\mathrm{RP}}{\mathrm{ZX}}$ (∵ Ratio of corresponding sides)
Given that ∠Y + ∠Z = 90°

In AXYZ, ∠X + ∠Y + ∠Z = 180°
∠X + 90° = 180°
∠X = 180° – 90° = 90°
∠X + ∠P = 90° and $\frac{\mathrm{XY}}{\mathrm{XZ}}$ = $\frac{3}{4}$
∴ ΔXYZ is a right angle triangle.
In a right angle triangle order of sides ratio may be 3 : 4 : 5.
* Remaining length YZ may be 5 units.
Ratio of XY : XZ : YZ = 3 : 4 : 5.
Similarly ratio of sides in ΔPQR is 3 : 4 : 5.

Question 13.
Find the value of ‘x’ in the given figure where ΔABC ~ ΔADE.

Solution:
ΔABC ~ ΔADE
⇒ $\frac{\mathrm{AB}}{\mathrm{AD}}$ = $\frac{\mathrm{BC}}{\mathrm{XE}}$ = $\frac{\mathrm{AC}}{\mathrm{AE}}$;
BC = x, DE = 5, AE = 3, AC = 9
By substituting $\frac{\mathrm{BC}}{\mathrm{DE}}$ = $\frac{\mathrm{AC}}{\mathrm{AE}}$
∴ $\frac{x}{5}$ = $\frac{9}{3}$ ⇒ x = $\frac{9 \times 5}{3}$ = 15
∴ x = 15 cm

Question 14.
It is given that ΔABC ~ ΔDEF. Is it true
to say that $\frac{\mathrm{BC}}{\mathrm{DE}}$ = $\frac{\mathrm{AB}}{\mathrm{EF}}$ ? Justify your answer.
Solution:
Given that ΔABC ~ ΔDEF
∴ $\frac{\mathrm{AB}}{\mathrm{DE}}$ = $\frac{\mathrm{BC}}{\mathrm{EF}}$ = $\frac{\mathrm{AC}}{\mathrm{DF}}$
(∵ Ratio of corresponding sides of similar triangles are equal)
But $\frac{\mathrm{BE}}{\mathrm{DE}}$ = $\frac{\mathrm{AB}}{\mathrm{EF}}$ (given)
∴ Given statement is wrong.

Question 15.
Srivani walks 12 m due East and turns left and walks another 5 m, how far is she from the place she started ?
Solution:
The distance of Srivani from the place she started
= $\sqrt{12^2+5^2}$ = $\sqrt{144+25}$ = $\sqrt{169}$ = 13 m

Question 16.
Write the similarity criterion by which the given pair of triangles are similar.

Solution:
$\frac{\mathrm{OA}}{\mathrm{OB}}$ = $\frac{3}{6}$ = $\frac{1}{2}$
$\frac{\mathrm{OC}}{\mathrm{OD}}$ = $\frac{2.5}{5}$ = 0.5 = $\frac{1}{2}$
$\frac{\mathrm{OA}}{\mathrm{OB}}$ = $\frac{\mathrm{OC}}{\mathrm{OD}}$ ⇒∠AOC = ∠BOD
Using SAS – Criterion ∴ ΔOAC ~ ΔOBD)

Question 17.
Madhavi said “All squares are simUar”. Do you agree yith her statement ? Justify your answer.
Answer:
Because two polygons with the same number of sides are similar if
i) All the corresponding angles are equal and
ii) All the lengths of the corresponding sides are in the same ratio (or in proportion)
Squares size may not be equal but their ratios of corresponding parts will always be equal.

Question 18.
In the given figure, ABC is a triangle. AD = 3 cm, DB = 5 cm, AE = 6 cm and EC – 10 cm. Is DE || BC ? Justify.

Solution:

AD = 3 cm, DB = 5 cm
AE = 6 cm, as EC = 10 cm
$\frac{\mathrm{AD}}{\mathrm{DB}}$ = $\frac{3}{5}$
$\frac{\mathrm{AE}}{\mathrm{EC}}$ = $\frac{6}{10}$ = $\frac{3}{5}$
$\frac{\mathrm{AD}}{\mathrm{DB}}$ = $\frac{\mathrm{AE}}{\mathrm{EC}}$ ⇒ DE || BC

Question 19.
The sides of a triangle measure 2√2, 4 and 2√6 units. Is it a right-angled triangle ? Justify.
Solution:
(2√2)² = 8
(4)² = 16
(2√6)² = 24
(2√2)² + (4)² – (2√6)²
∴ If sum of squares of two sides of a triangle is equal to the square of the third side, then the triangle is a right angled triangle. (Converse of Pythagorus theorem)
∴ Given triangle is a right-angled triangle.

Question 20.
In the given figure, if ABCD is a trapezium in which AB || CD || EF, then
AE BF prove that $\frac{\mathrm{AE}}{\mathrm{ED}}$ = $\frac{\mathrm{BF}}{\mathrm{FC}}$.

Solution:

Given : In a trapezium ABCD, AB || DC || EF.
Given: $\frac{\mathrm{AE}}{\mathrm{ED}}$ = $\frac{\mathrm{BF}}{\mathrm{FC}}$
Construction: Join AC which intersects EF at G.
Proof: ΔCAB, GF || AB
$\frac{\mathrm{FG}}{\mathrm{BF}}$ = $\frac{\mathrm{CG}}{\mathrm{AG}}$ (by BPT)
$\frac{\mathrm{BF}}{\mathrm{FC}}$ = $\frac{\mathrm{AG}}{\mathrm{CG}}$ …………. (1)
In ΔADC, EG || DC
$\frac{\mathrm{AE}}{\mathrm{ED}}$ = $\frac{\mathrm{AG}}{\mathrm{GC}}$ (by BPT) …… (2)
BY (1) & (2) $\frac{\mathrm{AE}}{\mathrm{ED}}$ = $\frac{\mathrm{BF}}{\mathrm{FC}}$.

Question 21.
In figure, if AD = 6 cm, DB = 9 cm, AE – 8 cm and EC = 12 cm and ∠ADE=48°, Find ∠ABC.

Solution:
In the figure AD = 6 cm
DB = 9cm
AE = 8 cm
EC = 12 cm
∠ADC = 48°
In ΔADE and ΔABC
$\frac{\mathrm{AD}}{\mathrm{AB}}$ = $\frac{6}{6+9}$ = $\frac{6}{15}$ = $\frac{2}{15}$
$\frac{\mathrm{AE}}{\mathrm{AC}}$ = $\frac{8}{8+12}$ = $\frac{8}{20}$ = $\frac{2}{5}$
∠A = ∠A (Common angle)
∴ by SAS criterion
ΔADE ~ ΔABC.
∠ADE = ∠ABC (by CPCT)
∴ ∠ABC = 48°

Question 22.
In the figure, altitudes AD and CE of ΔABC intersect each other at the point.
Show that
(i) ΔABD ~ ΔCBE
(ii) ΔPDC ~ ΔBEC

Solution:
i) Given ΔABD and ΔCBE
∠ADB = ∠CEB = 90°
∠B = ∠B (Common angle)
∴ By AA Similarity criterion
ΔADB ~ ΔCBE

ii) In ΔPDC and ΔBEC
∠PDC = ∠BEC = 90°
∠C = ∠C (Common angle)
∴ By AA similarity criterion
ΔPDC ~ ΔBEC.

10th Class Maths Triangles 4 Marks Important Questions

Question 1.
In ΔABC, P and Q are points on sides AB and AC respectively, such that PQ // BC. If AP = 2.4 cm, AQ = 2 cm, AC = 3 cm and BC = 6 cm. Find AB and PQ.
Solution:
Given in ΔABC, PQ // BC and AP = 2.4 cm, AQ = 2 cm, AC = 3 cm and BC = 6 cm
Let BP = x and PQ = y cm
By Thales theorem
$\frac{\mathrm{AP}}{\mathrm{BP}}$ = $\frac{\mathrm{AQ}}{\mathrm{CQ}}$

$\frac{2.4}{x}$ = $\frac{2}{3}$ and $\frac{\mathrm{AP}}{\mathrm{AB}}$ = $\frac{\mathrm{PQ}}{\mathrm{BC}}$
$\frac{2.4}{x}$ = $\frac{2}{3}$ ⇒ 2x = 2.4 × 3 ⇒ x = $\frac{2.4 \times 3}{2}$
BP = x = 3.6 cm
∴ AB = AP + BP = 2.4 + 3.6 = 6 cm
Now, $\frac{\mathrm{AP}}{\mathrm{AB}}$ = $\frac{\mathrm{PQ}}{\mathrm{BC}}$
$\frac{2.4}{6}$ = $\frac{P Q}{6}$
Therefore, PQ = 2.4 cm

Question 2.
In the given trapezium ABCD, AB // DC. Find the value of x.
Solution:
In the given figure,
ABCD is a trapezium AB // DC.
OA = 3x – 19, OB = x – 3, OC = x – 5 and DO = 3 cm

We know that diagonals of a trapezium divide each other proportionally,
$\frac{\mathrm{AO}}{\mathrm{CO}}$ = $\frac{\mathrm{BO}}{\mathrm{DO}}$
$\frac{3 x-19}{x-5}$ = $\frac{x-3}{3}$
3(3x – 19) = (x – 3) (x – 5)
9x – 19 × 3 = x² – 5x – 3x + 15
9x – x² + 8x – 57 – 15 = 0
– x² + 17x – 72 = 0
x² – 17x + 72 = 0
x² – 9x – 8x + 72 = 0
x(x – 9) – 8(x – 9) = 0
(x – 8) (x – 9) = 0
x – 8 = 0 (or) x – 9 = 0
Therefore, x = 8 (dr) x = 9

Question 3.
In the figure gIven $\frac{\mathrm{AO}}{\mathrm{OC}}$ = $\frac{\mathrm{BO}}{\mathrm{OD}}$ = $\frac{1}{2}$ and AB = 5cm. Find the value of DC.

Solution:
In quadrilateral ABCD, diagonals AC and BD are meeting at O.
$\frac{\mathrm{AO}}{\mathrm{OC}}$ = $\frac{\mathrm{BO}}{\mathrm{OD}}$ = $\frac{1}{2}$ and AB = 5cm.
In ΔAOB and ΔCOD
$\frac{\mathrm{AO}}{\mathrm{OC}}$ = $\frac{\mathrm{OB}}{\mathrm{OD}}$ and ∠AOB = ∠COD
by Side-Angle-Side similarity criterion
ΔAOB ~ ΔCOD
$\frac{\mathrm{OA}}{\mathrm{OB}}$ = $\frac{\mathrm{OB}}{\mathrm{OC}}$ = $\frac{\mathrm{AB}}{\mathrm{DC}}$
$\frac{1}{2}$ = $\frac{5}{\mathrm{DC}}$
DC = 2 × 5 = 10 cm
Therefore DC = 10 cm

Question 4.
In the given figure, $\frac{\mathrm{OA}}{\mathrm{OC}}$ = $\frac{\mathrm{OD}}{\mathrm{OB}}$ prove that ∠A = ∠C and ∠B = ∠D.

Solution:
Given in ΔAOD and ΔCOB
$\frac{\mathrm{OA}}{\mathrm{OC}}$ = $\frac{\mathrm{OD}}{\mathrm{OB}}$
and ∠AOD = ∠BOC (Vertically opposite angles)
by Side-Angle-Side similarity criterion ΔAOD ~ ΔCOB
If two triangles are similar then their corresponding angles are equal,
i.e., ∠A = ∠C and ∠D = ∠B
Therefore, ∠A = ∠C and ∠B = ∠D.

Question 5.
A girl of height 90 cm is walking away from the base of a lamppost at a speed of 1.2 m/sec. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Solution:

Let AB be the, height of lamp post
AB = 3.6 cm
and height of girl CD = 90 cm = 0.9 m length of the shadow of girl after 4 seconds = x m
length of BD = 1.2 × 4 = 4.8 cm
In ΔABE, ΔCDE
∠B = ∠D = 90° (Angle)
∠E = ∠E (Common angle)
By Angle-Angle similarity criterion
ΔABE ~ ΔCDE
then corresponding sides are in proportion.

4x = 4.8 + x
4x – x = 4.8
3x = 4.8
x = $\frac{4.8}{3}$ = 1.6 cm
Therefore length of shadow of the girls = 1.6 cm.

Question 6.
In the given figure, CD is the perpendicular bisector of AB. EF is perpendicular to CD. AE intersects CD
at G.Prove that $\frac{\mathrm{CF}}{\mathrm{CD}}$ = $\frac{\mathrm{FG}}{\mathrm{DG}}$.

Solution:
In ΔEFC and ΔBDC
∠EFC = ∠BDC = 90°
∠ECF = ∠BCD (Common angle)
by AA Similarity
ΔEFC ~ ΔBDC
by CPST $\frac{\mathrm{EF}}{\mathrm{BD}}$ = $\frac{\mathrm{FC}}{\mathrm{DC}}$.
In ΔEFG and ΔDGA
∠EGF = ∠DGA (VOA)
∠EFG = ∠GDA (= 90°)
by AA Similarity
Δ EFG ~ Δ ADG
we have AD = BD
$\frac{\mathrm{EF}}{\mathrm{BD}}$ = $\frac{\mathrm{FG}}{\mathrm{DG}}$.
$\frac{\mathrm{FG}}{\mathrm{DG}}$ = $\frac{\mathrm{CF}}{\mathrm{CD}}$.

Question 7.
In the given figure, ABCD is a parallelogram. BE bisects CD at M and intersects AC at L. Prove that EL = 2BL.

Solution:
Given ΔBMC and ΔEMD
∠BMC = ∠EMD (VOA)
Given MC = DM
∠BCM = ∠EDM (ALA)
∴ ΔBMC ≅ ΔEMD (by ASA)
by CPCT BC = DE
AE = AD + DE = BC + BC = 2BC
∴ ΔBLC ≅ ΔELA (by AA similarity)
$\frac{\mathrm{EL}}{\mathrm{BL}}$ = $\frac{\mathrm{AE}}{\mathrm{BC}}$
by CPST = $\frac{\mathrm{EL}}{\mathrm{BL}}$ = $\frac{\mathrm{2 BC}}{\mathrm{BC}}$ $\frac{\mathrm{EL}}{\mathrm{BL}}$ = $\frac{2}{1}$
EL = 2BL

Question 8.
In fig 6, if ΔABC ~ ADEF and their sides of lengths (in cm) are marked along them, then find the lengths of sides of each triangles.

Solution:
ΔABC ~ ΔDEF.

4x – 2 = 18
4x = 20
x = 5
AB = 2 × 5 – 1 = 9.
BC = 2 × 5 + 2 = 12
CA = 3 × 5 = 15
DE = 18
EF = 3 × 5 + 9 = 24.
FD = 6 × 5 = 30
AB = 9 cm, BC = 12 cm, EF = 24 cm
CA = 15 cm, DE = 18 cm, FD = 30 cm.

Question 9.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac{\mathrm{AO}}{\mathrm{BO}}$ = $\frac{\mathrm{CO}}{\mathrm{OD}}$.

Show that quadrilateral ΔABCD is a trapezium.
Solution:
Given: The diagonals of a quadrilateral ΔBCD intersect each other at the point ‘O’ such that
$\frac{\mathrm{AO}}{\mathrm{BO}}$ = $\frac{\mathrm{CO}}{\mathrm{OD}}$
To prove : ABCD is a trapezium.
Construction: Draw OE || DC such that E lies on BC.

Proof : In ΔBDC
By BPT, $\frac{\mathrm{BO}}{\mathrm{OD}}$ = $\frac{\mathrm{BE}}{\mathrm{EC}}$ ……… (1)
but $\frac{\mathrm{AO}}{\mathrm{CO}}$ = $\frac{\mathrm{BO}}{\mathrm{DO}}$ …….. (2)
from (1) & (2)
$\frac{\mathrm{AO}}{\mathrm{CO}}$ = $\frac{\mathrm{BE}}{\mathrm{EC}}$
by the converse of BPT.
OE || AB
Since AB || CE || DC
∴ AB || DC
∴ ΔBCD is a trapezium.

Question 10.
E is a point on the side AD produced of a parallelogram ABCD and BE interesects CD at F. Show that ΔABE ~ ΔCFB.
Solution:

In ΔABE and ΔCFB
∠ABE = ∠CFB (AIA)
∠BAE = ∠BCF (opposite angles of ||gm)
∴ by AA Similarity criterion
ΔABE ~ ΔCFB.

Question 11.

In the given figure, CM and RN are respectively the medians of ΔABC and ΔPQR. If ΔABC ~ ΔPQR, then prove that ΔAMC ~ ΔPNR.
Solution:
Given : ΔABC ~ ΔPQR
CM and RN are medians of ΔABC and ΔPQR respectively.
To prove : ΔAMC ~ ΔPNR
Proof : Given ΔABC ~ ΔPQR, ∠A = ∠P,
∠B = Q, ∠C = ∠R
$\frac{\mathrm{AB}}{\mathrm{PQ}}$ = $\frac{\mathrm{BC}}{\mathrm{QR}}$ = $\frac{\mathrm{CA}}{\mathrm{RP}}$
$\frac{\mathrm{AB}}{\mathrm{PQ}}$ = $\frac{x+y}{y+y}$ = $\frac{2 x}{2 y}$ = $\frac{x}{y}$
Let AM = MB = x
In ΔAMC and ΔPNR
$\frac{x}{y}$ = $\frac{\mathrm{AM}}{\mathrm{PN}}$
$\frac{x}{y}$ = $\frac{\mathrm{AB}}{\mathrm{PQ}}$ = $\frac{\mathrm{CA}}{\mathrm{RP}}$
$\frac{\mathrm{AM}}{\mathrm{PN}}$ = $\frac{\mathrm{CA}}{\mathrm{RP}}$
∠A = ∠P
by SAS similarity criterion.
ΔABC ~ ΔPNR.

Question 12.
In figure ∠1 = ∠2 and ΔNSQ ≅ ΔMTR, then prove that ΔPTS ~ ΔPRQ.

Solution:
Given ∠1 = ∠2
ΔNSQ ≅ ΔMTR
∴ PS = PT
by CPCT, SQ = TR
In ΔPTS & ΔPRQ
∠P = ∠P (Common angle)
∠Q = ∠R (PQ = PR)
by AA similarity
ΔPTS ~ ΔPRQ.

Question 13.
Observe the below figure.

In a ΔPQR, if XY // QR and PX – x – 2, XQ = x + 5, PY = x – 3 and YR = x + 3, then find the value of ‘x’.
Solution:
Given : In ΔPQR, XY // QR and
PX = x – 2, XQ = x + 5, PY = x – 3 and YR = x + 3.
By Basic Proportionality theorem,
If XY // QR then we should have
$\frac{\mathrm{PX}}{\mathrm{XQ}}$ = $\frac{\mathrm{PY}}{\mathrm{YR}}$
∴ $\frac{x-2}{x+5}$ = $\frac{x-3}{x+3}$
⇒ (x – 2) (x + 3) = (x – 3) (x + 5)
⇒ x² + 3x – 2x – 6 = x² + 5x – 3x – 15
⇒ x² + x – 6 = x² + 2x- 15
⇒ x – 6 = 2x- 15
⇒ 2x – x = 15 – 6
⇒ x = 9
∴ The value of x = 9 will make
XY // QR.

Question 14.
In a ΔABC, AD ⊥ BC and AD² = BD × CD, prove that ΔABC is a right – angled triangle.

Solution:

Given : In ΔABC, AD ⊥ BC and AD² = BD × CD
RTP : In ΔABC is a right – angled triangle
Proof : AD² = BD × CD
⇒ $\frac{\mathrm{AD}}{\mathrm{BD}}$ = $\frac{\mathrm{CD}}{\mathrm{AD}}$
∠ADB = ∠CDA = 90°
∴ Δ ADB – ΔCDA (∵ S.A.S Similarity)
In Δ ABD, ∠BAD + ∠ABD = 90° (∵ ∠ADB = 90°)
In Δ ACD, ∠DAC + ∠ACD = 90° (∵ ∠ADC = 90°)
∠ BAD = ∠ACD (∵ pair of corresponding angles)
∠ABD = ∠DAC (∵ pair of corresponding angles)
∴ ∠BAD + ∠DAC = 90°
∠BAC = 90°
Δ ABC is a right – angled triangle.

Question 15.
In ΔABC, $\overline{\mathbf{P Q}}$ || $\overline{\mathbf{B C}}$ and AP = 3x – 19, PB = x – 5, AQ = x + 3, QC = 3 cm. Find x.
Solution:
In Δ ABC PQ || BC

⇒ $\frac{\mathrm{AP}}{\mathrm{PB}}$ = $\frac{\mathrm{AQ}}{\mathrm{QC}}$ (∵ Basic proportionality theorem)
$\frac{3 x-19}{x-5}$ = $\frac{x-3}{3}$
9x – 57 = x² – 8x + 15
⇒ x² – 17x + 72 = 0
(x – 8) (x – 9) = 0
x = 8 or x = 9

Question 16.
If the ratio of areas of two equilateral triangles is 25 : 36, then find the ratio of heights of the triangles.
Solution:
For similar triangles
Ratio of area = Ratio of sides

h₁ : h₂ = 5 : 6
∴ The ratio of heights of the triangles = 5 : 6

Question 17.
In ΔABC, D and E are points on AB and AC respectively. If AB = 14 cm; AD = 3.5 cm, AE = 2.5 cm and
AC = 10 cm, show that DE || BC.
Solution:

Given AB = 14 cm, AD = 3.5 cm,
AE = 2.5 cm, AC = 10 cm.
BD = AB – AD = 14 – 3.5 = 10.5 cm
EC = AC – AE = 10 – 2.5 = 7.5 cm
$\frac{\mathrm{AD}}{\mathrm{DB}}$ = $\frac{3.5}{10.5}$ = $\frac{1}{3}$ ………… (1)
$\frac{\mathrm{AD}}{\mathrm{DB}}$ = $\frac{2.5}{7.5}$ = $\frac{1}{3}$ ……….. (2)
From (1) and (2), $\frac{\mathrm{AD}}{\mathrm{DB}}$ = $\frac{\mathrm{AE}}{\mathrm{EC}}$
∴ DE||BC (∵ using converse of BPT)

Question 18.
In a rectangle ABCD, AB = 2x – y, BC = 15, CD = 2 and DA = x + 3y, then find the values of x and y.
Solution:
Given that
In a rectangle ΔBCD

AB = 2x – y
BC = 15; CD = 2; OA = x + 3y
So,
In a rectangle opposite sides are equal
(1) → 2x – y = 2
(2) → x + 3y = 15

so the value of x is 3 and the value of y is 4.

Question 19.
Observe the below diagram and find the values of x and y.

Solution:
ΔABC ~ ΔEFC
$\frac{x}{6}$ = $\frac{9}{15}$ ⇒ x = $\frac{9 \times 6}{15}$ = $\frac{18}{5}$ = 3.6 cm
ΔBDC ~ ΔBEF
$\frac{x}{9.6}$ = $\frac{y}{15}$ ⇒ y = $\frac{3.6 \times 15}{9.6}$ = $\frac{45}{8}$ = 5.625 cm.

Question 20.
Give two different examples of pair of i) Similar figures, ii) Non – similar
figures.
Solution:
By keeping in view of the definition of similar figures.
(i) Similar figures :
For any two correct examples for similar figures.
(ii) Non similar figures :
For any two correct examples for similar figures.

Question 21.
The hypotenuse of a right triangle is 6m more than twice the shortest side. If the third side is 2m less than hypotenuse, find the sides of the triangle.
Solution:
Let the shortest side = ‘x’ m
then its hypotenuse = (2x + 6) m
Third side = 2x + 6 – 2 = (2x + 4) m
then from Pythagoras theorem, we have

x² + (2x + 4)² = (2x + 6)²
⇒ x² + 4x² + 16x + 16 = 4x² + 24x + 36
⇒ x² + 16x – 24x + 16 – 36 = 0
⇒ x² – 8x – 20 = 0
⇒ x² – 10x + 2x – 20 = 0
⇒ x(x – 10)+ 2(x – 10)= 0
⇒ (x – 10)(x + 2) = 0
⇒ x = 10 or x = -2
but x being the length cannot be negative.
So x ≠ -2, we consider x = 10 only
then the
shortest side = x = 10 m
third side 2x + 4 = 24m
hypotenuse = 2x + 6 = 26 m

Question 22.
A ladder 25 m long reaches a window of building 24 m above the ground. Determine the distance of the foot of the ladder from the building.
Solution:
In a ΔABC, ∠C = 90°

AB = length of ladder = 25m
AC = height of window = 24 m
AB² = AC² + BC²
BC² = AB² – AC²
= 25² – 24² = (25 + 24) (25 – 24)
= 49 × 1 = 7²
∴ BC = 7 m
∴ Distance of the foot of the ladder from the building = 7 m.

Question 23.
In figure EF || AB || DC, prove that $\frac{\mathrm{AE}}{\mathrm{ED}}$ = $\frac{\mathrm{BF}}{\mathrm{FC}}$ .

Solution:
Given in ABCD quadrilateral.
EF || AB || DC
In ΔADC, EP || DC.
By Basic Proportionality theorem,
$\frac{\mathrm{AE}}{\mathrm{ED}}$ = $\frac{\mathrm{AP}}{\mathrm{PC}}$ → (1)
In ΔCAB, FP || BA.
By Basic Proportionality theorem,
$\frac{\mathrm{BF}}{\mathrm{FC}}$ = $\frac{\mathrm{AP}}{\mathrm{PC}}$ → (2)
From (1) and (2) $\frac{\mathrm{AE}}{\mathrm{ED}}$ = $\frac{\mathrm{BF}}{\mathrm{FC}}$.

Question 24.
In figure, DE || BC and CD || EF. Prove that AD² = AB × AF.
Solution:

Given in ΔABC,
DE || BC and CD || EF.
In ΔABC, DE || BC
By Basic Proportionality theorem,
$\frac{\mathrm{AB}}{\mathrm{AD}}$ = $\frac{\mathrm{AC}}{\mathrm{AE}}$ → (1)
In ΔADC, FE || DC
By Basic Proportionality theorem,
$\frac{\mathrm{AD}}{\mathrm{AF}}$ = $\frac{\mathrm{AC}}{\mathrm{AE}}$ → (2)
From (1) and (2) $\frac{\mathrm{AB}}{\mathrm{AD}}$ = $\frac{\mathrm{AD}}{\mathrm{AF}}$ AD.AD = AB.AF
Therefore, AD² = AB.AF.

Question 25.
In a ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 2.5 cm, BD = 3.0 cm and AE = 3.75 cm. Find the length of AC.
Solution:
Given in ΔABC,
AD = 2.5 cm, BD = 3.0 cm and AE = 3.75 cm and DE || BC.
In ΔABC, DE || BC
By Basic Proportionality theorem,

∴ CE = 4.5 cm
AC = AE + CE = 3.75 + 4.5
Therefore, AC = 8.25 cm

Question 26.
If D and E are points on sides AB and AC respectively of a ΔABC such that DE || BC and BD = CE. Prove that ΔABC is isosceles.
Solution:
In ΔABC, DE || BC and BD = CE.
In ΔABC, DE || BC,
By Basic Proportionality theorem,
$\frac{\mathrm{AD}}{\mathrm{DB}}$ = $\frac{\mathrm{AE}}{\mathrm{EC}}$
$\frac{\mathrm{AD}}{\mathrm{AE}}$ = $\frac{\mathrm{DB}}{\mathrm{EC}}$ = 1
So, $\frac{\mathrm{AD}}{\mathrm{AE}}$ = 1

Therefore, AD = AE
So, AD + BD = AE + CE
AB = AC.
In ΔABC, AB = AC
Two sides are equal in ΔABC.
Therefore, ΔABC is an isosceles.

Question 27.
In the given figure AB || CD. If OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4, find x.
Solution:
In quadrilateral ABCD, AB || DC.
So, ABCD is a trapezium.
In trapezium diagonals divide each other proportionately.

$\frac{\mathrm{OA}}{\mathrm{OC}}$ = $\frac{\mathrm{OB}}{\mathrm{OD}}$
$\frac{3 x-19}{x-3}$ = $\frac{x-4}{4}$
⇒ 4(3x – 19) = (x – 3) (x – 4)
⇒ 12x – 76 = x² – 4x – 3x + 12
⇒ x² – 7x + 12x + 88 = 0
⇒ x² – 19x + 88 = 0
⇒ x² – 11x – 8x + 88 = 0
⇒ x (x – 11) – 8 (x – 11) = 0
⇒ (x – 8) (x – 11) = 0
Therefore, x = 8 (or) 11.

Question 28.
In figure, QA and PB are perpendiculars to AB. If AO = 10 cm, BO = 6 cm and PB = 9 cm. Find AQ.
Solution:
In ΔAOQ, ΔBOP
∠A = ∠B = 90°
∠AOQ = ∠BOP (Vertically opposite angles)
Angle-Angle criterion of similarity
ΔAOQ ~ ΔBOP.
So, corresponding sides are in proportion.

There, AQ = 15 cm.

Question 29.
In figure, considering triangles BEP and CPD. Prove that BP × PD = EP × PC.
Solution:

In ΔABC, BD ⊥ AC
and CE ⊥ AB.
In ΔEBP, ΔDCP
∠E = ∠D = 90°
∠EPB = ∠DPC (Vertically opposite angles)
By Angle-Angle criterion similarity
ΔEBP ~ ΔDCP.
So, corresponding sides are in proportion.
$\frac{\mathrm{BP}}{\mathrm{PC}}$ = $\frac{\mathrm{EP}}{\mathrm{PD}}$
Therefore, BP × PD = EP × PC.

Question 30.
In ΔABC, DE || BC, with D on AB and E on AC. If $\frac{\mathrm{AD}}{\mathrm{DB}}$ = $\frac{2}{3}$, find $\frac{\mathrm{BC}}{\mathrm{DE}}$.
Solution:
Given in ΔABC.DE || BC and $\frac{\mathrm{AD}}{\mathrm{DB}}$ = $\frac{2}{3}$
In ΔABC, DE || BC.
By Basic Proportionality theorem,

10th Class Maths Triangles 8 Marks Important Questions

Question 1.
In the given figure $\frac{\mathrm{BE}}{\mathrm{AC}}$ = $\frac{\mathrm{BC}}{\mathrm{BD}}$ and ∠1 = ∠2. Prove that ΔABD ~ ΔEBC.

Solution:
Given in ΔBCE,
∠1 = ∠2 and $\frac{\mathrm{BE}}{\mathrm{AC}}$ = $\frac{\mathrm{BC}}{\mathrm{BD}}$
$\frac{\mathrm{BE}}{\mathrm{BC}}$ = $\frac{\mathrm{AC}}{\mathrm{BD}}$ → (1)
In ΔMBC, ∠1 = ∠2 and then sides opposite to equal angles are equal.
That is AC = AB → (2)
From (1) and $\frac{\mathrm{BE}}{\mathrm{BC}}$ = $\frac{\mathrm{AB}}{\mathrm{BD}}$
In ΔABD and ΔEBC
∠ABD = ∠EBC = ∠B
and $\frac{\mathrm{BE}}{\mathrm{BC}}$ = $\frac{\mathrm{AB}}{\mathrm{BD}}$
So, by side-angle-side similarity criterion.
Therefore, ΔABD ~ ΔEBC

Question 2.
Given in the trapezium ABCD, AB // DC. If ΔAED ~ ΔBEC then prove that AD = BC.
Solution:

In trapezium ΔBCD, AB // DC and ΔAED ~ ΔBEC. A
In ΔEDC, ΔEBA
∠EDC = ∠EBA
∠1 = ∠2 (Alternate angles)
∠ECD = ∠EAB
∠3 = ∠4 (Alternate angles)
∠CED = ∠AEB (Vertically opposite angles)
By A – A similarity criterion
ΔEDC ~ ΔEBA
Corresponding sides are in proportion
$\frac{\mathrm{ED}}{\mathrm{EC}}$ = $\frac{\mathrm{BE}}{\mathrm{AE}}$ → (1)
But,.given ΔAED ~ ΔBEC
then the corresponding sides are in proportion $\frac{\mathrm{ED}}{\mathrm{EC}}$ = $\frac{\mathrm{AE}}{\mathrm{BE}}$ = $\frac{\mathrm{AD}}{\mathrm{BC}}$ → (2)
From (1) and (2)
$\frac{\mathrm{ED}}{\mathrm{EC}}$ = $\frac{\mathrm{BE}}{\mathrm{AE}}$ = $\frac{\mathrm{AE}}{\mathrm{BE}}$
BE² = AE²
BE = AE ⇒ $\frac{\mathrm{AE}}{\mathrm{BE}}$ = 1
Put BE = AE in (2)
$\frac{\mathrm{ED}}{\mathrm{EC}}$ = 1 = $\frac{\mathrm{AD}}{\mathrm{BC}}$
$\frac{\mathrm{AD}}{\mathrm{BC}}$ = 1
Therefore AD = BC.

Question 3.
If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse then prove that the triangle on each side, of the perpendicular are similar to each other and to the original triangle. Also prove that the square of the perpendicular is equal to the product of the length of the two parts of the hypotenuse. That is
i) ΔAOB ~ ΔBOC
ii) ΔADB ~ ΔABC
iii) ΔBOC ~ ΔABC
iv) BD² = AD.DC
v) AB² = AD.AC
vi) BC² = CD.AC
Solution:
Given in ΔABC, ∠B = 90° and BD ⊥ AC
i) ∠ABD + ∠DBC = 90° → (1)
In ΔBOC,
∠BDC + ∠DBC + ∠C = 180°
90° + ∠DBC + ∠C = 180°
∠DBC + ∠C = 180° – 90° = 90° → (2)
From (1) and (2) ∠ABD = ∠C.
In ΔADB, ∠BDC,
∠ABD = ∠C and ∠ADB = ∠BDC = 90°
by Angle-Angle similarity criterion
ΔADB ~ ΔBDC → (3)

ii) In ΔADB and ΔABC
∠ADB = ∠ABC = 90° (Angle)
and ∠A = ∠A (Common angle)
by Angle-Angle similarity criterion.
ΔADB – ΔABC → (4)

iii) In ΔBDC and ΔABC
∠BDC = ∠ABC = 90° (Angle)
∠C = ∠C (Common angle)
By Angle-Angle similarity criterion
ΔBDC ~ ΔABC → (5)
From (3), (4) and (5)
ΔADB ~ ΔBDC ~ ΔBDC – ΔABC

iv) We proved ΔADB ~ ΔBDC then corresponding sides are in proportion
$\frac{\mathrm{AD}}{\mathrm{BD}}$ = $\frac{\mathrm{BD}}{\mathrm{DC}}$
BD.BD = AD.DC
BD² = AD.DC

v) We proved ΔADB ~ ΔABC
$\frac{\mathrm{AD}}{\mathrm{AB}}$ = $\frac{\mathrm{AB}}{\mathrm{AC}}$
AB² = AD.AC

vi) We proved ΔBDC ~ ΔABC
$\frac{\mathrm{BC}}{\mathrm{AC}}$ = $\frac{\mathrm{DC}}{\mathrm{BC}}$
BC² = AC.DC

Question 4.
In ΔABC given AD and BE are perpendiculars to BC and AC respectively show that
i) ΔADC ~ ΔBEC
ii) CA.CE = CB.CD
iii) ΔABC ~ ΔDEC
iv) CD.AB = CA.DE
Solution:
In ΔABC, AD ⊥ BC and BE ⊥ AC

i) In ΔADC and ΔBEC
∠ADC = ∠BEC = 90° (Angle)
∠C = ∠C (Common angle)
By Angle-Angle similarity criterion
ΔADC – ΔBEC

ii) The corresponding sides are in proportion
$\frac{\mathrm{CA}}{\mathrm{CB}}$ = $\frac{\mathrm{CD}}{\mathrm{CE}}$ → (1)

iii) In ΔABC, ΔDEC
From(1) $\frac{\mathrm{AC}}{\mathrm{BC}}$ = $\frac{\mathrm{DC}}{\mathrm{EC}}$ (or) $\frac{\mathrm{AC}}{\mathrm{DC}}$ = $\frac{\mathrm{BC}}{\mathrm{EC}}$
By Side – Angle – Side similarity criterion ΔABC ~ ΔDEC.

iv) Corresponding sides are in proportion
$\frac{\mathrm{AB}}{\mathrm{DE}}$ = $\frac{\mathrm{AC}}{\mathrm{DC}}$
Therefore, AB.DC = AC.DE

Question 5.
i) State and prove the “Basic Proportionality theorem”.
ii) Using the theorem, find the length of AE, if AD = 1.8 cm, BD = 5.4 cm, EC = 7.2 cm.

Solution:
i) If a line is drawn parallel to one side of a triangle to intersecting the other two sides, in distinct points, then the other two sides are divided in the sgme ratio.
Given : A triangle ABC in which DE || BC and DE intersects AB in D and AC in E.

To prove: $\frac{\mathrm{AD}}{\mathrm{DB}}$ = $\frac{\mathrm{AE}}{\mathrm{EC}}$

Construction : Join BE, CD and draw EF ⊥ AB, DG ⊥ AC.
Proof : In ΔEAD and ΔEDB, here as EF is perpendicular to AB, therefore, EF is the height for both of triangles EAD and EDB.
Now,
Area of ΔEAD = $\frac{1}{2}$(base × height)
= $\frac{1}{2}$ × AD × EF
Area of ΔEDB = $\frac{1}{2}$(base × height)
= $\frac{1}{2}$ × DB × EF
$\frac{{Area}(\triangle \mathrm{EAD})}{\text { Area }(\triangle \mathrm{EDB})}$ =  = $\frac{\mathrm{AD}}{\mathrm{DB}}$ ………… (1)
Similarly, $\frac{{Area}(\triangle \mathrm{EAD})}{\text { Area }(\triangle \mathrm{ECD)}}$ = $\frac{\mathrm{AE}}{\mathrm{EC}}$ …….. (2)
∵ ΔDBE, ΔECD are on the same base DE and between the same parallels DE || BC, we have
Area of ΔDBE = Area of ΔECD
Hence, (1) = (2)
i.e., $\frac{\mathrm{AD}}{\mathrm{DB}}$ = $\frac{\mathrm{AE}}{\mathrm{EC}}$ (Q.E.D).

ii) Using the theorem to find the length of .

AD = 1.8 cm, BD = 5.4 cm, EC = 7.2 cm
From the theorem, $\frac{\mathrm{AD}}{\mathrm{BD}}$ = $\frac{\mathrm{AE}}{\mathrm{EC}}$
⇒ AE = $\frac{(\mathrm{AD})(\mathrm{EC})}{\mathrm{BD}}$ = $\frac{1.8 \times 7.2}{5.4}$ = 2.4
∴ AE = 2.4cm

Question 6.
ABC is a right angled triangle which is right angled at C. Let AB = c, BC = a, CA = b and let p be the length of perpendicular from C on AB. Prove that c = $\frac{\mathrm{ab}}{\mathrm{p}}$.
Solution:

CD ⊥ AB and CD = p.
Also area of ΔABC = $\frac{1}{2}$ × AB × CD
= $\frac{1}{2}$cp
Also area of ΔABC = $\frac{1}{2}$ × BC × AC
= $\frac{1}{2}$ap
$\frac{1}{2}$cp = $\frac{1}{2}$ap ⇒ cp = ap ⇒ c = $\frac{\mathrm{ab}}{\mathrm{p}}$

Question 7.
In given figure if PQ || BC and PR || CD. prove that
i) $\frac{\mathrm{AR}}{\mathrm{AD}}$ = $\frac{\mathrm{AQ}}{\mathrm{AB}}$
ii) $\frac{\mathrm{QB}}{\mathrm{AQ}}$ = $\frac{\mathrm{DR}}{\mathrm{AR}}$.

Solution:
In ΔABC, PQ || BC.
By Basic Proportionality theorem,
$\frac{\mathrm{AQ}}{\mathrm{AB}}$ = $\frac{\mathrm{AP}}{\mathrm{AC}}$ → (1)
In ΔACD, PR || CD
By Basic Proportionality theorem,
$\frac{\mathrm{AP}}{\mathrm{AC}}$ = $\frac{\mathrm{AR}}{\mathrm{AD}}$ → (2)

Question 8.
In a ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 4x – 3, AE = 8x – 7, BD = 3x – 1 and CE = 5x – 3. Find the value of x.

Solution:
In ΔABC, DE || BC
AD = 4x – 3, AE = 8x – 7, BD = 3x – 1 and CE = 5x – 3
In ΔABC, DE || BC
By Basic Proportionality theorem,
$\frac{\mathrm{AD}}{\mathrm{BD}}$ = $\frac{\mathrm{AE}}{\mathrm{CE}}$
⇒ $\frac{4 x-3}{3 x-1}$ = $\frac{8 x-7}{5 x-3}$
⇒ (4x – 3) (5x – 3) = (8x – 7) (3x – 1)
⇒ 20x² – 12x – 15x + 9 = 24x² – 8x – 21x + 7
⇒ – 24x² + 20x² – 27x + 29x + 9 – 7 = 0
⇒ – 4x² + 2x + 2 = 0
⇒ – 2 (2x² – x – 1) = 0
⇒ 2x² – x – 1 = 0
⇒ 2x² – 2x + x – 1 = 0
⇒ 2x(x- 1) + 1 (x- 1) = 0
⇒ (2x + 1) (x – 1) = 0
⇒ 2x + 1 = 0 and x – 1 = 0
x = – $\frac{1}{2}$ (or) 1
Length cannot be negative.
Therefore, x = 1

Question 9.
ABCD is a trapezium, in which AB || DC and its diagonals intersect each other at point ‘O’. Show that $\frac{\mathrm{OA}}{\mathrm{OB}}$ = $\frac{\mathrm{OC}}{\mathrm{OD}}$.
Solution:

In Δ OAB, Δ OCD
∠AOB = ∠COD (∵ pair of vertically Opp. Angles)
∠OAB = ∠OCD (∵ Pair of alternate angles)
∠OB A = ∠ODC (∵ Pair of alternate angles)
∴ Δ OAB ~ Δ OCD (∵ AAA similarity)
∴ $\frac{\mathrm{OA}}{\mathrm{OC}}$ = $\frac{\mathrm{OB}}{\mathrm{OD}}$ ⇒ $\frac{\mathrm{OA}}{\mathrm{OB}}$ = $\frac{\mathrm{OC}}{\mathrm{OD}}$

Question 10.
Construct an equilateral triangle XYZ of side 5 cm and construct another triangle similar to ΔXYZ, such that each of its sides is $\frac{4}{5}$ of the sides of ΔXYZ.
Solution:

Steps of construction :

Draw an equilateral triangle XYZ with side 5 cm.
Draw a ray $\overrightarrow{X A}$ such that ∠YXA is an acute angle.
Draw X₁, X₂, X₃, X₄, X₅ arcs on $\overrightarrow{X A}$ such that XX₁ = X₁X₂ = ……… = X₄X₅
Join X₅ and Y.
Draw a parallel line to X₅Y through X₄ to meet XY at Y’.
Draw a parallel line to YZ through Y’ to meet XZ at Z’.
Δ XY’ Z’ is required similar triangle.

AP 10th Class Maths Important Questions Chapter 6 Similar Triangles

Question 1.
Is a square similar to a rectangle ? Justify your answer.
Solution:
In a square and rectangle, the corresponding angles are equal. But the corresponding sides are not proportional.

∴ A square and a rectangle are not similar

Question 2.
In ΔABC LM||BC and $\frac{\mathbf{A L}}{\mathbf{L B}}=\frac{2}{3}$, AM = 5 cm, find AC.
Solution:
In ΔABC, LM // BC

Given AM = 5 cm and AM : MC = 2:3
$\frac{\mathrm{AM}}{\mathrm{MC}}=\frac{2}{3} \Rightarrow \frac{5}{\mathrm{MC}}=\frac{2}{3}$
⇒ MC = $\frac{5 \times 3}{2}=\frac{15}{2}$ = 7.5
∴ AC = AM + MC = 5 + 7.5 = 12.5 cm

Question 3.

In the figure, ∠BAC = ∠CED, then verify whether the value of ‘x’ is 3 or not.
Solution:
Given: In ΔABC and ΔECD
∠A = ∠E
∠ACB = ∠ECD [∵ Vertically opposite]
∴ LB ¿D [∵ Angle sum property]
∴ ΔABC ~ ΔEDC
⇒ $\frac{\mathrm{AB}}{\mathrm{ED}}=\frac{\mathrm{BC}}{\mathrm{DC}}=\frac{\mathrm{AC}}{\mathrm{EC}} \Rightarrow \frac{36}{12}=\frac{9}{\mathrm{x}}$
⇒ 36x= 108 ⇒ x= $\frac{108}{36}$ = 3

Question 4.
In ΔABC, DE || BC and AC = 5.6 cm, AE = 2.1 cm, then find AD : DB.
Solution:

In ΔABC DE || BC,
AE = 2.1 cm
EC = AC – AE
= 5.6 – 2.1 = 3.5 cm
As per B.P.T, AD : DB = AE : EC
= 2.1 : 3.5 = 3:5

Question 5.
In the given figure, $\overline{\mathbf{A B}} \| \overline{\mathbf{Q R}}$ and PA = 2 cm, AQ = 3 cm, then find the ratio of areas of ΔPQR and ΔPAB.

Solution:
Given AB || QR; PA = 2, AQ = 3
then, APQR ~ APAB
∴ PQ = 2 + 3 = 5
∴ Ratio of area of similar triangles is equal to square of ratio of their corre-sponding sides.

Question 6.
Give two different examples of pair of i) Similar figures, ii) Non – similar figures. Mar. ’17
By keeping in view of the definition of similar figures.
Solution:
By keeping in view of the definition of similar figures.
(i) Similar figures :

For any two correct examples for similar figures.
(ii) Non similar figures :

For any two correct examples for non similar figures.

Question 7.
Define A.A. and S.S.S. rules for similarity of two triangles in your own words.
Solution:
A.A. similarity : If two angles of a triangle are equal to two corresponding angles of another triangle, then the two triangles are similar.
S.S.S. similarity : If in two triangles, sides of one triangle are proportional to the sides of other triangle, then their corresponding angles are equal and hence the two triangles are similar.

Question 8.
The hypotenuse of a right triangle is 6m more than twice the shortest side. If the third side is 2m less than hypotenuse, find the sides of the triangle.
Solution:
Let the shortest side = ‘x’m
then its hypotenuse = (2x + 6) m
Third side = 2x + 6 – 2 = (2x + 4) m then from Pythagoras theorem, we have

x² + (2x + 4)² = (2x + 6)²
⇒ x² + 4x² + 16x + 16 = 4x² + 24x + 36
⇒ x² + 16x – 24x + 16 – 36 = 0
⇒ x² – 8x – 20 = 0
⇒ x² – 1 Ox + 2x – 20 = 0
⇒ x(x- 10) + 2 (x- 10) = 0
⇒ (x – 10) (x + 2) = 0
⇒ x = 10 or x = -2
but ‘x’ being the length cannot be nega¬tive.
So x ≠ -2, we consider x = 10 only then the
shortest side = x = 10m
third side = 2x + 4 = 24 m
hypotenuse = 2x + 6 = 26m

Question 9.
Draw a line segment of length 7.2 cm. and divide it in the ratio 5 : 3 using compasses and ruler.
Solution:

Question 10.
Construct a triangle of sides 4.2 cm, 5.1 cm and 6 cm. Then construct a tri¬angle similar to it, whose sides are 2/3 of corresponding sides of the first tri¬angle.
Solution:

Draw a triangle ABC, with sides AB = 4.2cm,BC = 5.1 cm,CA = 6cm.
Draw a ray BX making an acute angle with BC on the side opposite to vertex A.
Locate 3 points B,, B2, B3 on BX so that BB₁ = B₁B₂ = B₂B₃.
Join B₃,C an draw a line through B₂ parallel to B₃C intersecting BC at C¹.
Draw a line through .C¹ parallel to CA intersect AB at A¹.
A¹BC¹ is required triangle.

Question 11.
Construct a triangle PQR, where QR = 5.5cm,∠Q= 65°andPQ = 6cm. Then draw another triangle, whose sides are $\frac{2}{3}$ times of the corresponding sides of ΔPQR.
Solution:
Construction Steps:

Draw a triangle ΔPQR with sides QR = 5.5cm, ∠Q = 65°and PQ = 6cm.
Draw a ray OX making an acute angle with QR on the side opposite to vertex P
Locate 3 points Q₁ Q₂ Q₃ on QX. So that QQ₁ = Q₁Q₂ = Q₂Q₃
Join Q₃,R and draw a line through Q₂ parallel to Q₃R intersecting QR at R.
Draw a line through ‘R’ parallel to PR intersect PQ at P.
ΔP’QR’ is required triangle.

Question 12.
i) State and prove the “Basic Propor-tionality theorem”.
ii) Using the theorem, find the length of AE, if AD = 1.8 cm, BD .= 5.4 cm, EC = 7.2 cm.

Solution:
If a lines is drawn parallel to one side of a triangle to intersecting the other two sides, in distinct points, then the other two sides are divided in the same ratio.
Given : A triangle ABC in which DE || BC and DE intersects AB in D and AC in E.
To prove : $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$

Construction : Join BE, CD and draw EF ⊥ AB, DG ⊥ AC.
Proof : In ∆EAD and ∆EDB, here as EF is perpendicular to AB, therefore, EF is the height for both of triangles EAD and EDB.
Now,
Area of ∆EAD = $\frac { 1 }{ 2 }$ (base x height)
= $\frac { 1 }{ 2 }$ x AD x EF
Area of ∆EDB = $\frac { 1 }{ 2 }$(base x height)
= $\frac { 1 }{ 2 }$ x DB x EF

∵ ∆DBE, ∆ECD are on the same base DE and between the same parallels DE || BC, we have
Area of ∆DBE = Area of ∆ECD
Hence, (1) = (2)
i.e., $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$ (Q.E.D)

ii) Using the theorem, to find the length of $\overline{\mathrm{AE}}$.

AD = 1.8 cm, BD = 5.4 cm, EC = 7.2 cm
From the theorem, $\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{EC}}$
⇒ AE = $\frac{(\mathrm{AD})(\mathrm{EC})}{\mathrm{BD}}=\frac{1.8 \times 7.2}{5.4}$ = 2.4 cm
∴ AE = 2.4 cm

Question 13.
Draw ΔABC with sides 4.3 cm, 5.2 cm and 6.5 cm and then construct a tri-angle similar to ΔABC, whose sides are $\frac { 3 }{ 5 }$th of the corresponding sides.
Solution:
Constructing a similar triangle whose sides are $\frac { 3 }{ 5 }$ of corresponding sides.
Steps of construction:
1) Draw a triangle ΔABC with sides AB = 4.3 cm, BC = 5.2 cm and AC = 6.5 cm
2) Draw a ray $\overline{\mathrm{BX}}$ making an acute angle with BC on the side opposite to vertex ‘A’.

3) Locate 5 points B₁; B₂, B₃, B₄, B₅ on the ray $\overline{\mathrm{BX}}$ Such that
$\overline{\mathrm{BB}_{1}}=\overline{\mathrm{B}_{1} \mathrm{~B}_{2}}=\overline{\mathrm{B}_{2} \mathrm{~B}_{3}}=\overline{\mathrm{B}_{3} \mathrm{~B}_{4}}=\overline{\mathrm{B}_{4} \mathrm{~B}_{5}}$
4) Now join $\overline{\mathrm{B}_{5} \mathrm{C}}$ and draw another line through the point ‘B₃‘ such that it is parallel to $\overline{\mathrm{B}_{5} \mathrm{C}}$ which meets $\overline{\mathrm{BC}}$ at the point ‘C’.
5) Now draw another line from the point (C’) to $\overline{\mathrm{AB}}$ such that it is par¬allel to $\overline{\mathrm{AC}}$, which meets $\overline{\mathrm{AB}}$ at the point (A’).
6) Now ΔA’B’C’ is required similar tri-angle.

Question 14.
Draw a triangle ABC in which AB = 4 cm, BC = 6 cm and AC = 4 cm. Draw a triangle similar to ΔABC with its sides $\frac { 3 }{ 4 }$ times of the corresponding sides of ΔABC.
Solution:

Question 15.
Divide the line segment AB = 6 cm in the ratio of 3 : 2. Explain the con-struction procedure.
Solution:
Steps to construct:

1) Draw a line segment AB = 6 cm
2) Draw a line AX such that ∠BAX is an acute.
3) Make 5 equal parts on AX. (3 + 2)
4) Now join B with last part A₅.
5) Then draw a line from A₃ to AB which meets at ‘C’ such that it is parallel to BA₅.
6) So ‘C’ is the point which divides AB in the ratio of 3 : 2

Question 16.
In the given ΔABC, E and F are two points on AB and AC respectively. Then verify whether EF || BC or not in the following cases.
Solution:
From the converse of Thales theorem
EF || BC if $\frac{\mathrm{AE}}{\mathrm{EB}}=\frac{\mathrm{AF}}{\mathrm{FC}}$

Case I) AE = 3.9, EB = 3,
AF = 3.6, CF = 2.4
Then $\frac{\mathrm{AE}}{\mathrm{EB}}=\frac{3.9}{3}=\frac{1.3}{1} ; \frac{\mathrm{AF}}{\mathrm{FC}}=\frac{3.6}{2.4}=\frac{3}{2}$
So $\frac{A E}{E B} \neq \frac{A F}{F B}$ then EF ∦ BC
Case II) AE = 4, EB = 4.5;
AF = 8 and CF = 9
So $\frac{A E}{E B}=\frac{4}{4.5}=\frac{8}{9}$ and $\frac{\mathrm{AF}}{\mathrm{FB}}=\frac{8}{9}$
So then $\frac{\mathrm{AE}}{\mathrm{EB}}=\frac{\mathrm{AF}}{\mathrm{FB}}$
hence EF || BC.

Question 17.
In the given figure BC || DE and AD = DB = 3.4 and AC = 14.

So find AE and EC.
Solution:
In the given ΔABC, DE || BC is given.
So from Thales theorem,
$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$
In this problem, AD = DB = 3.4 is given.
And also AC = 14 is given.
So $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$
⇒ $\frac{3.4}{3.4}=\frac{\mathrm{AE}}{\mathrm{EC}}$
⇒ AE = EC
But AC = 14 is given.
AC = AE + EC = 2AE = 14
⇒ AE = EC = 7 cm

Question 18.
In given ΔABC points D and E are mid points of AB and AC and also BC = 6 cm then find DE. .
Solution:
As shown in the figure D, E are mid points of AB and AC.

Then AD = BD, AE = EC
∴ $\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{EC}}$ hence DE || BC (From the Converse of Thales theorem)
$\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{DE}}{\mathrm{BC}}=\frac{\mathrm{AE}}{\mathrm{EC}}$
⇒ $\frac{A D}{2 A D}=\frac{D E}{6}$
⇒ 2DE = 6 and DE = 3 cm
∴ DE = 3 cm

Question 19.
In the given figure, AB || CD || EF given AB = 73cm,DC=ycm,EF 4.5cm, BC = x cm. Calculate the values of x and y.

Solution:
Given that, AB || CD || EF and
AB = 7.5 cm, DC = y cm
According to Thales theorem
Δ𝜏CDE ~ Δ𝜏ABE

From (1) and (2)
$\frac{7.5-y}{7.5}=\frac{y}{4.5}$
7.5 y = (7.5) (4.5)-4.5 y
12y = (7.5) (4.5) = 33.75
y = $\frac{33.75}{12}$ = 2.8125
From (2)
$\frac{y}{4.5}=\frac{x}{x+3} ; \frac{2.8}{4.5}=\frac{\dot{x}}{x+3}$
4.5 x = (2.8) x + 8.4
4.5 x – (2.8) x = 8.4
1.7 x = 8.4
x = $\frac{8.4}{1.7}$ = 5 (approx)

AP 10th Class Maths Chapter 7 Important Questions Coordinate Geometry

August 14, 2024August 14, 2024 by Mahesh

AP 10th Class Maths Important Questions Chapter 7 Coordinate Geometry

These AP 10th Class Maths Chapter Wise Important Questions 7th Lesson Coordinate Geometry will help students prepare well for the exams.

7th Lesson Coordinate Geometry Class 10 Important Questions with Solutions

10th Class Maths Coordinate Geometry 1 Mark Important Questions

Question 1.
Write the distance between two points (x₁, y₁) and (x₂y, ₂.
Solution:
Distance between two points = $\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}$

Question 2.
Write the distance from the origin to (x, y).
Solution:
Distance from origin to (x, y) = $\sqrt{x^2+y^2}$

Question 3.
Find the distance between the points (-6, 7) and (-1, -5).
Solution:
Let A(-6, 7) and B(-1, -5) are points. Distance between the two points
= $\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}$
x₁ = – 6, x₂ = -1, y₁ = 7 and y₂ = – 5
AB = $\sqrt{(-1-(-6))^2+(-5-7)^2}$
= $\sqrt{(-1+6)^2+(-12)^2}$
= $\sqrt{5^2+144}$ = $\sqrt{169}$ = 13 units
Distance AB = 13 units

Question 4.
Find the distance (4, 3) from the origin.
Solution:
Let A(4, 3) be any point.
Distance from origin to (x, y) = $\sqrt{x^2+y^2}$
x = 4, y = 3
OA = $\sqrt{4^2+3^2}$ = $\sqrt{16+9}$ = $\sqrt{25}$ = 5 units
Therefore, distance from the origin to A(4, 3) is 5 units.

Question 5.
Write the section formula.
Solution:
P(x, y) = ($\frac{m_1 x_2+m_2 x_1}{m_1+m_2}$, $\frac{m_1 y_2+m_2 y_1}{m_1+m_2}$)

Question 6.
Write the midpoint of (x₁, y₁) and (x₂, y₂).
Solution:
Mid point = ($\frac{x_1+x_2}{2}$, $\frac{y_1+y_2}{2}$)

Question 7.
If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are vertices of AABC, then write its cen-troid.
Solution:
Centroid of the ΔABC = ($\frac{x_1+x_2+x_3}{3}$, $\frac{y_1+y_2+y_3}{3}$)

Question 8.
If end points of a diameter of a circle are (-5, 4) and (1, 0) then find the radius of the circle.
Solution:

Centre of circle = Midpoint of diameter
O = ($\frac{1-5}{2}$, $\frac{0+4}{2}$)
O = (-2, 2)
Radius = $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
= $\sqrt{(-5+2)^2+(4-2)^2}$
= $\sqrt{9+4}$
= $\sqrt{13}$ units.

Question 9.
Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:

Assertion (A) : If the points A(4, 3) and B(x, 5) lie on a circle with centre 0(2, 3), then the value of x is 2.
Reason (R) : Centre of a circle is the mid-point of each chord of the circle.

A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion
(A).
B) Both Assertion (A) and Reason (R) . are true, but Reason (R) is the correct explanation of the Assertion (A).
C) Assertion (A) is true, but Reason (R) is false.
D) Assertion (A) is false, but Reason (R) is true.
Solution:
C) Assertion (A) is true, but Reason (R) is false.

A : A(4, 3), B(x, 5), Centre O(2, 3)
AO = BO
AO² = BO²
(2 – 4)² + (3 – 3)² = (2 – x)² + (3 – 5)²
4 = (2 – x)² + 4
R is false 2 – x = 0
x = 2

Question 10.
Find the distance of the point (-6, 8) from origin.
Solution:
(0, 0), (-6, 8)
(x₁ y₁), (x₂ y₂)
Distance = $\sqrt{(-6)^2+8^2}$ = $\sqrt{36+64}$
= $\sqrt{100}$ = 10

Question 11.
Find the distance of the point (-1, 7) from X-axis.
Solution:

Distance = 1 unit

Question 12.
If (3, √3), B(0, 0)and C(3, k) are the three vertices of an equilateral triangle ABC, then find the value of k.
Solution:
A(3, √3), B(0, 0), C(3, k)
ΔABC is an equilateral triangle
AB = BC = AC
AB = $\sqrt{3^2+(\sqrt{3})^2}$ = $\sqrt{9+3}$ = $\sqrt{12}$
BC = $\sqrt{3^2+k^2}$ = $\sqrt{9+k^2}$
AC = $\sqrt{(3-3)^2+(k-\sqrt{3})^2}$ = k – √3
$\sqrt{12}$ = $\sqrt{9+k^2}$
12 = 9 + k² ⇒ k² = 12 – 9 ⇒ k² = 3
k = ±√3

Question 13.
Three vertices of a parallelogram ABCD are A(1, 4), B(-2, 3) and C(5,8). Find the ordinate of the fourth vertex D.
Solution:

Midpoint of AC = Midpoint of BD
($\frac{1+5}{2}$, $\frac{4+8}{2}$) = ($\frac{-2+x}{2}$, $\frac{y+3}{2}$)
(3, 6) = ($\frac{-2+x}{2}$, $\frac{y+3}{2}$)
$\frac{-2+x}{2}$ = 3, $\frac{y+3}{2}$ = 6
6 = -2 + x, 12 = y + 3
x = 8 ⇒ y = 9
Ordinate = 9

Question 14.
Points A(-1, y) and B(5, 7) lie on a circle with centre O(2, -3y). What are the values of y ?
Solution:

OA² = OB²
(2 + 1)² + (- 3y – y)² = (2 – 5)² + (- 3y – 7)²
9 + 16y² = 9 + (3y + 7)²
16y² = 9y² + 42y + 49
9y² + 42y + 49 – 16y² = 0
-7y² + 42y + 49 = 0
-7(y² – 6y – 7) = 0
y² – 6y – 7 = 0
y² – 7y + y – 7 =
y(y – 7) + 1(y – 7) = 0
(y + 1) (y – 7) = 0
y = -1, 7

Question 15.
If A(4, -2), B(7, -2) and C(7, 9) are the vertices of a ΔABC, then ΔABC is
A) equilateral triangle
B) isosceles triangle
C) right angled triangle
D) isosceles right angled triangle
Solution:
C) right angled triangle

A(4, -2), B(7, -2), C(7, 9)
AB = $\sqrt{(7-4)^2+(-2+2)^2}$ = $\sqrt{3^2}$ = 3
BC = $\sqrt{(7-7)^2+(9+2)^2}$ = $\sqrt{11^2}$ = 11
CA = $\sqrt{(7-4)^2+(9+2)^2}$ = $\sqrt{9+121}$ = $\sqrt{130}$
11² + 3² = 121 + 9 = 130 = ($\sqrt{130}$)²
∴ ΔABC is an right angled triangle.

Question 16.
The line segment joining the points P(-3, 2) and Q(5, 7) is divided by the Y-axis in the ratio
A) 3 : 1
B) 3 : 4
C) 3 : 2
D) 3 : 5
Solution:
D) 3 : 5
P(-3, 2), Q(5, 7)
PQ is divided by Y-axis then ($\frac{m_1 x_2+m_2 x_1}{m_1+m_2}$) = 0
m₁x₂ = -m₂x₁
$\frac{\mathrm{m}_1}{\mathrm{~m}_2}$ = $\frac{-\mathrm{x}_1}{\mathrm{~x}_2}$ = $\frac{3}{5}$ = 3 : 5

Question 17.
Find the distance between the points (a cos0+b sin6,0) and (0, asin6-bcos6).
Solution:
A (a cos θ + b sin θ, 0)
B (0, a sin θ – b cos θ)

Question 18.
What is the value of p, for the points A(3, 1), B(5,p) and C(7, -5) are collinear?
Solution:
A(3, 1), B(5, p), C(7, -5)
AB slope = BC slope
$\frac{y_2-y_1}{x_2-x_1}$ = $\frac{y_2-y_1}{x_2-x_1}$
$\frac{p-1}{5-3}$ = $\frac{-5-p}{7-5}$
$\frac{p-1}{2}$ = $\frac{-5-p}{2}$
p – 1 = -5 – p
2p = -4
P = -2

Question 19.
Find the point p on X-axis equidistant from the points A(-1, 0) and B(5, 0).
Solution:
A(-1, 0), B(5, 0)
Let P(x, 0)
AP² = BP²
(-1, 0) (x, 0) = (5, 0) (x, 0)
(x+ 1)² = (x – 5)²
x² + 2x + 1 = x² – 10x + 25
2x + 10x = 25 – 1
12x = 24
x = 2
∴ Point P = (2, 0)

Question 20.
Find the co-ordinates of the point which is reflection of point (-3, 5) in X-axis.
Answer:
(3, -5)

Question 21.
If the point p(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3 : 1, then find the value of y.
Solution:
A(6, 5), B (4, y), P(6, 2)
m₁ : m₂ = 3 : 1
($\frac{m_1 x_2+m_2 x_1}{m_1+m_2}$, $\frac{m_1 y_2+m_2 y_1}{m_1+m_2}$) = (6, 2)
$\frac{m_1 y_2+m_2 y_1}{m_1+m_2}$ = 2
3(y) + 1(5) = 2m₁ + 2m₂
3y + 5 = 2(3) + 2(1)
3y + 5 = 8 ⇒ 3y = 3
⇒ y = 1

Question 22.
Find the distance between the points (3, -2) and (-3, 2).
Solution:
A(3, -2), B(-3, 2)
(x₁, y₁), (x₂, y₂)
AB = $\sqrt{(-3-3)^2+(2+2)^2}$ = $\sqrt{36+4}$
= $\sqrt{40}$ = 2$\sqrt{10}$ units

Question 23.
If A(3, √3 ), B(0, 0) and C(3, k) are the three vertices of an equilateral triangle ABC, then find the value of k.
Solution:
A(3, √3), B(0, 0), C(3, k)
AB² = BC² = AC²
9 + 3 = 9 + k² = 0 + (k – √3)²
12 = 9 + k²
12 – 9 = k²
k = ±√3

Question 24.
A point (x, 1) is equidistant from (0,0) and (2, 0). Find the value of x.
Solution:
A(x, 1), B(0, 0), C(2, 0)
AB² = AC²
(0 – x)² + (0 – 1)² = (2 – 0)² + 0
x² + 1 = 4
x² = 3
x = ±√3

Question 25.
What is the ratio in which the point (4, 0) divides the line segment joining the points (4, 6) and (4, -8)?
Solution:
(4, 6), (4, -8) and p(4, 0)
(x₁, y₁), (x₂, y₂), (4, 0)
m₁ : m₂
$\frac{m_1 y_2+m_2 y_1}{m_1+m_2}$ = 0
m₁y₂ = -m₂y₁
$\frac{\mathrm{m}_1}{\mathrm{~m}_2}$ = $\frac{-y_1}{y_2}$ = $\frac{-6}{-8}$ = $\frac{3}{4}$
m₁ : m₂ = 3 : 4

Question 26.
The vertices of a triangle OAB are 0(0, 0), A(4, 0) and B(0,6). The median AD is drawn on OB. Find the length AD.

Solution:
D = Midpoint of OB = ($\frac{0+0}{2}$, $\frac{6+0}{2}$) = (0, 3)
AD = $\sqrt{4^2+3^2}$ = $\sqrt{25}$ = 5

Question 27.
The origin divides the line segment AB joining the points A(1, -3) and B(-3, 9) in the ratio.
A) 3 : 1
B) 1 : 3
C) 2 : 3
D) 1 : 1
Solution:
B) 1 : 3
A(1, -3), B(-3, 9)
(x₁, y₁), (x₂, y₂)
m₁ : m₂
($\frac{m_1 x_2+m_2 x_1}{m_1+m_2}$, $\frac{m_1 y_2+m_2 y_1}{m_1+m_2}$) = (0, 0)
m₁x₂ = -m₂x₂
$\frac{\mathrm{m}_1}{\mathrm{~m}_2}$ = $\frac{-x_1}{x_2}$ = $\frac{-1}{-3}$ = $\frac{1}{3}$
m₁ : m₂ = 1 : 3

Question 28.
The perpendicular bisector of a line segment A(-8, 0) and B(8, 0) passes through a point (0, k). Find the value of ok.
Solution:

∴ k = 0 only

Question 29.
A circle of radius 3 units is centered at (0, 0). Which of the following points lie outside the circle ?
A) (-1, -1)
B) (0, 3)
C) (1, 2)
D) (3, 1)
Solution:
D) (3, 1)

AO = $\sqrt{1+1}$ = √2 < 3
BO = $\sqrt{0+3^2}$ = 3 = 3
CO = $\sqrt{1^2+2^2}$ = √ 5 < 3
DO = $\sqrt{9+1}$ = $\sqrt{10}$ > 3
∴ (3, 1) lies outside of circle.

Question 30.
Find the value of a, for which point p($\frac{a}{3}$, 2) is the mid-point of the line
segment joining the points Q(-5, 4) and R(-1, 0).
Solution:
Midpoint of QR = P($\frac{a}{3}$, 2)
Q(-5, 4), R(-1, 0)
($\frac{-5+2}{2}$ + $\frac{4+0}{2}$) = ($\frac{a}{3}$, 2)
-3 = $\frac{a}{3}$
a = -9

Question 31.
Assertion : The point (0, 4) lies on Y-axis.
Reason (R): The x-coordinate of a point on Y-axis is zero.
Choose the correct answer.
A) Assertion (A) is true and Reason (R) is true and R is the correct explanation of A.
B) A is true and R is true, but R is not the correct explanation of A.
C) A is true, R is false.
D) A is false, R is true.
Answer:
A) Assertion (A) is true and Reason (R) is true and R is the correct explanation of A.

Question 32.
What is the perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0)?
Solution:
A(0, 4), B(0, 0), C(3, 0)
AB = $\sqrt{4^2}$ = 4
BC = $\sqrt{3^2}$ = 3
AC = $\sqrt{9+16}$ = 5
∴ Perimeter = 3 + 4 + 5 = 12

Question 33.
Assertion : The ratio in which the line segment joining (2, -3) and (5, 6) internally divided by X-axis is 1 : 2.
Reason (R): As formula for the internal division is ($\frac{m x_2+n x_1}{m+n}$, $\frac{m y_2+n y_1}{m+n}$)

Choose the correct answer.
A) Assertion (A) is true and Reason (R) is true and R is the correct explanation of A.
B) A is true and R is true, but R is not the correct explanation of A.
C) A is true, R is false.
D) A is false, R is true.
Solution:
A) Assertion (A) is true and Reason (R) is true and R is the correct explanation of A.
A(2, -3), B(5, 6)
(x₁, y₁), (x₂, y₂)
X-axis divides AB
$\frac{m_1 y_2+m_2 y_1}{m_1+m_2}$ = 0
m₁y₂ : -m₂y₁
$\frac{\mathrm{m}_1}{\mathrm{~m}_2}$ = $\frac{-y_1}{y_2}$ = $\frac{3}{6}$ = 1 : 2
‘A’ is true, ‘R’ is true.

Question 34.
Find the co-ordinate of the point p dividing the line segment joining the poins A(1, 3) and B(4, 6) internally in the ratio 2 : 1.
Solution:
A(1, 3), B(4, 6)
(x₁, y₁), (x₂, y₂)
m₁ : m₂ = 2 : 1
($\frac{m_1 x_2+m_2 x_1}{m_1+m_2}$, $\frac{m_1 y_2+m_2 y_1}{m_1+m_2}$)
= ($\frac{2(4)+1(1)}{2+1}$, $\frac{2(6)+1(3)}{2+1}$)
= ($\frac{9}{3}$, $\frac{15}{3}$) = (3, 5)

Question 35.
The mid-point of (3p, 4) and (-2, 2q) is (2, 6). Find the value of p + q.
Solution:

p + q = 2 + 4 = 6

Question 36.
Find the mid point of the line segment joining the points (-5, 7) and (-1, 3).
Solution:
Midpoint = ($\frac{x_1+x_2}{2}$, $\frac{y_1+y_2}{2}$)
= ($\frac{-5-1}{2}$, $\frac{7+3}{2}$) = (-3, 5)

Question 37.
The coordinate of a point A on Y-axis is 5 and B has coordinates (-3, 1) from length of AB = _______ units.
Solution:
Let A(0, 5), B(-3, 1)
AB = $\sqrt{(-3)^2+(1-5)^2}$ = $\sqrt{9+16}$
= $\sqrt{25}$ = 5

Question 38.
Find the midpoint of the line segment joining the point A(cos²α, 0), B(0, sin²α).
Answer:
($\frac{\cos ^2 \alpha}{2}$, $\frac{\sin ^2 \alpha}{2}$)

Question 39.
Assertion (A) : The midpoint of the line segment joining the points A(-1, 1), B(1, 3) is (9, 3).
Reason (R): Midpoint of (a, b) and (c, d) = ($\frac{a+c}{2}$, $\frac{b+d}{2}$)

Question 40.
Assertion (A) : The distance between the points (1, 0) and (-1, 0) is 2 units.
Reason (R) : Any point on Y-axis will be (0, 0).

Question 41.
Assertion (A) : If (6, 1), (4, 4) and (-1, 3) are three vertices of a parallelogram respectively then the fourth vertex is (0, 0).
Reason (R) : The diagonals of a para-llelogram are not equal.

Choose the correct answer.
A) Assertion (A) is true and Reason (R) is true and R is the correct explanation of A.
B) A is true and R is true; but R is not the correct explanation of A.
C) A is true, R is false.
D) A is false, R is true.
Answer:
C) A is true, R is false.

Question 42.
Assertion (A) : A point on Y-axis which is equidistant from the points (5, -2) and (-3, 2) is (0, -2).
Reason (R) : All sides of an equilateral triangle are equal.

Question 43.
Identify the distance between A(8, 3) and B (4, 3) from the following :
A) 12
B) 0
C) 4
D) 6
Answer:
C) 4

Question 44.
Write the nearest point to origin,
i) (2, -3)
ii) (5, 0)
iii) (0, -5)
iv) (1, 3)
Answer:
(1, 3)

Question 45.
The distance of a point (3, 4) from the origin is how many units ?
Answer:
5 units.

Question 46.
Find the distance between the points (x₁, y₁) and (x₂, y₂) which are on the line parallel to Y – axis.
Answer:
|y – y₂| or |y₂ – y₁|

Question 47.
If the distance between the points (3, k) and (4, 1) is $\sqrt{10}$, then find the value of k.
Answer:
4 (or) – 2.

Question 48.
Write a formula to the coordinates of the point which divides the line joining (x₁, y₁) and (x₂, y₂) in the ratio m : n internally.
Answer:
P = ($\frac{\mathrm{mx}_2+\mathrm{nx}_1}{\mathrm{~m}+\mathrm{n}}$, $\frac{\mathrm{my}_2+\mathrm{ny}_1}{\mathrm{~m}+\mathrm{n}}$)

Question 49.
Assertion : (0, 2) is a point on Y-axis.
Reason : Every point on Y-axis is at a distance of zero units from the Y-axis.

Now, choose the correct answer.
(A) Both Assertion and Reason are true. Reason is supporting the Assertion.
(B) Both Assertion and Reason are true, but Reason is not supporting the Assertion.
(C) Assertion is True, but Reason is False.
(D) Assertion is False, but Reason is True.
Answer:
A) Both Assertion and Reason are true.
Reason is supporting the Assertion.

Question 50.
Mid point of a line joining the two points (0, 0) and (4, 6) is ……………… .
Solution:
Mid point of line joining (0, 0) and (4, 6)
= ($\frac{x_1+x_2}{2}$, $\frac{y_1+y_2}{2}$)
= ($\frac{0+4}{2}$, $\frac{0+6}{2}$) = (2, 3)

Question 51.
Identify the distance between A(8, 3) and B (4, 3) from the following :
A) 12
B) 0
C) 4
D) 6
Answer:
C) 4

10th Class Maths Coordinate Geometry 2 Marks Important Questions

Question 1.
If the point A(4, 3) and B(x, 5) are on the circle with centre origin, find value of x.
Solution:
Given A(4, 3) and B(x, 5) are on the circle with centre O(0,0)
then radius OA = OB

x² + 25 = 25
x² = 25 – 25 = 0
Therefore, x = 0

Question 2.
Find a point on the y-axis which is equidistant from the point A(6, 5) and B(-4, 3).
Solution:
Let the point on the y-axis is P(0, y)
Given A(6, 5) and B(-4, 3) are equidistant from the point P(0, y)
That is AP = BP
Distance AP = Distance BP
Distance between two points

y² – 10y + 61 = y² – 6y + 25
– 10y + 6y = 25 – 61
– 4y = -36
y = $\frac{36}{4}$ = 9
Therefore midpoint on y-axis P is (0, 9).

Question 3.
Find the midpoint (5, 3) and (-1, 4).
Solution:
Given points A(5, 3) and B(-1, 4)

Question 4.
Find the coordinates of the centroid of a triangle whose vertices are (0,6), (8,12) and (1,0).
Solution:
Given A(0, 6), B(8, 12) and C(1, 0) are the vertices of ΔABC.
If (x₁, y₁), (x₂, y₂) and (x₃, y₃) are the vertices of a triangle, then its centroid
G = ($\frac{x_1+x_2+x_3}{3}$, $\frac{y_1+y_2+y_3}{3}$)
x₁ = 0, x₂ = 8, x₃ = 1, y₁ = 6, y₂ = 12 and y₃ = 0
Centroid (G) = ($\frac{0 + 8 + 1}{3}$, $\frac{6 + 12 + 0}{3}$)
= ($\frac{9}{3}$, $\frac{18}{3}$)
= (3, 6)
Therefore, centroid of ΔABC is (3, 6).

Question 5.
Find the coordinates of the point which divides the join of A(-1, 7) and B(4, -3) in the ratio 2 : 3.
Solution:
Given points
A(-1, 7), B(4, -3)
(x₁, y₁), (x₂, y₂)
Ratio = 2 : 3
= m₁ : m₂
P = ($\frac{\mathrm{m}_1 \mathrm{x}_2+\mathrm{m}_2 \mathrm{x}_1}{\mathrm{~m}_1+\mathrm{m}_2}$, $\frac{\mathrm{m}_1 \mathrm{y}_2+\mathrm{m}_2 \mathrm{y}_1}{\mathrm{~m}_1+\mathrm{m}_2}$)
P = ($\frac{2(4)+3(-1)}{2+3}$, $\frac{2(-3)+3(7)}{2+3}$)
P = ($\frac{8 – 3}{5}$, $\frac{- 6 + 21}{5}$)
P = ($\frac{5}{5}$, $\frac{15}{5}$) P = (1, 3)

Question 6.
If the points A(2, 3), B(-5, 6), C(6, 7) and D(p, 4) are the vertices of a parallelogram ABCD, find the value of P.
Solution:
The diagonals of a parallelogram will bisect each other.
Midpoint of AC = Midpoint of BD

8 = -5 + P ⇒ P 8 + 5 ⇒ P = 13

Question 7.
A line intersects Y-axis and X-axis at point P and Q, respectively. If R(2, 5) is the mid-point of the line segment PQ, then find the coordinates of P and Q.
Solution:
Let P(x, 0), Q(0, y), R(2, 5)
Midpoint of PQ = R(2, 5)
($\frac{x + 0}{2}$, $\frac{0 + y}{2}$) = (2, 5)
$\frac{x}{2}$ = 2, $\frac{y}{2}$ = 5
x = 4, y = 10
∴ P = (4, 0), Q = (0, 10)

Question 8.
Find the coordinates of the point which divides the line segment joining the points (7, -1) and (-3, 4) internally in the ratio 2 : 3.
Solution:
Let
(7, -1) B(-3, 4)
(x₁, y₁), (x₂, y₂)
Ratio = 2 : 3
= m₁ : m₂
P = ($\frac{\mathrm{m}_1 \mathrm{x}_2+\mathrm{m}_2 \mathrm{x}_1}{\mathrm{~m}_1+\mathrm{m}_2}$, $\frac{\mathrm{m}_1 \mathrm{y}_2+\mathrm{m}_2 \mathrm{y}_1}{\mathrm{~m}_1+\mathrm{m}_2}$)
P = ($\frac{2(-3)+3(7)}{2+3}$, $\frac{2(4)+3(-1)}{2+3}$)
P = ($\frac{-6 + 21}{5}$, $\frac{8 – 3}{5}$)
P = ($\frac{15}{5}$, $\frac{5}{5}$)
⇒ P = (3, 1)
∴ Required point (3, 1)

Question 9.
Find the value(s) of y for which the distance between the points A(3, -1) and B(11, y) is 10 units.
Solution:
Let A(3, -1), B(11, y)
given AB = 10
AB² = 10² = 100
(11 – 3)² + (y + 1)² = 100
64 + (y + 1)² = 100
(y + 1)² = 100 – 64
⇒ (y + 1)² = 36
y + 1 = $\sqrt{36}$
⇒ y + 1 = ± 6
y + 1 = 6 (or) y + 1 = -6
y = 5 (or) y = -7

Question 10.
Find the value(s) of ‘x’ so that PQ = QR, where the coordinates of P, Q and R are (6, -1), (1, 3) and (x, 8) respectively.
Solution:
Let P(6, -1), Q(1, 3), R(x, 8)
PQ = QR
PQ² = QR²
(1 – 6)² + (3 + 1)² = (x – 1)² + (8 – 3)²
25 + 16 = (x – 1)² + 25
x – 1 = $\sqrt{16}$
⇒ x – 1 = ± 4
x – 1 = 4 (or) x – 1 = – 4
x = 5 (or) x = -3

Question 11.
The vertices of a triangle are (-2. 0), (2, 3) and (1, -3). Is the triangle equilateral, isosceles or scalene ?
Solution:
Let the vertices of triangle are A(-2, 0), B(2, 3), C(1, -3)
AB = $\sqrt{(2+2)^2+(3-0)^2}$ = $\sqrt{16 – 9}$ = 5
BC = $\sqrt{(1-2)^2+(-3-3)^2}$ + $\sqrt{1 + 36}$ = $\sqrt{37}$
AC = $\sqrt{(1+2)^2+(-3)^2}$ + $\sqrt{9 + 9}$ = $\sqrt{18}$
AB ≠ BC ≠ AC
∴ ΔABC is a scalene triangle.

Question 12.
Find the coordinates of the point p which divides the join of A(-2, 5) and B(3, -5) in the ratio 2 : 3.
Solution:
Given points
A(-2, 5), B(3, -5)
(x₁, y₁), (x₂, y₂)
Ratio = m₁ : m₂ = 2 : 3
P($\frac{\mathrm{m}_1 \mathrm{x}_2+\mathrm{m}_2 \mathrm{x}_1}{\mathrm{~m}_1+\mathrm{m}_2}$, $\frac{\mathrm{m}_1 \mathrm{y}_2+\mathrm{m}_2 \mathrm{y}_1}{\mathrm{~m}_1+\mathrm{m}_2}$)
P = ($\frac{2(3)+3(-2)}{2+3}$, $\frac{2(-5)+3(5)}{2+3}$)
P = ($\frac{6 – 6}{5}$, $\frac{- 10 + 15}{5}$)
⇒ P = (0, 1)

Question 13.
If (1, $\frac{p}{3}$) is the mid-point of the line segment joining the points (2, 0) and (0, $\frac{2}{9}$) then show that the line 5x + 3y + 2 = 0 passes through the point (-1, 3p).
Solution:
Let A (2, 0), B(0, $\frac{2}{9}$)
Let C(1, $\frac{p}{3}$) in the midpoint of AB

Question 14.
P and Q are the points with co-ordinates (2, -1) and (-3, 4). Find the co-ordinates of the point R such that
PR is $\frac{2}{5}$ of PQ.
Solution:
Let
P(2, -1), Q(-3, 4)
(x₁, y₁), (x₂, y₂)
PR = $\frac{2}{5}$ of PQ.

Question 15.
Show that A(1, 2), B(5, 4), C(3, 8) and D(-1, 6) are vertices of parallelogram ABCD.
Solution:
Given points A(1, 2) B(5, 4) C(3, 8) D(-1, 6)
Distance between two points

AB = BC = CD = AD and AC = BD
ABCD is a square.
Every square is a parallelogram
∴ ABCD is a parallelogram.

Question 16.
Show that the points A(3, 0), B(6, 4) and C(-1, 3) are vertices of a right-angled triangle.
Solution:
Given A(3, 0), B(6, 4), C(-1, 3)
Distance between two points

50 = 25 + 25 ⇒ 50 = 50
∴ The given points are vertices of right angled triangle.

Question 17.
Find a relation between x and y such that the point p(x, y) is equidistant from the points A(-5, 3) and B(7, 2).
Solution:
P(x, y), A(-5, 3) and B(7, 2)
Given PA = PB
PA² = PB²
(x + 5)² + (y – 3)² = (x – 7)² + (y – 2)²
x² + 10x + 25 + y² – 6y + 9 = x² – 14x + 49 + y² – 4y + 4
10x – 6y + 34 = -14x – 4y + 53
10x + 14x – 6y + 4y = 53 – 34
24x – 2y = 19
∴ The required relation is 24x – 2y – 19 = 0

Question 18.
Find the value of k if the point P(0, 2) is equidistant from (3, k) and (k, 5).
Solution:
Let P(0, 2), A(3, k) B(k, 5)
Given PA = PB
PA² = PB²
(3 – 0)² + (k – 2)² = (k – 0)² + (5 – 2)²
9 + k² – 4k + 4 = k² + 9
– 4k + 4 = 0 ⇒ -4k = -4 ⇒ k = 1

Question 19.
Find the quadrant in which the line segment joining the points (7, -6) and (3, 4) is the ratio 1 : 2 internally.
Solution:
Given points
(7, -6), (3, 4)
(x₁, y₁), (x₂, y₂)
m₁ : m₂ = 1 : 2
P = ($\frac{\mathrm{m}_1 \mathrm{x}_2+\mathrm{m}_2 \mathrm{x}_1}{\mathrm{~m}_1+\mathrm{m}_2}$, $\frac{\mathrm{m}_1 \mathrm{y}_2+\mathrm{m}_2 \mathrm{y}_1}{\mathrm{~m}_1+\mathrm{m}_2}$)
P = ($\frac{1(3)+2(7)}{1+2}$, $\frac{1(4)+2(-6)}{1+2}$)
P = ($\frac{3 + 14}{2}$, $\frac{4 – 12}{3}$)
P = ($\frac{17}{3}$, $\frac{-8}{3}$) ∈ Q₄
∴ The point lies in Q₄.

Question 20.
Find the fourth vertex of a parallelogram PQRS whose three vertices are P(-2, 3), Q(6, 7) and R(8, 3).
Solution:

In a parallelogram, diagonals bisect each other.
∴ Midpoint of PR = Midpoint of QS
($\frac{-2 + 8}{2}$, $\frac{3 + 3}{2}$) = ($\frac{6 + x}{2}$, $\frac{7 + y}{2}$)
($\frac{6}{2}$, $\frac{6}{2}$) = ($\frac{6 + x}{2}$, $\frac{7 + y}{2}$)
6 = 6 + x
x = 6 – 6
x = 0

6 = 7 + y
7 + y = 6
y = 6 – 7 = -1
∴ Fourth vertex = (0, -1)

Question 21.
Find the distance between the points (0, 0) and (a, b).
Solution:
Given points are (0, 0) and (a, b). Distance lormula
$\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
Here x₁ = – 0, x₂ = a, y₁ = 0 and y₂ = b.
$\sqrt{(a-0)^2+(b-0)^2}$ = $\sqrt{a^2+b^2}$
∴ Required distance is $\sqrt{a^2+b^2}$.

Question 22.
Find the mid point of the line segment formed by the points (- 5, 5) and (5, -5).
Solution:
Formula for the mid point of line segment formed bv (x₁, y₁) and (x₂, y₂) is [$\frac{x_1+x_2}{2}$, $\frac{y_1+y_2}{2}$] here x₁ = -5, y₁ = 5 and x₂ = 5, y₂ = -5
∴ Mid point = [$\frac{- 5 + 5}{2}$, $\frac{5 – 5}{2}$] = [$\frac{0}{2}$, $\frac{0}{2}$] = [0, 0]

Question 23.
A (0, 3), B (k, 0) and AB = 5. Find the positive value of k.
Solution:
Distance = $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
5 = $\sqrt{(k-0)^2+(0-3)^2}$
25 = k² + 9
k² = 16
⇒ k = ±4
∴ Positive value of k is 4.

Question 24.
Find the distance between the points (1, 5) and (5, 8).
Solution:
(1, 5) (5, 8)
Distance = $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
= $\sqrt{(5-1)^2+(8-\dot{5})^2}$
= $\sqrt{(4)^2+(3)^2}$
= $\sqrt{16 + 9}$ = $\sqrt{25}$
= 5 units.

Question 25.
What is the other end of the diameter of the Circle, whose centre is (1, 2) and one end point of the diameter is (3, 4) ?
Solution:
Let the other end point of diameter be (x, y).
($\frac{3 + x}{2}$, $\frac{4 + y}{2}$) = (1, 2)

$\frac{3 + x}{2}$ = 1
3 + x = 2
x = 2 – 3 = – 1

$\frac{4 + y}{2}$ = 2
4 + y = 4
y = 4 – 4 = 0
∴ The required point is (-1, 0).

Question 26.
Find the co-ordinates of the point, which divides the line segment joining (2, 0) and (0, 2) in the ratio 1 : 1.
Solution:
x₁ = 2 ; x₂ = 0 ; y₁ = 0 ; y₂ = 2
Point divides the line segment joining (2, 0) and (0, 2) in the ratio 1 : 1
or mid point = ($\frac{x_1+x_2}{2}$, $\frac{y_1+y_2}{2}$)
= ($\frac{2 + 0}{2}$, $\frac{0 + 2}{2}$)
= (1, 1)

Question 27.
Find the distance between (a cos θ, 0) and (0, a sin θ).
Solution:
To find the distance between the points (a cos θ, 0) and (0, a sin θ) we use the formula.
= $\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}$
Where x₁ = 0, y₁ = a sin θ and x₂ = a cos θ, y₂ = 0
∴ The distance between above given two points
= $\sqrt{(a \cos \theta-0)^2+(0-a \sin \theta)^2}$
= $\sqrt{a^2 \cos ^2 \theta+a^2 \sin ^2 \theta}$
= $\sqrt{a^2\left(\cos ^2 \theta+\sin ^2 \theta\right)}$
= $\sqrt{\mathrm{a}^2(1)}$ = $\sqrt{\mathrm{a}^2}$ = a units
(∵ sin² θ + cos² θ = 1)

Question 28.
If A(4, 0), B(0, y) and AB = 5, find the possible values of y.
Solution:
A(4,0), B(0, y) and AB = 5
$\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$ = 5
$\sqrt{16+y^2}$ = 5 ⇒ 16 + y² = 25
y² = 25 – 16 = 9
y = ± √9 = ± 3
Possible values of y are 3 or – 3.

Question 29.
Find the radius of the circle with centre (3, 2) and passes through (4, – 1).
Solution:

Radius = AB
Distance formula = $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
Radius ‘r’ = $\sqrt{(4-3)^2+(-1-2)^2}$
= $\sqrt{(1)^2+(-3)^2}$
= $\sqrt{1 + 9}$ = $\sqrt{10}$ units.

Question 30.
Find the radius of the circle whose cen-tre is (3,2) and passes through (- 5, 6).
Solution:
Given : A circle with centre A (3, 2) passing through B (- 5, 6).

Radius = AB
[∵ Distance of a point from the centre of the circle]
Distance formula = $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
Radius ‘r’ = $\sqrt{(-5-3)^2+(6-2)^2}$
= $\sqrt{64 + 16}$ = $\sqrt{80}$
= $\sqrt{16 \times 5}$ = 4√5 units

Question 31.
What is the distance between (0, – sin x) and (- cos x, 0) ?
Solution:
Distance between (0, – sin x) and (- cos x, 0)
= $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
= $\sqrt{(-\cos x-0)^2+(0+\sin x)^2}$
= $\sqrt{\cos ^2 x+\sin ^2 x}$
= √1
= 1 unit

Question 32.
Find the mid point of the line segment joining the points (3, 0) and (-1, 4).
Solution:
Mid point of (x₁ y₁), (x₂, y₂)
= ($\frac{x_1+x_2}{2}$, $\frac{y_1+y_2}{2}$)
∴ Mid point of (3, 0) and (-1, 4)
= ($\frac{3 – 1}{2}$, $\frac{4 + 0}{2}$) = ($\frac{2}{2}$, $\frac{4}{2}$)
= (1, 2)

Question 33.
Find the midpoint of the line segment joining the A(- 2, 3), B(0, 7).
Solution:
Given
A(- 2, 3) B(0, 7)
x₁, y₁ x₂, y₂
Mid-point of AB = ($\frac{x_1+x_2}{2}$, $\frac{y_1+y_2}{2}$)
= ($\frac{- 2 + 0}{2}$, $\frac{3 + 7}{2}$)
= (-1, 5)

Question 34.
Find the distance between the two points (7, 8) and (- 2, 3).
Solution:
A(x₁ y₁) = (7, 8) B(x₂, y₂) = (-2, 3)
Distance between two points = $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
AB = $\sqrt{(-2-7)^2+(3-8)^2}$
= $\sqrt{(-9)^2+(-5)^2}$
= $\sqrt{81 + 25}$ = $\sqrt{106}$

Question 35.
The base BC of an equilateral ΔABC lies on the y-axis. The coordinates of C are (0, -3). If the origin is the mid-point of the base BC, what are the co-ordinates of A and B ?
Solution:

To satisfy mid point B(0, 3) is taken.
BC = 6 units
Let A(x, 0)
AB = $\sqrt{x^2+9}$
BC = $\sqrt{(-3-3)^2}$ = $\sqrt{36}$
AB² = BC²
x² + 9 = 36 ⇒ x² = 27 ⇒ x = ±3√3
A(±3√3, 0) , B(0, 3)

Question 36.
If the point p(k, 0) divides the line segment joining the points A(2, -2) and B(-7, 4) in the ratio 1 : 2 then find the value of k.
Solution:
P(k, 0) divides line joining
A(2, -2), B(-7, 4)
(x₁ y₁)> (x₂, y₂)
m₁ : m₂ = 1 : 2
$\frac{m_1 y_2+m_2 y_1}{m_1+m_2}$ = 0
m₁y₂ + m₂y₁ = 0 ⇒ m₁y₂ = -m₂y₁
$\frac{\mathrm{m}_1}{\mathrm{~m}_2}$ = $\frac{-y_1}{y_2}$ = $\frac{2}{4}$ = $\frac{1}{2}$
$\frac{m_1 x_2+m_2 x_1}{m_1+m_2}$ = k
$\frac{1(-7)+2(2)}{1+2}$ = k ⇒ $\frac{- 7 + 4}{3}$ = k
$\frac{-3}{3}$ = -1 = k
∴ k = -1

10th Class Maths Coordinate Geometry 4 Marks Important Questions

Question 1.
Show that A(6, 4), B(5, -2) and C(7, -2) are the vertices of an isosceles triangle. Also find the length of the median through A.
Solution:
Given A(6, 4), B(5, -2) and C(7, -2) are the vertices of an isosceles triangle ABC.

Distance between two points

AB = AC = $\sqrt{37}$ units
Therefore ΔABC is isosceles.
Let D be the mid-point of BC.

Therefore, length of the median AD = 6 units.

Question 2.
If a point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), then find the value of p.
Solution:
Given A(0,2) is the equidistant from the points B(3, p) and C(p, 5).
That is AB = AC
AB² = AC²
Distance between two points

⇒ -4p = 9 – 13
⇒ -4p = -4 ⇒ p = $\frac{-4}{-4}$ = 1
Therefore p = 1

Question 3.
Determine the ratio in which the line 3x + y – 9 = 0 divides the segment join-ing the points (1, 3) and (2, 7).
Solution:
Let C(x, y) divides the line segment join¬ing the points A(l, 3) and B(2, 7) in the ratio k : 1 and C lies on 3x + y – 9 = 0. Section formula

= ($\frac{2 \mathrm{k}+1}{\mathrm{k}+1}$, $\frac{7 \mathrm{k}+3}{\mathrm{k}+1}$)
But given C(x, y) lies on 3x + y – 9 = 0
3 = ($\frac{2 \mathrm{k}+1}{\mathrm{k}+1}$, $\frac{7 \mathrm{k}+3}{\mathrm{k}+1}$) – 9 = 0
⇒ $\frac{3(2 k+1)+(7 k+3)-9(k+1)}{k+1}$ = 0
⇒ 6k + 3 + 7k + 3 – 9k – 9 = 0
⇒ 13k – 9k + 6 – 9 = 0 ⇒ 4k – 3 = 0
∴ k = $\frac{3}{4}$
⇒ k : 1 = $\frac{3}{4}$ : 1 = 3 : 4
Therefore dividing ratio is 3 : 4.

Question 4.
If A(-2, -1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram. Find the values of a and b.
Solution:
Given A(-2, -1), B(a, 0), C(4, b) and D(1,2) are the vertices of a parallelogram ABCD.
In parallelogram diagonals bisect each other.
That is mid-point of AC = mid-point of BD.
Midpoint = ($\frac{x_1+x_2}{2}$, $\frac{y_1+y_2}{2}$)
AC mid-point = BD mid-point

Therefore a = 1 and b = 3

Question 5.
Find the ratio in which die point (2, y) divides the line segment joining the points A(-2, 2), B(3, 7). Also find the value of y.
Solution:
Let P(2, y) can divide the line segment joining by the points A(-2,2), B(3, 7) in the ratio nij: m2.
Section formula

3m₁ – 2m₂ = 2(m₁ + m₂)
3m₁ – 2m₂ = 2m₁ + 2m₂
1m₁ = 4m₂
$\frac{m_1}{m_2}$ = $\frac{4}{1}$ = 4 : 1
Therefore, m₁ : m₂ = 4 : 1
Put m₁ : m₂ = 4 : 1 in
$\frac{7 m_1+2 m_2}{m_1+m_2}$ = y ⇒ $\frac{7(4)+2(1)}{4+1}$ = y
⇒ $\frac{28+2}{5}$ = y ⇒ y = $\frac{30}{5}$ = 6
∴ y = 6.

Question 6.
If A and B are (1, 4) and (5, 2) respectively, find the coordinates of P
when
$\frac{\mathrm{AP}}{\mathrm{BP}}$ = $\frac{3}{4}$.
Solution:
Let P(x, y) can divide the line joining by the points A(1, 4), B(5, 2) in the ratio AP : BP = 3 : 4
Section formula
= ($\frac{m_1 x_2+m_2 x_1}{m_1+m_2}$, $\frac{m_1 y_2+m_2 y_1}{m_1+m_2}$)
P = ($\frac{3.5+4.1}{3+4}$, $\frac{3.2+4.4}{3+4}$)
= ($\frac{15+4}{7}$, $\frac{6+16}{7}$)
Therefore coordinates are P(x, y) = ($\frac{19}{7}$, $\frac{22}{7}$)

Question 7.
Do the points A(3, 2), B(-2, -3) and C(2, 3) form alriangle ? If so, name the type of triangle formed.
Solution:
Given points are A(3, 2), B(-2, -3) and C(2, 3)
Distance between two points

From the above, AB + BC > AC, AC + BC > AB and AB + AC > BC
Therefore, A, B and C form a triangle.

So, AB² + AC² = BC²
Therefore, ΔABC is right angle triangle and ∠A is the right angle.

Question 8.
Show that the points (1, -1), (5, 2) and (9, 5) are collinear.
Solution:
Let A(1,-1), B(5, 2) and C(9, 5) are given points.
Distance between two points

So, 10 = 5 + 5
AC = AB + BC
Therefore. A, B, C are collinear points.

Question 9.
Find a point on the x-axis which is equidistant from the points (7, 6) and (-3, 4).
Solution:
Let a point on the x-axis is P(x, 0) which is equidistant from the points A(7, 6) and B(-3, 4)
that is AP = BP
AP² = BP²
Distance between two points

x = 3
Therefore required point is P(3, 0).

Question 10.
Find the centre of the circle passing through (6, -6), (3, -7) and (3, 3).
Solution:
Let A(6, -6), B(3, -7) and C(3, 3) are the points on the circle which are the vertices of ΔABC.
Let P(x, y) be the centre of the circle.

then AP = BP = CP
that is AP² = BP² = CP²
Distance between two points
= $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
AP² = BP²

+ 14y + 58 = – 6y + 18
+ 14y + 6y = 18 – 58 = – 40
+ 20y + – 40
y = $\frac{-40}{20}$
Put y = 2 in (1) ⇒ 6x + 2 (-2) = 14
⇒ 6x = 14 + 4
⇒ 6x = 18
⇒ x = $\frac{18}{6}$ = 3
Therefore centre of the circle P(x, y) = (3, -2)

Question 11.
The coordinates of one end of the diamter of a circle is (4, -1) and the co-ordinates of centre of the circle is (1, -3). Find the coordinates of the other end of the diameter.
Solution:
Let the two end of the diameter of a circle are A(4, -1) and B(x, y).

Centre of the circle O(1, -3)
Midpoint of AB = Centre of the circle.
Mid-point = (($\frac{x_1+x_2}{2}$, $\frac{y_1+y_2}{2}$))
= ($\frac{4+x}{2}$, $\frac{-1+y}{2}$) = (1, -3)
by equating the x and y-coordinates

$\frac{4+x}{2}$ = 1
4 + x = 2 × 1
x = 2 – 4 = -2

$\frac{-1+y}{2}$ = -3
– 1 + y = -3 × 2
y = – 6 + 1 = 5
Therefore other end of the diameter B(x, y) = (-2, 5).

Question 12.
Two vertices of a triangle are (3, -5) and (-7, 4). It Its centroid is (2, -1) then find the third vertex.
Solution:
Let the vertices of ΔABC are A(-3, 5), B(-7, 4) and C(x, y) and its centroid G = (2, -1)
Centroid of the triangle
= ($\frac{x_1+x_2+x_3}{3}$, $\frac{y_1+y_2+y_3}{3}$)
Centroid of ΔABC
= ($\frac{3+(-7)+x}{3}$, $\frac{-5+4+y}{3}$) = G = (2, -1)
= ($\frac{3-7+x}{3}$, $\frac{-1+\mathrm{y}}{3}$) = (2, -1)
By equating the x, y- coordinates
$\frac{-4+x}{3}$ = 2
-4 + x = 2 × 3
x = 6 + 4 = 10

$\frac{-1+y}{3}$ = -1
-1 + y = -1 × 3
y = -3 + 1 = -2
Therefore, third vertex of the triangle C(x, y) = (10, -2).

Question 13.
Determine the ratio in which the point P(a, +2) divides the line segment joining the points A(-4, 3) and
B(2, -4). Also find the value of ‘a’.
Solution:

Question 14.
In the given figure, in ΔABC points D and E are mid-points of sides BC and AC respectively. If given vertices are A(4, -2), B(2, -2) and C(-6, -7), then verify the result DE = $\frac{1}{2}$AB.

Solution:
Let A(4, -2), B(2, -2), C(-6, -7)
D – Midpoint of BC

Question 15.
If the point C(-1, 2) divides internally the line segment joining A(2, 5) and B(x, y) in the ratio 3 : 4, find the coordinates of B.
Solution:
Let C(-1, 2)

Question 16.
If the mid-point of the line segment joining the points A(3, 4) and B(k, 6) is p(x, y) and x + y – 10 = 0, find the value of k.
Solution:
Given points A(3, 4) and B(k, 6)
Midpoint of AB = P(x, y)
($\frac{3+\mathrm{k}}{2}$, $\frac{4+6}{2}$) = (x, y)
$\frac{3+\mathrm{k}}{2}$ = x ⇒ $\frac{10}{2}$ = y ⇒ y = 5
Given x + y – 10 = 0
$\frac{3+\mathrm{k}}{2}$ + 5 – 10 = 0 ⇒ $\frac{3+\mathrm{k}}{2}$ – 5 = 0
$\frac{3+\mathrm{k}}{2}$ = 5 ⇒ 3 + k = 10 ⇒ k = 10 – 3
∴ k = 7

Question 17.
Krishna has an apple orchard which has a 10 m × 10 m sized kitchen garden attached to it. She divides it into a 10 × 10 grid and puts soil and manure into it. She grows a lemon plant at A, a coriander plant at B, an onion plant at C and a tomato plant at D. Her husband Ram praised her kitchen garden and points out that on joining A, B, C, D they may form a parallelogram. Look at the below figure carefully and answer the following questions:

i) Write the coordinates of the points A, B, C and D, using the 10 × 10 grid as coordinate axes.
ii) Find whether ABCD is a parallelogram or not.
Solution:
i) Let A(2, 2), B(5, 4), C(7, 7), D(4, 5)

All sides equal, diagonals not equal.
∴ ABCD is a rhombus.
∴ ABCD is a parallelogram.

Question 18.
Find the ratio in which the line segment joining the points A(6, 3) and B(-2, -5) is divided by X-axis.
Solution:
Let the points A(6, 3), B (-2, -5)
(x₁, y₁) (x₂, y₂)
AB is divided by X-axis
The y-coordinate of that point is 0
$\frac{m_1 y_2+m_2 y_1}{m_1+m_2}$ = 0
m₁y₂ + m₂y₁ = 0
m₁y₂ = -m₂y₁
$\frac{m_1}{m_2}$ = $\frac{-y_1}{y_2}$ = $\frac{-3}{-5}$
m₁ : m₂ = 3 : 5
∴ Required ratio = 3 : 5

Question 19.
Show that the points A(1, 7), B(4, 2), C(-1, -1) and D(-4, 4) are vertices of the square ABCD.
Solution:
Let A(1, 7), B(4, 2), C(-1, -1), D(-4, 4)

All sides are equal
AB = BC = CD = AD
Two diagonals are equal.
AC = BD
∴ ABCD is a square.

Question 20.
Prove that the points A(-1, 0), B(3, 1), C(2, 2) and D(-2, 1) are the vertices of a parallelogram ABCD. Is it also a rectangle ?
Solution:
Let A(-1, 0), B(3, 1), C(2, 2) and D(-2, 1).

AC = $\sqrt{(2+1)^2+(2-0)^2}$ = $\sqrt{9+4}$ = $\sqrt{13}$
BD = $\sqrt{(-2-3)^2+(1-1)^2}$ = $\sqrt{25}$ = 5
Appropriate sides are equal and diagonals are not equal.
∴ ABCD is only a parallelogram.

Question 21.
Find the co-ordinate of the points of trisection of the line-segment joining the points (5, 3) and (4, 5).
Solution:
Given points A(5, 3), B(4, 5)
(x₁, y₁) (x₂, y₂)
In points of section the ratio will be 1 : 2 (or) 2 : 1
1 : 2 = m₁ : m₂

Question 22.
In what ratio does the X-axis divide the line segment joining the points (-4, -6) and (-1, 7) ? Find the coordinates of the point of division.
Solution:
Given points A(-4, -6) and B(-1, 7)
(x₁, y₁) (x₂, y₂)
X-axis divides AB
Let the ratio = m₁ : m₂
The y-coordinate of that point is 0

Question 23.
The vertices of a triangle are A(-1, 3), B(1, -1) and C(5, 1). Find the length of the median through the vertex.
Solution:

∴ Length of median from vertex C is 5 units.

Question 24.
Read the following passage and answer the questions that follows: In a classroom, four students. Sita, Gita, Rita and Anita are sitting at A(3, 4), B(6, 7), C(9, 4), D(6, 1) respectively, then a new student Anjali joins the class.

i) Teacher tells Anjali to sit in the middle of the four students. Find the coordinates of the position where she can sit.
ii) Calculate the distance between Sita and Anitha
iif) Which two students are equidistant from Gita ?
Solution:
i) Let A(3, 4), B(6, 7), C(9, 4), D(6, 1)
Midpoint of AC = ($\frac{3+9}{2}$, $\frac{4+4}{2}$)
= ($\frac{12}{2}$, $\frac{8}{2}$) = (6, 4)

ii) Sita A(3, 4), Anita D (6, 1)
AD = $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
AD = $\sqrt{(6-3)^2+(1-4)^2}$
AD = $\sqrt{9+9}$ = $\sqrt{18}$ = $\sqrt{9 \times 2}$
= 3√2 units

iii) Sita and Rita.

Question 25.
If the point C(-l, 2) divides internally the line segment joining the points A(2, 5) and B(x, y) is the ratio 3 : 4 find the value of x² + y².
Solution:

Question 26.
If the coordinate of the points A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that AP = $\frac{3}{7}$ AB. Where P lies on the line segment AB.
Solution:

Question 27.
Find the value of k, for which the points (7, 2), (5, 1) and (3, k) are collinear.
Solution:
The area of the triangle formed by those points = 0
Area of the triangle
$\frac{1}{2}${x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)} = 0
$\frac{1}{2}$|7(1 – k) + 5(k – 2) + 3(2 – 1)| = 0
$\frac{1}{2}$|-2k) = 0 .-. k = 0

Question 28.
If the distance between two points (x, 1) and (-1, 5) is ‘5’, find the value of V.
Solution:
Given points (x, 1) and (-1, 5)
Let A (x, 1) and B (-1, 5)
Distance between
$\overline{\mathrm{AB}}$ = $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
= $\sqrt{(-1-x)^2+(5-1)^2}$
= $\sqrt{(-1-x)^2+4^2}$ = 5
Now, squaring on both sides
(-1 – x)² + 16 = 25
(-1 – x)² = 25 – 16 = 9
(-1 – x)² = 32 ⇒ -1 – x = 3
-1 – 3 = x ⇒ x = -4

Question 29.
Verify whether the following points are collinear or not.
(1, -1), (4, 1), (-2, -3)
Solution:
To show that three points are collinear the area formed by the triangle is zero. Given points are (1, -1), (4, 1), (-2, -3) Formula for area of triangle
Δ = $\frac{1}{2}$|x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃ (y₁ – y₂)|
= $\frac{1}{2}$|1(1 + 3) + 4(- 3 + 1) – 2(- 1 – 1)|
= $\frac{1}{2}$|4 – 8 + 4| = $\frac{1}{2}$|8 – 8|
= $\frac{1}{2}$|0| = 0
So the given three points are collinear.

Question 30.
Find out whether the points (1, 5), (2, 5) and (-2, -1) are collinear using the distance formula.
Solution:
A(1, 5), B(2, 5), C(-2, -1)
Distance formula,

Question 31.
Find the value of k for which the points A (1, 2), B (-1, k), C (-3, -4) are collinear.
Solution:
Given A(1, 2), B(-1, k), C(-3, -4) are collinear.
∴ Area of ΔABC = 0
⇒ $\frac{1}{2}$|x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃ (y₁ – y₂)| = o
⇒ $\frac{1}{2}$|1[k – (- 4) + (-1)[- 4 – 2] (- 3) [2 – k] | = 0
⇒ $\frac{1}{2}$|k + 4 + 6 – 6 + 3k| = 0
⇒ $\frac{1}{2}$|4k + 4| = 0 ⇒ 4k + 4 = 0
⇒ 4k = -4 ⇒ k = $\frac{-4}{4}$ = – 1
(OR)
If A, B, C are collinear, then Slope of AB = Slope of BC
Now Slope of AB = $\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}$ = $\frac{\mathrm{k}-2}{-2}$
Slope of AC = $\frac{-4-2}{-3-1}$ = $\frac{-6}{-4}$ = $\frac{3}{2}$
So $\frac{\mathrm{k}-2}{-2}$ = $\frac{3}{2}$ ⇒ 2k – 4 = -6
⇒ 2k = – 2 .-. k = – 1

Question 32.
Show that the points A = (4, 2), B (7, 5) and C(9, 7) are collinear.
Solution:
To show that three points are collinear the area formed by the triangle is zero. Area of triangle
Δ = $\frac{1}{2}$|x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃ (y₁ – y₂)|
Here x₁ = 4, y₁ = 2, x₂ = 7, y₂ = 5, x₃ = 9, y₃ = 7
Now Δ = $\frac{1}{2}$|4(5 – 7) + 7(7 – 2) + 9(2 – 5)|
= $\frac{1}{2}$|4(- 2) + 7(5) + 9 (-3)|
= $\frac{1}{2}$|- 8 + 35 – 27|
= $\frac{1}{2}$|- 35 + 35| = $\frac{1}{2}$|0| = 0
So the above three are collinear.

Question 33.
If the distance between die two points (8, x) and (x, 8) is 2√2 units, then find the value of ‘x’.
Solution:
Given points are (8, x) and (x, 8) distance between them is 2√2 units.
d = $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
2√2 = $\sqrt{(x-8)^2+(8-x)^2}$
2√2 = $\sqrt{2(x-8)^2}$
2√2 = √2 $\sqrt{(x-8)^2}$
⇒ ± 2 = x – 8
x – 8 = 2 or x – 8 = -2
⇒ x = 10 or x = 6

Question 34.
Find the coordinates of point which divides the segment joining (2, 3) and (- 4, 0) in 1 : 2.
Solution:

10th Class Maths Coordinate Geometry 8 Marks Important Questions

Question 1.
The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(2, -5) and R(-3, 6), then find the coordinates of P.
Solution:
Let the point P(x, y) be the point equidistant from Q(2, -5) and R(-3, 6) and
given x = 2y. That is PQ = PR
Distance between two points = $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
Distance PQ
= $\sqrt{(x-2)^2+(y-(-5))^2}$
= $\sqrt{(x-2)^2+(y+5)^2}$
= $\sqrt{x^2-4 x+4+y^2+25+10 y}$

2y + 29 = 45
2y = 45 – 29
2y = 16
y = $\frac{16}{2}$ = 8
x = 2y = 2 × 8 = 16
Therefore P(x, y) = (16, 8)

Question 2.
Show that four points (0, -1), (6, 7), (-2, 3) and (8, 3) are the vertices of a rectangle also find its area.
Solution:
Let A(0, -1), B(6, 7), C(-2, 3) and D(8, 3) be the given points.

CD = $\sqrt{(8-(-2))^2+(3-3)^2}$
= $\sqrt{(8+2)^2+0^2}$ = $\sqrt{10^2}$
CD = $\sqrt{100}$ = 10 units
AB = CD = 10 units
In the given quadrilateral AD = BC, BD = AC and diagonals AB = CD.
So, ADBC is a rectangle.
Area of the rectangle = AD × AC = 4√5 × 2√5
Therefore area of the rectangle = 8 × 5 = 40 sq. units

Question 3.
If the vertices of an equilateral triangle (0,0), (3, √3), find the third vertex.
Solution:
Let O(0, 0), A(3, √3 ) and B(x, y) be the vertices of an equilateral triangle OAB.

That is OA = OB = AB.
OA² = OB² = AB²
Distance between two points = $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
OA = $\sqrt{(0-3)^2+(0-\sqrt{3})^2}$
OA² = (-3)² + (-√3)²
= 9 + 3 = 12 units
OB = $\sqrt{(0-x)^2+(0-y)^2}$
= $\sqrt{(-x)^2+(-y)^2}$
OB² = x² + y² units
AB = $\sqrt{(x-3)^2+(y-\sqrt{3})^2}$

(x, y) = (3, – √3)
Therefore, the third vertex B are (0, 2√3) or(3,- √3).

Question 4.
Prove that the points A(2, 3), B(-2, 2), C(-1, -2) and D(3, -1) are the vertices of square ABCD.
Solution:
Given A(2, 3), B(-2, 2), C(-1, -2) and D(3, -1) are the vertices of quadrilateral ABCD.

If in a quadrilateral all sides are equal and diagonals are equal, then it is said to be square.
Therefore, ABCD is a square.

Question 5.
Prove that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right angled triangle.
Solution:
Let A(7, 10), B(-2, 5), C(3, -4) are the vertices of ΔABC.
Distance between two points = $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
AB = $\sqrt{(-2-7)^2+(5-10)^2}$
= $\sqrt{(-9)^2+(-5)^2}$
= $\sqrt{81+25}$
= $\sqrt{106}$ units

AB² + BC² = ($\sqrt{106}$)² + ($\sqrt{106}$)² = 106 + 106 = 212
AB² + BC² = AC²
So, given points are vertices of an isos-celes right angled triangle.

Question 6.
Point X divides the line segment joining the points A(-1, 3) and B(9, 8) such that $\frac{\mathrm{AD}}{\mathrm{DB}}$ = $\frac{\mathbf{k}}{\mathbf{l}}$. If x lies on the line x – y + 2 = 0, find the value of k.
Solution:
Let X(x, y) divides the line segment joining the points A(-1, 3) and B(9, 8) in the ratio AX : BX = k : 1 and X(x, y) lies on x – y + 2 = 0 Section formula = ($\frac{m_1 \mathbf{x}_2+m_2 x_1}{m_1+m_2}$, $\frac{m_1 \mathbf{y}_2+m_2 y_1}{m_1+m_2}$)
x₁ = -1, y₁ = 3; x₂ = 9, y₂ = 8;
m₁ = k, m₂ = 1
X(x, y) = ($\frac{k \cdot 9+1 \cdot(-1)}{k+1}$, $\frac{k \cdot 8+1 \cdot 3}{k+1}$)
= ($\frac{9 \mathrm{k}-1}{\mathbf{k}+1}$, $\frac{8 \mathrm{k}+3}{\mathbf{k}+1}$)
⇒ But, given X(x, y) lies on x – y + 2 = 0
($\frac{9 \mathrm{k}-1}{\mathbf{k}+1}$, $\frac{8 \mathrm{k}+3}{\mathbf{k}+1}$) + $\frac{1}{2}$ = 0
⇒ $\frac{9 k-1-(8 k+3)+2(k+1)}{k+1}$ = 0
⇒ 9k – 1 – 8k – 3 + 2k + 2 = 0
⇒ 11k – 8k – 4 + 2 = 0
⇒ 3k – 2 = 0 => k = $\frac{2}{3}$
⇒ k : 1 = – : 1 = 2 : 3
AX : BX = k : 1 = 2 : 3
Therefore, dividing ratio is 2 : 3.

Question 7.
Find the ratio in which the line segmentjoining (-2, -3) and (5,6) divided by i) x-axis ii) y-axis.
Also find the coordinates of the point of division in each case.
Solution:
i) Let P(x, 0) is the point on x-axis which divides the line joining the points A(-2, -3) and B(5, 6) in the ratio m₁ : m₂

By comparing x and y coordinates

Therefore, P(x, 0) = P($\frac{1}{3}$, 0) divides in the ratio 1 : 2 internally,

ii) Let Q(0, y) is the point on y-axis which divides the line joining the points A(-2, -3) and B(5, 6) in the ratio m₁ : m₂
x₁ = -2
x₂ = 5,
y₁ = -3
y₂ = 6
Q(0, y) = $\frac{m_1 \cdot 5+m_2(-2)}{m_1+m_2}$, $\frac{m_1 \cdot 6+m_2(-3)}{m_1+m_2}$)
By comparing x and y coordinates
$\frac{5 m_1-2 m_2}{m_1+m_2}$ = 0, $\frac{6 m_1-3 m_2}{m_1+m_2}$ = y → (2)
5m₁ – 2m₂ = 0 ⇒ 5m₁ – 2m₂ ⇒ $\frac{m_1}{m_2}$ = $\frac{2}{5}$
m₁ : m₂ = 2 : 5
put m₁ : m₂ 2 : 5 in (2)
⇒ $\frac{6(2)-3(5)}{2+5}$ = y
⇒ y = $\frac{12 – 15}{7}$ = $\frac{-3}{7}$
∴ Q(0, y) (0, $\frac{-3}{7}$)
Therefore, Q(0, y) = Q (0, $\frac{-3}{7}$) divides in the ratio 2 : 5 internally.

Question 8.
If the coordinates of the mid-points of the sides of a triangle are (3, 4), (4, 6) and (5, 7), find the vertices.
Solution:
Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of ΔABC.

D(3, 4), E(4, 6) and F(5, 7) are the mid-points of AB, BC and AC respectively.
Mid-point = ($\frac{x_1+x_2}{2}$, $\frac{y_1+y_2}{2}$)
Mid-point of AB is D = ($\frac{x_1+x_2}{2}$, $\frac{y_1+y_2}{2}$) = (3, 4)
By comparing x and y coordinates
$\frac{x_1+x_2}{2}$ = 3
x₁ + x₂ = 6 → (1)

$\frac{y_1+y_2}{2}$ = 4
y₁ + y₂ = 8 → (1)

Put y₂ = 3 in (4)
⇒ 3 + y₃ = 12
⇒ y₃ = 12 – 3 = 9
Put y₂ = 3 in (2)
⇒ y₁ + 3 = 8
⇒ y₁ = 8 – 3 = 5
A(x₁, y₁) = (4, 5)
B(x₂, y₂ = (2, 3)
C(x₃, y₃) = (6, 9) are the vertices of ΔABC.

Question 9.
The line segment joining the points (3, -4) and (1, 2) is trisected at the points X and Y. If the coordinates of X and Y are (p, -2) and ($\frac{5}{3}$, q) respectively. Find the value of p and q.
Solution:
Let A(3, -4) and B(1, 2) are the points of line segment AB.
Given X(p, -2) and Y ($\frac{5}{3}$, q) are the trisecting points which divides in the ratio X is 2 : 1 and Y is 1 : 2.
Section formula = ($\frac{m_1 x_2+m_2 x_1}{m_1+m_2}$, $\frac{m_1 y_2+m_2 y_1}{m_1+m_2}$)
Trisecting point X(p, -2) can divide A(3, -4) and B(1, 2) in the ratio 1 : 2
x₁ = 3, y₁ = -4, m₁ = 1
x₂ = 1, y₂ = 2, m₂ = 2
X = ($\frac{1 \cdot 1+2 \cdot 3}{1+2}$, $\frac{1 \cdot 2+2(-4)}{1+2}$) = (p, -2)
⇒ ($\frac{1+6}{3}$, $\frac{2-8}{3}$) = (p, -2)
By comparing x, y coordinates $\frac{7}{3}$ = p
Trisecting point Y ($\frac{5}{3}$, q) can divide
A(3, -4) and B(1, 2) in the ratio 2 : 1
x₁ = 3, y₁ = -4, m₁ = 2
x₂ = 1, y₂ = 2, m₂ = 1
Y = ($\frac{2 \cdot 1+1 \cdot 3}{2+1}$, $\frac{2 \cdot 2+1(-4)}{2+1}$) = ($\frac{5}{3}$, q)
⇒ ($\frac{2+3}{3}$, $\frac{4-4}{3}$) = ($\frac{5}{3}$, q)
⇒ ($\frac{5}{3}$, 0) = ($\frac{5}{3}$, q)
By comparing x and y-coordinates q = 0.
Therefore p = $\frac{7}{3}$ and q = 0.

Question 10.
Find the points of trisection of the line segment joining the points A(5, -6) and B(-7, 5).
Solution:
Given AB is the line joining the points A(5, -6) and B(-7, 5).

Trisecting point: A point which divides the line joining by the points in the ra¬tio 1:2 (or) 2:1 is called trisecting point.

Let P divides 1 : 2 and Q divides AB are the trisecting points.

Q (x, y) = (-3, $\frac{4}{3}$)
Therefore, trisecting points are P(1, $\frac{7}{3}$) AND (-3, $\frac{4}{3}$)

Question 11.
If the points A(6, 1), B(8, 2), C(9, 4) and D(x, y) are the vertices of a parallelogram taken in order, then find the value of x and y.
Solution:
Given A(6, 1), B(8, 2), C(9, 4) and D(x, y) are the vertices of parallelogram ABCD.
In parallelogram diagonals bisect each other that is mid points of diagonals are same.
So, midpoint of diagonal BD = midpoint of diagonal AC

Therefore, fourth vertex D(x, y) = (7, 3).

Question 12.
In what ratio does the point (-4, 6) divide the line segment joining the points A(-6, 10) and B(3, -8) ?
Solution:
Let P(-4, 6) can divide the line joining by the points A(-6, 10) and B(3, -8) in the ratio m₁ : m₂.

= ($\frac{3 m_1-6 m_2}{m_1+m_2}$, $\frac{-8 m_1+10 m_2}{m_1+m_2}$)
= (-4, 6)
By equating the x, y-coordinates
$\frac{3 m_1-6 m_2}{m_1+m_2}$ = -4
$\frac{-8 m_1+10 m_2}{m_1+m_2}$ = 6
3m₁ – 6m₂ = – 4(m₁ + m₂)
3m₁ – 6m₂ = – 4m₁ – 4m₂
3m₁ + 4m₁ = – 4m₂ + 6m₂
7m₁ = 2m₂
$\frac{m_1}{m_2}$ = $\frac{1}{2}$
m₁ : m₂ = 2 : 7
Therefore, ratio = 2 : 7

Question 13.
Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).
Solution:
Let P(x, y); A(7, 1); B(3, 5)
The distance between points (x₁, y₁) and (x₂, y₂) is

50 – 14x + x² – 2y + y² = 34 – 6x + x² – 10y + y²
50- 14x – 2y – 34 + 6x + 10y = 0
-8x + 8y + 16 = 0
-x + y + 2 = 0
∴ x = y + 2

Question 14.
If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, then
i) find the value of p.
ii) find the area of ABCD.
Solution:
i) A, B, C, D are the vertices of a parallelogram
Mid point of AC = Mid point of BD

ii) ar ΔABC = $\frac{1}{2}$ |x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)
= $\frac{1}{2}$ |6(2 – 4) + 8(4 – 1) + 9(1 – 2)|
= $\frac{1}{2}$ |-12 + 24 – 9|
= $\frac{1}{2}$ |3| = $\frac{3}{2}$ sq. units.
∴ Area of parallelogram ABCD
= 2 × ar ΔABC
= 2 × ar ΔABC = 2 × $\frac{1}{2}$
= 3 sq. units.

Question 15.
Find the value of ‘k’ for which the points (3k – 1, k – 2), (k, k – 7) and (k – 1, -k -2) are collinear.
Solution:
Given points are collinear
Let A (3k – 1, k – 2), B (k, k – 7) and C (k- 1, -k -2) are on the same line $\overline{\mathrm{ABC}}$.
Then slope of $\overline{\mathrm{AB}}$ and slope of $\overline{\mathrm{AC}}$ should be same (∵ they are collinear)

Now (1) = (2)
⇒ $\frac{-5}{1-2 k}$ = 1 ⇒ 1 – 2k = -5
⇒ 1 + 5 = 2k
⇒ 2k = 6
∴ k = 3

Question 16.
Find the co-ordinates of the points of trisection of the line segment joining the points A(2, 1) and B(5, – 8).
Solution:
Let P and Q be the points of trisection of AB, i.e.f AP = PQ = QB.

Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordi¬nates of P are (by applying the section formula)

Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are

There fore, the coordinates of the points of trisection of the line segment are P(3, – 2) and Q(4, – 5).

Question 17.
If A(-5, 7), B(-4, -5), C(-1, -6) and D(4, 5) are the vertices of a quadrilateral, then find the area of the quadrilateral ABCD.
Solution:

By joining B to D, you will get two triangles ABD and BCD.
The area of ΔABD
= $\frac{1}{2}$ |- 5 (- 5 – 5) + (- 4) (5 – 7) + 4 (7 + 5)|
= $\frac{1}{2}$|50 + 8 + 48| = $\frac{106}{2}$ = 53 sq. units.
Also, the area of ΔBCD
= $\frac{1}{2}$|- 4 (- 6 – 5) – 1 (5 + 5) + 4 (- 5 + 6)|
= $\frac{1}{2}$|44 – 10 + 41 = 19 sq. units.
So, the area of quadrilateral
= Area of ΔABD + Area of ΔBCD
= 53 + 19 = 72 sq. units.

Question 18.
Find the trisection points of the line segment joined by the points (-3, 3) and (3, -3).
Solution:
The points which divide the line segment by 1 : 2 and 2 : 1 ratio (internally) are called trisection points.
Formula for the points of trisection of the line segment joined by (x₁, y₁) and (x₂, y₂) are
= [$\frac{\mathrm{mx}_2+n \mathrm{x}_1}{\mathrm{~m}+\mathrm{n}}$, $\frac{\mathrm{my}_2+n \mathrm{y}_1}{\mathrm{~m}+\mathrm{n}}$]
Where m = 2 and n = 1.
Here x₁ = -3, y₁ = 3 and x₂ = 3,
y₂ = -3 then the point in the ratio 2 : 1 is
($\frac{2(3)+1(-3)}{2+1}$, $\frac{2(-3)+1(3)}{2+1}$)
= $\frac{6-3}{3}$, $\frac{-6+3}{3}$ = ($\frac{3}{3}$, $\frac{-3}{3}$) = (1, -1)
So (1, -1) is one trisection point. The point which is at 1 : 2 ratio is another trisection point.
So m = 1, n = 2, x₁ = -3, y₁ = 3, x₂ = 3 and y₂ = -3

P, Q are trisection points.

Question 19.
If the points P(-3, 9), Q(a, b) and R(4, -5) are collinear and a + b = 1, then find the values of a and b.
Solution:
Points P, Q, R are colliner ⇒ area of ΔPQR = 0
P(-3, 9), Q(a, b), R(4, – 5)
Area of triangle
= $\frac{1}{2}$ |x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) 0 = $\frac{1}{2}$ |-3(b + 5) + a(- 5 – 9) + 4(9 – b)|
after simplifications,
we get 2a + b = 3 ______ (1)
given equation a + b = 1 ______ (2)
Solving equations (1) and (2), we obtain
a = 2 and b = – 1

Question 20.
The area of the triangle is 18 sq. units, whose vertices are (3, 4), (-3, -2) and (p, -1); then find the value of ‘p’.
Solution:
Given points are (3, 4) (-3, -2) (p, -1)
x₁ = 3, y₁ = 4, x₂ = -3, y₂ = -2, x₃ = p, y₃ = -1
Area of triangle = $\frac{1}{2}$ |x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)
18 = $\frac{1}{2}$ |(3 (-2 + 1) – 3 (- 1 – 4) + p(4 + 2)|
⇒ |- 3 + 15 + 6p| = 36
⇒ 6p + 12 = 36 (or) 6p + 12 = -36
P = $\frac{36-12}{6}$, P = $\frac{-36}{-12}$
P = $\frac{24}{6}$ = 4, P = $\frac{-48}{6}$ = -8
∴ P = 4 or -8

Question 21.
Find the points of tri-section of the line segment joining the points (-2, 1) and
a 4). 1
Solution:
Given points (-2, 1) (7, 4)
x₁ = – 2
y₁ = 1

x₂ = 7
y₂ = 4

Points of trisection means the points which divide it in the ratio 1 : 2 and 2 : 1
Section formula = ($\frac{m_1 x_2+m_2 x_1}{m_1+m_2}$, $\frac{m_1 y_2+m_2 y_1}{m_1+m_2}$)
Point which divides it in the ratio 1 : 2 is
= ($\frac{1(7)+2(-2)}{1+2}$, $\frac{1(4)+2(1)}{1+2}$)
= ($\frac{7-4}{3}$, $\frac{4+2}{3}$) = ($\frac{3}{3}$, $\frac{6}{3}$)
= (1, 2)

Point which divides it in the ratio 2 : 1 is
= ($\frac{2(7)+1(-2)}{2+1}$, $\frac{2(4)+1(1)}{2+1}$)
= ($\frac{14-2}{3}$, $\frac{8+1}{3}$) = ($\frac{12}{3}$, $\frac{9}{3}$)
= (4, 3)
∴ Points of trisection are (1,2) and (4, 3).

Question 22.
Find the ratio in which X-axis divides the line segment joining the points (2, – 3) and (5, 6). Then find the intersecting point on X-axis.
Solution:
Let X – axis divides the line segment joining points
(2, – 3) and (5, 6) in the ratio m₁ : m₂
x₁ = 2
y₁ = -3

x₂ = 5
y₂ = 6

Co-ordinate of point
p = ($\frac{m_1 x_2+m_2 x_1}{m_1+m_2}$, $\frac{m_1 y_2+m_2 y_1}{m_1+m_2}$)
= ($\frac{m_1(5)+m_2(2)}{m_1+m_2}$, $\frac{m_1(6)+m_2(-3)}{m_1+m_2}$)
But this is a point on X-axis, so its y co-ordinate is zero.
$\frac{m_1(6)+m_2(-3)}{m_1+m_2}$ = 0
6m₁ – 3m₂ = 0
6m₁ = 3m₂ ⇒ $\frac{m_1}{m_2}$ = $\frac{3}{6}$ = $\frac{1}{2}$
∴ Required ratio = 1 : 2
point P on X – axis = ($\frac{1(5)+2(2)}{1+2}$, 0)
= ($\frac{9}{3}$, 0) = (3, 0)

Question 23.
Show that the points A(-1, -2), B(4, 3), C(2, 5) and D(-3, 0) in that order form a rectangle.
Solution:
Given points are A (-1, -2), B (4, 3)
C (2, 5), D (-3, 0)
Mid point of AC = ($\frac{-1+2}{2}$, $\frac{-2+5}{2}$)
= ($\frac{1}{2}$, $\frac{3}{2}$)

Diagonal are equal and bisect each other. So given vertices form a rectangle.

Question 24.
Find the area of a rhombus ABCD, whose vertices taken in order, are A (-1, 1), B(1, -2), C(3, 1) andD(1, 4).
Solution:
Area of a rhombus = $\frac{1}{2}$ × (Product of its diagonals)
Let the vertices be = $\frac{1}{2}$ × AC × BD
A (-1, 1), B(1, -2), C(3, 1), D (1, 4)

∴ Area of Rhombus = $\frac{1}{2}$ × d₁ × d₂
= $\frac{1}{2}$ × 4 × 6
= 12 sq. units.

Question 25.
If A(-2, 2), B(a, 6), C(4, b) and D(2( -2) are the vertices of a parallelogram ABCD, then find the values of a and b. Also find the lengths of its sides.
Solution:
Vertices of parallelogram ABCD are
A(-2, 2), B(a, 6), C(4, b), D(2, -2)

Since it is a parallelogram, mid-points of diagonals coincide.

AB = DC = $\sqrt{20}$ [∵ opposite sides of a parallelogram are equal ]
AD = BC = $\sqrt{32}$

Question 26.
The three vertices of a parallelogram ABCD are A(-1, -2), B(4, -1) and C (6, 3). Find the coordinates of vertex D and find the area of parallelogram ABCD.
Solution:
Mid point of AC
= ($\frac{x_1+x_2}{2}$, $\frac{y_1+y_2}{2}$)
= ($\frac{-1+6}{2}$, $\frac{-2+3}{2}$)
= ($\frac{5}{2}$, $\frac{1}{2}$)

Area of ΔABD = $\frac{1}{2}$|x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)
= $\frac{1}{2}$| (-1) (- 1 – 2) + 4[2 – (- 2)] + 1[- 2 -(- 1)]|
= $\frac{1}{2}$|(-1) (-3) + 4 (4) + 1(-1)|
= $\frac{1}{2}$|3 + 16 – 1|
= $\frac{1}{2}$ × 18
= 9 Square units
Area of parallelogram ABCD
= 2 × Δ ABD area
= 2 × 9
= 18 Square Units

AP 10th Class Maths 7th Lesson Important Questions and Answers Coordinate Geometry

Question 1.
What do you mean by centroid of a triangle ?
Solution:
“The concurrent point of medians of a triangle is called centroid of the triangle”.

Question 2.
Find the co-ordinates of the point, which divides the line segment joining (2, 0) and (0, 2) in the ratio 1:1.
Solution:
x₁ = 2 ; x₂ = 0 ; y₁ = 0 ; y₂ = 2
Point divides the line segment joining (2,0) and (0, 2) in the ratio 1 : 1

Question 3.
Find the distance between (a cos θ, 0) and (0, a sin θ).
Solution:
To find the distance between the points (a cos θ, 0) and (0, a sin θ) we use the formula.
$\sqrt{\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)^{2}}$
Where x₁ = 0, y₁ = a sin θ and
x₂ = a cos θ, y₂ = 0
∴ The distance between above given two points

Question 4.
If A(4,0), B(0, y) and AB = 5, find the possible values of y. d
Solution:
A(4,0), B(0, y) and AB = 5
 = 5
 = 5
16 + y² = 2
y² = 25 – 16 = 9
y = ± $\sqrt{9}$ = ± 3
Possible values of y are 3 or – 3.

Question 5.
Find the radius of the circle with cen¬tre (3, 2) and passes through (4, – 1).
Solution:

Question 6.
Let the 3 vertices of a triangle ABC are A(3, -2), B(-5, 4) and C(2, – 2). What do you observe for the centroid of this triangle ?
Solution:
centroid of this triangle
= $\left(\frac{\mathbf{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}{3}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}\right)$
= $\left(\frac{3+(-5)+2}{3}, \frac{-2+4+(-2)}{3}\right)$ = (0,0)
Observed that the centroid is the origin.

Question 7.
Find the centroid of a triangle, whose vertices are (6, 2), (0, 0) and (4, -5).
Solution:

Given vertices are (6, 2) (0, 0) and (4, -5) then centroid

Question 8.
Find the radius of the circle whose cen-tre is (3, 2) and passes through (- 5, 6).
Solution:
Given : A circle with centre A (3, 2) passing through B (- 5, 6).

Radius = AB
[ ∵ Distance of a point from the centre of the circle]
Distance formula

Question 9.
What is the distance between (0, – sin x) and (- cos x, 0) ?
Solution:
Distance between (0, – sin x) and (- cos x, 0)

Question 10.
Where do the points (0, -3) and (-8, 0) lie on co-ordinate axis ?
Solution:
The point (0, – 3) lie on OY
∵ Its x co-ordiante is zero and
y – coordinate is negative.
and the point (-8, 0) lie on OX’
∵ Its y – cordinate is zero and
x-cordinate is negative.

Question 11.
Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).
Solution:
Let P(x, y); A(7, 1); B(3, 5)
The distance between points (x₁, y₁) and (x₂, y₂) is

by the sum, PA = PB

50 – 14x + x² – 2y + y²
= 34 – 6x + x² – 10y + y²
50 – 14x – 2y – 34 + 6x 4- 10y = 0
-8x + 8y + 16 = 0
-x + y + 2 = 0 ,
∴ x = y + 2

Question 12.
Find the value of k, for which the points (7, 2), (5, 1) and (3, k) are collinear.
Solution:
The area of the triangle formed by those points = 0
Area of the triangle
$\frac { 1 }{ 2 }${x₁(y₂-y₃) + x₂(y₃-y₁)
+ x₃(y₁ – y₂)} = 0
$\frac { 1 }{ 2 }$|7(1 – k) + 5(k – 2) + 3(2 – 1)| = 0
$\frac { 1 }{ 2 }$ |-2k| = 0
∴ k = 0

Question 13.
Find the centroid of the triangle, whose vertices are (-4, 4), (-2, 2) and (6,-6).
Solution:
Centroid of the triangle

= (0,0)

Question 14.
If the distance between two points (x, 1) and (-1,5) is ‘5’, find the value of V.
Solution:
Given points (x, 1) and (-1, 5)
Let A (x, 1) and B (-1, 5)
Distance between

Now, squaring on both sides (-1 -x)² + 16 = 25
(-1 – x)² = 25- 16 = 9
(-1 – x)² = 3²
-1 – x = 3
-1 – 3 = x ⇒ x = -4

Question 15.
Verify whether the following points are collinear or not.
(1,-1), (4, 1), (-2,-3)
Solution:
To show that three points are collinear the area formed by the triangle is zero.
Given points are (1, -1), (4, 1), (-2, -3) Formula for area of triangle
A = $\frac { 1 }{ 2 }${x₁(y₂-y₃) + x₂(y₃-y₁)
= $\frac { 1 }{ 2 }$|1(1 +3) + 4(-3 + 1)
-2 (-1-1)1
= $\frac { 1 }{ 2 }$|4-8 + 4|.
= $\frac { 1 }{ 2 }$ |8 – 8| = $\frac { 1 }{ 2 }$|0| = 0
So the given three points are collinear.

Question 16.
Find the area of a triangle, whose sides are 5 cm, 12 cm and 13 cni, by using Heron’s formula.
Solution:
Let a = 5 cm; b = 12 cm; :c =13 cm
s= $\frac{a+b+c}{2}=\frac{5+12+13}{2}$ = 15
Area of triangle
(Δ) = $\sqrt{s(s-a)(s-b)(s-c)}$
= $\sqrt{15(15-5)(15-12)(15-13)}$
= 30 cm²

Question 17.
Find out whether the points (1,5), (2, 5) and (-2,-1) are collinear using the distance formula.
Solution:
A(1, 5), B(2, 5), C(- 2,-1)
Distance formula,

Question 18.
Read the following graph and answer the questions given below.

i) Write the coordinates of the points A and B.
ii) What is the slope of the line $\overline{\mathbf{A B}}$ ?
Solution:

i) Coordinates of
point ‘A’ = (0, 2)
point B = (-3, 0)

ii) Slope = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
= $\frac{0-2}{-3-0}=\frac{-2}{-3}=\frac{2}{3}$

Question 19.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).
Solution:

Question 20.
Check whether the points (3, 0), (6,4) and (—1, 3) are the vertices of a right – angled isosceles triangle or not. Also find the area of the triangle.
Solution:
Distance between A(3, O), B(6, 4)

∴ AB² = 25, BC² = 50, CA²= 25
BC² = AB² + CA² and AB = CA
∴ ∆ ABC is an Isosceles right angled triangle
∴ Area of ∆ABC = $\frac { 1 }{ 2 }$ x AB x AC
= $\frac { 1 }{ 2 }$ x 5 x 5 = 12.5sq.u

Question 21.
Find the area of the triangle formed by the points (2, 3), (- 1, 3)and (2,-1) using Heron’s formula.
Solution:
To find the area of the triangle formed by (2, 3) (- 1, 3) and (2,-1) using Heron’s formula.
Let the co-ordinates of A = (2, 3) ; B = (- 1, 3); C = (2, – 1) then the sides of AABC are represented by as follows

AB = c, BC = a, CA = b then the formula of the triangle using Heron’s formula
= $\sqrt{s(s-a)(s-b)(s-c)}$
where s = $\frac{a+b+c}{2}$
Now, we find the sides of ∆ABC, using the formula $\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}$
∴ CB = a = distance between the points (2, – 1) and (-1,3)

⇒ CB = a = 5 ………………..(1) .
and AB = c the distance between (-1,3) and (2, 3)

∴ AB = c = 3 …………………..(2)
and AC = b the distance between A(2, 3) and C(2, – 1)

AC = b = 4 …………..(3)

∴ Area of above triangle = 6 sq.units.

Question 22.
If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a paral-lelogram, taken in order, then
i) find the value of p.
ii) find the area of ABCD.
Solution:
i) A, B, C, D are the vertices of a parallelogram
Mid point of AC =- Mid point of BD

A(6, 1) = (x₁, y₁); C(9, 4) – (x₂, y₂)

ii) ar ΔABC = $\frac { 1 }{ 2 }$| x₁ (y₂ – y₃) + x₂(y₃ – y₁)
+ x₃ (y₁ -y₂)|
= $\frac { 1 }{ 2 }$|6(2 – 4) + 8(4 – 1) + 9(1 – 2)|
= $\frac { 1 }{ 2 }$ |-12+24-9|
= $\frac { 1 }{ 2 }$ |3
= $\frac { 3 }{ 2 }$|sq. units.
∴ Area of parallelogram ABCD = 2 x ar AABC 3
= 2 × $\frac { 3 }{ 2 }$ = 3 sq. units.

Question 23.
Find the value of ‘k’ for which the points (3k – 1, k – 2), (k, k – 7) and (k – 1, -k -2) are collinear.
Solution:
Given points are collinear
Let A (3k – 1, k – 2), B (k, k – 7) and
C (k – 1, -k – 2) are on the same line $\overleftrightarrow{L N}$
Then slope of $\overline{\mathrm{AB}}$ and slope of $\overline{\mathrm{AC}}$ should be same (∵ they are collinear) Formula for slope

Now (1) = (2)
⇒ $\frac{-5}{1-2 k}$ = 1
⇒ 1 – 2k = -5
⇒ 1 + 5 = 2k ⇒ 2k = 6
∴ k = 3

Question 24.
Find the co-ordinates of the points of trisection of the line segment joining the points A(2, 1) and B(5, – 8).
Solution:
Let P and Q be the points of trisection of AB, i.e., AP = PQ = QB.

Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordi¬nates of P are (by applying the section formula)

Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are

Therefore, the coordinates of the points of trisection of the line segment are P(3, – 2) and Q(4, – 5).

Question 25.
Find the distance between the points (5, 7) and (7, 5).
Solution:
Formula for the distance between the points (x₁, y₁) and (x₂, y₂) is $\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}$
Here (x₁, y₁)=(5,7) and (x₂, y₂) = (7, 5)
∴ The distance between (5, 7) and (7, 5) is

Question 26.
Find the distance between the points (5, 7) and (7, 5) by plotting them in co-ordinate plane with the help of a right angled triangle.
Solution:
A (5, 7), B (7, 5) then O (5, 5)

In the ΔAOB, ∠O = 90
∴ AB is hypotenuse.
AO =(7 – 5)= 2
OB = 7 – 5 = 2
Then from Pythagorus theorem

Hence the distance between (5, 7) and (7, 5) = 2$\sqrt{2}$ units

Question 27.
Are the points (5, 7) and (7, 5) equal ?
Solution:
No, the given points (5, 7) and (7, 5) are not equal.
Why because, the above two points represent two different points in the co-ordinate plane.
So they are not equal.
For (x₁, y₁) = (x₂, y₁) then x₁ must be equal to x₂ and y₁ must be equal to y₂.
So to become (5, 7) = (7, 5)
5 should be equal to 7, which is impossible.
So (5, 7) will not bp equal to (7, 5).

Question 28.
Find the point on X-axis which is equi distant from the points (5, 7) and (7,5).
Solution:
Let the point (p, q) i$ on X-axis, which is equi distant from the points (5, 7) and (7, 5).
As this point (p, q) is on X-axis, its y-coordinate q = 0. Then the point is
(p, 0).
The distance between (5, 7) and (p, 0) is $\sqrt{(p-5)^{2}+(0-7)^{2}}$ …………… (1)
And the distance between (7, 5) and (p,0) is $\sqrt{(p-7)^{2}+(0-5)^{2}}$……………. (2)
The above (1) and (2) are equal.

= (p – 5)² + 7²
= (p – 7)² + 5²
= p² – 10p + 25 + 49 = p² – 14p + 49 + 25
∴ -10p + 14p = 0
⇒ 4p = 0
∴ P = o
So (p, 0) = (0, 0) is the point on X-axis, which is equidistant from (5, 7) and (7, 5).

Question 29.
To which quadrants do the following points belong ?
(i) (5, 7)
(ii) (5, – 7)
(iii) (-5, 7)
(iv) (-5, -7)
Solution:
(5, 7) belongs to first quadrant q₁
(5, -7) belongs to fourth quadrant q₄.
(-5, 7) belongs to second quadrant q₂.
(-5, -7) belongs to third quadrant q₃.

Question 30.
What will be the lengths of line seg-ments that are parallel to X-axis, as shown in figure ?

Solution:
In the above figure, length of the line segment above the X-axis is 4 – 2 = 2
And length of line segment below X-axis is 4 – 1 = 3

AP 8th Class Telugu Important Questions and Answers 2024-2025

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