AP 9th Class Maths Chapter 7 Triangles Important Questions

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:

1. The sum of any two sides of a triangle is greater than the third side.
2. 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.