These AP 9th Class Maths Important Questions 8th Lesson Quadrilaterals will help students prepare well for the exams.

## AP Board Class 9 Maths 8th Lesson Quadrilaterals Important Questions

### 9th Class Maths Quadrilaterals 2 Marks Important Questions

Question 1.

The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Solution:

Let ABCD be a quadrilateral in which ∠A : ∠B : ∠C: ∠D = 3 : 5 : 9 : 13

Sum of the ratios = 3 + 5 + 9 +13 = 30

Also, ∠A + ∠B + ∠C + ∠D = 360° (Sum of the angles of a quadrilateral is 360°)

∴ ∠A = \(\frac{3}{30}\) × 360° = 36°

∠B = \(\frac{5}{30}\) × 360° = 60°

∠C = \(\frac{9}{30}\) × 360° = 108°

and ∠D = \(\frac{13}{30}\) × 360° = 156°

Question 2.

If three angles of a quadrilateral are v 70°, 80° and 100°, find its fourth angle.

Solution:

Let the fourth angle be x°. Then,

70° + 80° + 100° + x° = 360° (∵ The sum of the angles of a quadrilateral is 360°)

⇒ 250° + x° = 360°

⇒ x° = 360° – 250° = 110°

Hence, the fourth angle of the quadrilateral is 110°.

Question 3.

In the given figure, ABCD is a parallelogram. If ∠A = 65°, then find the value of (∠B + ∠D).

Solution:

∵ Sum of two adjacent angles of a parallelogram is 180°.

∴ ∠A + ∠B = 180°

⇒ 65° + ∠B = 180°[∵ ∠A = 65° (Given)]

⇒ ∠B = 180° – 65° = 115° …………. (1)

∵ Opposite angles of a parallelogram are equal.

∴ ∠D = ∠B

⇒ ∠D = ∠115° ………………. (2) [From (1)]

Now, ∠B + ∠D = 115° + 115° = 230° [From (1) and (2)]

Question 4.

In the given figure, ABCD is a parallelogram. What is the sum of the angles x, y and z?

Solution:

∵ ABCD is a parallelogram.

∴ ∠B = ∠D (Opposite angles of a parallelogram are equal)

⇒ ∠B = z [∵ ∠D = z (given)]

Now, sum of the angles x, y and z

= x + y + z = 180° (∵ The sum of the angles of a triangle is 180°)

Question 5.

In a parallelogram ABCD, diagonals AC and BD intersect at O and AC = 6.8 cm and BD = 5.6 cm. Find the measures of OC and OD.

Solution:

∵ Diagonals of a parallelogram bisect each other.

∴ OC = \(\frac{1}{2}\)AC = \(\frac{1}{2}\) × 6.8 = 3.4 cm

and OD = \(\frac{1}{2}\)BD = \(\frac{1}{2}\) × 5.6 = 2.8 cm

Question 6.

In the given figure, ABCD is a rhombus, AO = 4 cm and DO = 3 cm. Then, find the perimeter of the rhombus.

Solution:

∵ Diagonals of a rhombus intersect at right angles.

∴ ∠ AOD = 90°

∴ By Pythagoras Theorem, in Δ AOD,

AD^{2} = AO^{2} + DO^{2} = 4^{2} + 3^{2}

= 16 + 9 = 25

⇒ AD = 5 cm

∴ Perimeter of the rhombus ABCD = AB + BC + CD + DA

= AD + AD + AD + AD (∵ Sides of a rhombus are equal)

= 4 AD

= 4 × 5 = 20 cm

Question 7.

In Δ ABC, AB = 13 cm, BC = 16 cm and AC = 8 cm. Find the perimeter of the triangle formed by joining the midpoints of the sides of the triangle.

Solution:

Let D, E and F be the mid-points of the sides BC, CA and AB of Δ ABC. Then by Mid-Point Theorem,

DE = \(\frac{1}{2}\)AB ; EF = \(\frac{1}{2}\)BC ; FD = \(\frac{1}{2}\)CA

Adding we get

DE + EF + FD = \(\frac{1}{2}\) (AB + BC + CA)

= \(\frac{1}{2}\) (13 + 16 + 8)

= \(\frac{37}{2}\) = 18.5 cm

⇒ Perimeter of Δ DEF = 18.5 cm

Question 8.

In ΔABC, AD is the median and DE || AB, such that E is a point on AC. Prove that BE is another median.

Solution:

∵ AD is a median of Δ ABC.

∴ D is the mid-point of BC.

Also, DE || AB (Given)

∴ By converse of Mid-Point Theorem, E is the mid-point of CA.

⇒ BE is a median of Δ ABC.

Thus, BE is another median of Δ ABC.

Question 9.

The angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6. Find the measure of the smallest angle of the quadrilateral.

Solution:

Let ABCD be a quadrilateral in which ∠A : ∠B : ∠C : ∠D = 3 : 4 : 5 : 6

Sum of the ratios = 3 + 4 + 5 + 6 = 18

Also, ∠A + ∠B + ∠C + ∠D = 360° (∵ The sum of the angles of a quadrilateral is 360°)

∴ Measure of the smelliest angle (∠A) of the quadrilateral

= \(\frac{3}{18}\) × 360° = 3 × 20° = 60°

Question 10.

The sides BA and DC of a quadrilateral ABCD are produced as shown in the figure. Prove that a + b = x + y.

Solution:

In Δ ABD,

Ext. ∠b = ∠ABD + ∠ADB …………….. (1) (∵ An exterior angle of a triangle is equal to the sum of its two interior opposite angles)

In Δ BCD,

Ext. ∠a = ∠CBD + ∠CDB ……………. (2)

(∵ An exterior angle of a triangle is equal to the sum of its two interior opposite angles)

Adding (1) and (2), we get

b + a = (∠ABD + ∠CBD) + (∠ADB + ∠CDB)

⇒ a + b = ∠ ABC + ∠ADC

∴ a + b = x + y

Question 11.

The perimeter of a parallelogram is 32 cm. If the longer side measures 9.5 cm, then what is the measure of the shorter side ?

Solution:

Let ABCD be a parallelogram in which AB is the longer side and BC is the – shorter side.

∴ Perimeter of parallelogram ABCD

= AB + BC + CD + DA

= AB + BC + AB + BC (∵ Opposite sides of a parallelogram are equal)

= 2(AB + BC)

According to the question,

2(AB + BC) = 32

⇒ AB + BC = 16

⇒ 9.5 + BC=16 [∵ AB = 9.5 cm (Given)]

∴ BC = 6.5 cm

Question 12.

ABCD is a quadrilateral in which P, Q, R and S are midpoints of the sides AB, BC, CD and DA. Length of the diagonal AC is 10.2 cm then find the length of SR and PQ.

Solution:

In ΔADC, S, R are the midpoints of sides AD and DC respectively.

∴ By the midpoint theorem SR // AC and

SR = \(\frac{1}{2}\) AC.

∴ SR = \(\frac{1}{2}\) × 10.2 = 5.1 cm

Similarly PQ = 5.1 cm

### 9th Class Maths Quadrilaterals 4 Marks Important Questions

Question 1.

Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Solution:

Given : ABCD is a quadrilateral whose diagonals AC arid BD bisect each other at right angles at O.

To Prove : Quadrilateral ABCD is a rhombus.

Proof : In Δ AOB and ∠AOD,

AO = AO (Common)

OB = OD (given)

∠AOB = ∠AOD (Each = 90°)

∴ ΔAOB ≅ ΔAOD (SAS Congruent Rule)

∴ AB = AD ……………. (1) (CPCT)

Similarly, we can prove that

AB = BC …………. (2)

BC = CD …………….. (3)

CD = AD ……………… (4)

In view of (1), (2), (3) and(4), we obtain

AB = BC = CD = DA

∴ Quadrilateral ABCD is a rhombus.

Question 2.

ABCD is a quadrilateral in which the bisectors of ∠A and ∠C meet DC and BA produced at X and Y respectively.

Prove that ∠X + ∠Y = (\(\frac{1}{2}\)∠A + ∠C).

Solution:

In Δ YCB,

∠Y + ∠3 + ∠B = 180° (The sum of the angles of a triangle is 180°)

⇒ ∠Y + \(\frac{1}{2}\)∠C + ∠B = 180° …………… (1)

(∵ CY is the bisector of ∠C)

In Δ XDA.

∠X + ∠D + ∠1 = 180° (∵ The sum of the angles of a triangle is 180°)

⇒ ∠X + ∠D + \(\frac{1}{2}\)∠A = 180° ……………….. (2)

(∵ AX is the bisector of ∠A)

Adding (1) and (2), we get

∠Y + \(\frac{1}{2}\)∠C + ∠B +∠X + ∠D + \(\frac{1}{2}\)∠A

∠X + ∠Y + \(\frac{1}{2}\)∠A + \(\frac{1}{2}\)∠C + ∠B + ∠D = 360° …………… (3)

But, ∠A + ∠B + ∠C + ∠D = 360° ………….. (4)

(∵ The sum of the angles of a quadrilateral 360°)

From (3) and (4), we get

∠X + ∠Y + \(\frac{1}{2}\)∠A + \(\frac{1}{2}\)∠C + ∠B + ∠D

= ∠A + ∠B + ∠C + ∠D

⇒ ∠X + ∠Y = \(\frac{1}{2}\)(∠A + ∠C)

Question 3.

AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B. Show that the line PQ is the perpendicular bisector of AB.

Solution:

Given : AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B.

To Prove : Line PQ is the perpendicular bisector of AB.

Proof : In Δ APQ and Δ BPQ,

AP = BP (Given)

AQ = BQ (Given)

PQ = PQ (Common)

∴ Δ APQ ≅ Δ BPQ (SSS Rule)

∴ ∠ APQ = ∠BPQ (CPCT) …………… (1)

Now, in Δ APM and Δ BPM,

∠APM = ∠BPM [From (1)]

AP = BP (Given)

PM = PM (Common)

∴ Δ APM ≅ Δ BPM (SAS Rule)

∴ AM = BM (CPCT) ……………. (2)

and ∠AMP = ∠BMP (CPCT) ……………. (3)

But ∠AMP + ∠BMP = 180° …………… (4)

(Linear Pair Axiom)

From (3) and (4),

∠AMP = ∠BMP = 90° ………….. (5)

In view of (2) and (5),

PQ is the perpendicular bisector of AB.

Question 4.

In a quadrilateral ABCD, the line segments bisecting ∠C and ∠D meet at E. Prove that ∠A + ∠B = 2 ∠CED.

Solution:

Given: In a quadrilateral ABCD, the line segments bisecting ∠C and ∠D meet at E.

To Prove : ∠A + ∠B = 2 ∠CED

Proof : In quadrilateral ABCD,

∠A + ∠B + ∠C + ∠D = 360° ……………(1)

(∵ The sum of the angles of a quadrilateral is 360°)

In Δ CED,

∠CED + ∠ EDC + ∠ ECD = 180° (∵ The sum of the angles of a triangle is 180°)

∠CED + ∠ EDC + ∠ ECD = 180°

⇒ ∠CED + \(\frac{1}{2}\)∠D + \(\frac{1}{2}\)∠C = 180° (∵ DE and CE are the bisectors of ∠D and ∠C respectively)

⇒ 2 ∠CED + ∠D + ∠C = 360° ……………. (2)

From (1) and (2),

∠A + ∠B + ∠C + ∠D

= 2∠CED + ∠D + ∠C

⇒ ∠A + ∠B = 2 ∠CED.

Question 5.

ABCD is a parallelogram and line segments AX, CY bisect the angles A and C respectively. Show that AX || CY.

Solution:

Given : ABCD is a! parallelogram and line segments AX, CY bisect angles A and C respectively.

To Prove : AX || CY

Proof : ∵ ABCD is a parallelogram.

∴ ∠A = ∠C (Opposite angles of a parallelogram are equal)

⇒ \(\frac{1}{2}\)∠A = \(\frac{1}{2}\)∠C (Halves of equals are equal)

⇒ ∠1 = ∠2 …………. (1)

(∵ AX is the bisector of ∠A and CY is the bisector of ∠C)

Now, AB || DC

(Opposite sides of a parallelogram are parallel and a transversal CY intersects them)

∠2 = ∠3 ……………. (2)

(Alternate Interior Angles)

From (1) and (2), we get

∠1 =∠3

But these angles form a pair of equal corresponding angles.

∴ AX || CY

Question 6.

In a triangle ABC, points M and N on sides AB and AC respectively are taken so that

AM = \(\frac{1}{4}\)AB and AN = \(\frac{1}{4}\)AC.

Prove that MN = \(\frac{1}{4}\) BC.

Solution:

Given : In triangle ABC, points M and N on sides AB and AC respectively are taken so that

AM = \(\frac{1}{4}\)AB and AN = \(\frac{1}{4}\)AC.

To Prove : MN = \(\frac{1}{4}\)BC

Construction : Join EF where E and F are the mid-points of AB and AC respectively.

Proof : In ΔABC,

∵ E is the mid-point of AB and F is the mid-point of AC.

(By mid-point theorem)

Now, AE = \(\frac{1}{2}\)AB

and AM = \(\frac{1}{4}\)AB

∴ AM = \(\frac{1}{2}\)AE

Similarly, AN = \(\frac{1}{2}\)AF

⇒ M and N are the mid-points of AE and AF respectively.

∴ MN || EF

and MN= \(\frac{1}{2}\)EF (By Mid-Point Theorem)

∴ MN = \(\frac{1}{2}\)(\(\frac{1}{2}\)BC) [From (1)]

= \(\frac{1}{4}\)BC

Question 7.

ABCD is a parallelogram in which P is the mid-point of DC and Q is a point onAC such that CQ = \(\frac{1}{4}\)AC. If PQ produced meets BC at R, prove that R is the mid-point of BC.

Solution:

Given : ABCD is a parallelogram in which P Is the mid-point of DC and Q is a point on AC such that CQ \(\frac{1}{4}\)AC. PQ produced meets BC at R.

To Prove : R is the mid-point of BC.

Construction : Join BD to intersect AC at O.

Proof : ∵ ABCD is a parallelogram and the diagonals of a parallelogram bisect each other.

∴ AO = OC = \(\frac{1}{2}\)AC ……….. (1)

Now CQ = \(\frac{1}{4}\)AC

= \(\frac{1}{4}\)(2 OC) [From(1)]

= \(\frac{1}{2}\)OC

⇒ Q is the mid-point of CO.

In Δ CDO,

∵ P is the mid-point of DC and Q is the mid-point of CO.

∴ PQ || DO (By Mid-Point Theorem)

⇒ PR || DB

⇒ QR || OB

Again, in Δ COB.

∵ Q is the mid-point of CO and QR || OB.

∴ R is the mid-point of BC. (By Converse of Mid-Point Theorem)

Question 8.

In quadrilateral ACBD, AC = AD and AB bisects ∠A. Show that ΔABC ≅ ΔABD.

Solution:

AC = AD (Given)

∠CAD = ∠DAB (AB is the angle bisector)

AB is common

∴ By SAS criterion ΔABC ≅ ΔABD

Question 9.

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively.

Based on the above information, answer the following questions :

Then,

i) Quadrilateral ABED is a parallelogram because,

A) both pairs of opposite sides are equal.

B) both pairs of opposite sides are parallel.

C) same pair of opposite sides are equal as well as parallel.

D) diagonals bisect each other.

Answer:

C) same pair of opposite sides are equal as well as parallel.

ii) Quadrilateral BEFC is a parallelogram because,

A) both pairs of opposite sides are equal.

B) same pair of opposite sides are equal as well as parallel.

C) both pairs of opposite sides are parallel.

D) diagonals bisect each other.

Answer:

B) same pair of opposite sides are equal as well as parallel.

iii) Which of the following is true ?

A) AD is parallel to CF but AD is not equal to CF.

B) AD is not parallel to CF but AD and CF are equal.

C) AD and CF are neither parallel nor equal.

D) AD || CF and AD = CF.

Answer:

D) AD || CF and AD = CF.

iv) Which of the following is true ?

A) quadrilateral ACFD is not a parallelogram.

B) quadrilateral ACFD is a parallelogram.

C) quadrilateral ACFD may or may not be a parallelogram.

D) quadrilateral ACFD is always a rectangle.

Answer:

C) quadrilateral ACFD may or may not be a parallelogram.

v) Δ ABC and Δ DEF are congruent by which criteria

A) SSS

B) SAS

C) ASA

D) RHS

Answer:

A) SSS

### 9th Class Maths Quadrilaterals 8 Marks Important Questions

Question 1.

Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Solution:

Given : The diagonals AC and BD of a quadrilateral ABCD are equal and bisect each other night angles.

To Prove : Quadrilateral ABCD is a square.

Proof : In Δ OAD and Δ OCB,

OA = OC (Given)

OD = OB (Given)

∠AOD = ∠COB (Vertically Opposite Angles)

∴ Δ OAD ≅ Δ OCB (SAS Congruence Rule)

∴ AD = CB (C.P.C.T.)

∠ODA = ∠OBC (C.P.C.T)

∴ ∠BDA = ∠DBC

∴ AD || BC

Now, ∵ AD = CB and AD || CB

∴ Quadrilateral ABCD is a || gm. (A quadrilateral is a parallelogram if a pair of opposite sides are parallel and equal.)

In Δ AOB and Δ AOD,

AO = AO (Common)

OB = OD (Given)

∠AOB = ∠AOD [Each = 90° (Given)]

∴ ΔAOB ≅ ΔAOD (SSS Congruence Rule)

∴ AB = AD (C.P.C.T)

Now, ∵ ABCD is a parallelogram and

∴ AB = AD

∵ ABCD is a rhombus.

Again, in Δ ABC and Δ BAD,

AC = BD (Given)

BC = AD (∵ ABCD is a rhombus)

AB = BA (Common)

∴ Δ ABC ≅ Δ BAD (SSS Congruence Rule)

∴ ∠ABC = ∠BAD (C.P.C.T)

∵ AD || BC (Opp. sides of || gm ABCD and transversal AB intersects them)

∴ ∠ABC + ∠BAD = 180° (Sum of consecutive interior angles on the same side of a transversal is 180°)

∴ ∠ABC = ∠BAD = 90°

Similarly, ∠ BCD = ∠ADC = 90°

∴ ABCD is a square.

Question 2.

In ΔABC, D, E and F are the midpoints of sides AB, BC and CA respectively. Show that ΔABC is divided into four congruent triangles, when the three midpoints are joined to each other. (ΔDEF is called medial triangles).

Solution:

Proof : D, E are midpoints of \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{BC}}\) of triangle ABC respectively.

So by Mid-point theorem,

DE || AC

Similarly DF || BC and EF || AB.

Therefore ADEF, BEFD, CFDE are all parallelograms.

In the parallelogram ADEF, DF is the diagonal

So ΔADF ≅ ΔDEF (Diagonal divides the parallelogram into two congruent triangles)

Similarly, ΔBDE ≅ ΔDEF

and ΔCEF ≅ ΔDEF

So, all the four triangles are congruent.

We have shown that a triangle ABC is divided into four congruent triangles by joining the midpoints of the sides.

Question 3.

In the rhombus ABCD of the given figure ∠PAQ is an obtuse angle. Two equilateral triangles ABP and ADQ are drawn outside the rhombus. Prove that ACPQ is also an equilateral triangle.

Solution:

Given : ABCD is a rhombus. ∠PAQ is an obtuse angle. Two equilateral triangles ABP and ADQ are drawn outside the rhombus.

To Prove : ACPQ is also an equilateral triangle.

Proof : In rhombus ABCD,

Let ∠ABC = x

Then, ∠ADC = x (Opposite angles of a rhombus are equal)

∵ Sum of any two adjacent angles of a rhombus is 180°.

∴ ∠ADC + ∠BCD = 180°

⇒ x + ∠BCD = 180° ⇒ ∠BCD = 180° – x

∠PBC = ∠PBA + ∠ ABC = 60° + x (∵ ΔPBA is an equilateral triangle)

∴ ∠PBA = 60°

∠QDC = ∠QDA + ∠ADC – 60° + x (∵ ΔQAD is an equilateral triangle)

∴ ∠QDA = 60°

∴ ∠PBC = ∠QDC ………….. (1)

In Δ PBC and Δ CDQ,

∠PBC = ∠QBC [from (1)]

∴ Δ PBC ≅ Δ QDC (SAS Rule)

∴ CP = CQ (CPCT) ……….. (2)

In Δ PBC,

PB = BC ⇒ ∠PCB = ∠BPC …………… (3)

(Angles opposite to equal sides of a triangle are equal)

In Δ PBC,

∠PCB + ∠BPC + ∠PBC = 180° (∵ The sum of the angles of a triangle is 180°)

⇒ ∠PCB + ∠PCB + (60° + x) = 180° [From (3)]

⇒ 2∠PCB = 120° – x

⇒ ∠PCB = 60° – \(\frac{x}{2}\) …………. (4)

Similarly, we can prove that

∠QCD= 60° – \(\frac{x}{2}\) …………….. (5)

∴ ∠PCQ = ∠BCD – ∠PCB – ∠QCD

= (180°- x) – (6o° – \(\frac{x}{2}\)) (60° – \(\frac{x}{2}\))

= 60° …………. (6)

In Δ PCQ,

CP = CQ [from (2)]

∴ ∠CQP = ∠CPQ ……………… (7)

(Angles opposite to equal sides of a triangle are equal)

In Δ PCQ, ∠PCQ + ∠CQP + ∠CPQ = 180° (∵ The sum of the angles of a triangle is 180°)

60° + ∠CQP + ∠CPQ = 180° (from (6)

∠CQP + ∠CPQ = 120° …………….. (8)

From (7) and (8)

∠CQP = ∠CPQ = 60° …………….. (9)

From (6) and (9)

∠PCQ = ∠CPQ = ∠CQP = 60°

∴ Δ CPQ is an equilateral triangle.

**AP 9th Class Maths Chapter 8 Quadrilaterals Important Questions**

Question 1.

The four angles of quadrilateral are in the ratio 2 : 4 : 5 : 7. Find it’s angles.

Solution:

Ratio of angles of quadrilateral = 2 : 4 : 5 : 7

Let the four angles be 2x, 4x, 5x and 7x.

Sum of them = 2x + 4x + 5x + 7x = 360°

⇒ 18x = 360° ⇒ x = 20°

∴ Angles in a quadrilateral = 40°, 80°, 100°, 140°

Question 2.

Draw a diagram representing the following data.

“In a triangle ABC, AD is the median drawn on to the side BC is produced to E such that AD = ED to form a ABEC parallelogram”.

Solution:

Question 3.

The ratio of consecutive angles of a parallelogram is 2 : 3. Find the angles.

Solution:

Ratio of angles = 2 : 3

Let the angles are 2x and 3x.

Sum of adjacent angles in a parallelogram is 180°.

∴ 2x + 3x = 180°

5x = 180° ⇒ x° = 36°

∴ Angles are = 2x = 2 × 36° = 72°

= 3x = 3 × 36° = 108°

Question 4.

D, E and F are midpoints of the sides of triangle ABC respectively, if AB = 8 cm., BC 7.2 cm, and AC = 6 cm. then find the perimeter of ΔDEF.

Solution:

DF = BC = × 7.2 = 3.6 cm

DE = \(\frac{1}{2}\) AC = \(\frac{1}{2}\) × 6 = 3cm

EF = \(\frac{1}{2}\) AB = \(\frac{1}{2}\) × 8 = 4 cm

∴ Permiter of ΔDEF = 3.6 + 3 + 4 = 10.6 cm

Question 5.

ABCD is a parallelogram and ∠D = 150°. The side AB is produced to E. Then find the measure of ∠CBE.

Solution:

From figure ∠B = ∠D {ABCD is a parallelogram}

∠B = 150°

∠CBA + ∠CBE = 180° {Linear pair} 150° + ∠CBE = 180°

∠CBE = 180° – 150° = 30°

Question 6.

ABCD is a rhombus and ∠A = 60°, find the measure of ∠C and ∠B.

Solution:

ABCD is a rhombus in which ∠A = 60° then ZC = 60° [ ∵ ∠A, ∠C are the opposite angles, which are equal in a rhombus]

∠A + ∠B = 180° [Sum of adjacent angles of a rhombus are supplementary]

60° + ∠B = 180°

∴ ∠B = 180° – 60° = 120°

Question 7.

Write whether the following statements are true (or) false. Justify your answer.

i) A rhombus is a parallelogram,

ii) For any real number x, x^{2} ≥ 0.

iii) The sum of the interior angles of a quadrilateral is 350°.

iv) Square numbers can be written as the sum of two odd numbers.

Solution:

i) A Rhombus is a parallelogram. This statement is true. Because squares, – rectangles and rhombuses are all parallelograms.

ii) For any real number x, x^{2} ≥ 0. This statement is true. Because square of a real number cannot be negative and it can be equal or greater than zero.

iii) The sum of the interior angles of a quadrilateral is 350°. This statement is false. Because of the sum of the interior angles of a quadrilateral is 360°.

iv) Square numbers can be written as the sum of two odd numbers. This statement is true. Square number will always be even. The sum of two odd numbers is always an even number.

Question 8.

In the given figure CD || BE || AF. Prove that ar(ΔAEC) = ar (ΔDBF).

Solution:

Area of triangles between same base and same parallel lines are equal.

∴ ΔCBE area = ΔDBE area

ΔBAE area = ΔBEF area

ΔCBE + ΔBAE = ΔDBE + ΔBEF

∴ Area of ΔAEC = area of ΔDBF

Question 9.

The angles of a quadrilateral are in the ratio 1 : 2 : 3 : 3. Find the measure of each angle of the quadrilateral.

Solution:

Ratio of angles = 1 : 2 : 3 : 3

Angles of quadrilateral x, 2x, 3x and 3x.

x + 2x + 3x + 3x = 360°

9x = 360°

x = \(\frac{360}{9}\) = 40°

First angle = x = 40°

Second angle = 2x = 80°

Third angle = 3x = 120°

Fourth angle = 3x = 120°

Question 10.

The opposite angles of a parallelogram are (3x – 2)° and (x + 48)°. Find the measure of each angle of the parallelogram.

Solution:

Opposite angles of a parallelogram are given.

They are (3x – 2)°, (x + 48)°

⇒ 3x – 2 = x + 48° [ ∵ opposite angles of a parallelogram are equal]

⇒ 3x – x = 48 + 2 => 2x = 50

∴ x = 25°

A = 3x – 2° = 3 x 25° -2° = 75° -2° = 73°

∠A = ∠C = 73°

∠B = 180° – ∠A [ v adjacent angles of a parallelogram are supplementary]

= 180° – 73° = 107°

∴ ∠B = ∠D = 107°

Question 11.

i) ABCD is a quadrilateral such that AB = BD, AC = CD. Show that ΔABC ≅ ΔDBC.

ii) ABCD is a quadrilateral such that AD = BC and ∠DAB = ∠CBA.

Show that ΔABD ≅ ΔBAC.

ABCD is a quadrilateral in which

AB = BD

AC = CD

from ΔABC & ΔDBC

AB = DB (S) (given)

AC = DC (S) given

BC = BC (S) (common side)

By side – side – side congruency rule ΔABC ≅ ΔDBC

Proof : ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA from A ABD & Δ BAC

AD = BC (S) (given)

∠DAB = ∠CBA (A) given

AB = AB (S) (common side)

∴ By S.A.S congruency rule ΔABD ≅ ΔBAC