## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.2

Question 1.

In the given figure, ABCD is a parallelogram. Complete each statement along with the definition or property used.

(i) AD = ………..

(ii) DC = ………..

(iii) ∠DCB = ………..

(iv) ∠ADC = ………..

(v) ∠DAB = ………..

(vi) OC = ………..

(vii) OB = ………..

(viii) m∠DAB + m∠CDA = ………..

Solution:

In paralleloram ABCD

(i) AD = 6 cm (Opposite sides of parallelogram)

(ii) DC = 9 cm (Opposite sides of parallelogram)

(iii) ∠DCB = 60° (∵ ∠DCB + ∠CBA = 180°)

(iv) ∠ADC = ∠ABC = 120°

(v) ∠DAB = ∠DCB = 60°

(vi) OC = AO = 7 cm

(vii) OB = OD = 5 cm

(viii) m∠DAB + m∠CDA = 180°

Question 2.

Consider the following parallelograms. Find the values of x, y, z in each.

Solution:

(i) ABCD is a parallelogram.

Side BC is produced to E

∠DCE = 120°

But ∠DCE + ∠DCB = 180° (Linear pair)

⇒ 120° + ∠DCB = 180°

⇒ ∠DCB = 180° – 120° = 60°

But ∠A = ∠C

⇒ x = 60°

∠DCE = ∠ABC (Corresponding angles)

∴ y = 120°

But z = y (Opposite angle of a ||gm)

⇒ z = 120°

Hence x = 60°, y = 120°, z° = 120°

(ii) In parallelogram ABCD, diagonals bisect each other at O.

∠DAC = 40°, ∠CAB = 30°, ∠DOC = 100°

∠ACB = ∠DAC = 40° (Alternate angles)

∴ z = 40°

∠ACD = ∠CAB (A ltemate angles)

⇒ ∠ACD = 30°

In ∆OCD,

∠DOC + ∠CDO + ∠OCD = 180° (Angles of a triangle)

⇒ 100° + x + 30° = 180°

⇒ x + 130° = 180°

⇒ x = 180°- 130° = 50°

Ext. ∠COD = y + z

100° = y + 40°

⇒ y = 100° – 40° = 60°

∴ x = 50°, y = 60°, z = 40°

(iii) In parallelogram ABCD, AC is its diagonal.

∠B = 120°, ∠DAC = 35°

∠DAB + ∠ABC = 180° (Co-interior angles)

35° + z + 120°= 180°

⇒ 155°+ z = 180°

⇒ z = 180° – 155° = 25°

But x = z (Alternate angles)

∴ x = 25°

y = ∠B (Opposite angles of a ||gm)

y = 120°

Hence x = 25°, y = 120°, z = 25°

(iv) In parallelogram ABCD

∠B = 70°, ∠DAC = 67°

∠D = ∠B (Opposite angles of a ||gm)

⇒ z = 70°

In ∆DAC

Ext. DCE = ∠D + ∠DAC

y = z + 67°

y = 70° + 67°= 137°

and ∠DCA + ∠DCE = 180° (Linear pair)

∠DCA + 137°= 180°

⇒ ∠DCA = 180° – 137° = 43°

But ∠CAB = ∠DCA (Alternate angles)

∴ x = 43°

∴ x = 43°, y = 137°, z = 70°

Question 3.

Two adjacent sides of a parallelogram are in the ratio 5 : 7. If the perimeter of parallelogram is 72 cm, find the length of its sides.

Solution:

In ||gm ABCD

AD : AB = 5 : 7

Perimeter of ||gm = 72 cm

⇒ 2(DA + AB) = 72 cm

∴ DA + AB = \(\frac{72}{2}\) = 36 cm

Let DA = 5x and AB = 7x

5x + 7x = 36

⇒ 12x = 36

⇒ x = \(\frac{36}{12}\) = 3

∴ AB = 7x = 7 × 3 = 21 cm

AD = 5x = 5 × 3 = 15 cm

Question 4.

The measure of two adjacent angles of a parallelogram are in the ratio 4 : 5. Find the measure of each angle of the parallelogram.

Solution:

In ||gm ABCD

∠A : ∠B = 4 : 5

Let ∠A = 4x, ∠B = 5x

But ∠A + ∠B = 180° (Cointerior angle)

∴ 4x + 5x = 180° ⇒ 9x= 180°

⇒ x = \(\frac{180^{\circ}}{9}\) = 20°

∴ ∠A = 4x = 4 × 20° = 80°

∠B = 5x = 5 × 20° = 100°

But ∠C = ∠A = 80° and ∠D = ∠B = 100°

(Opposite angles of a ||gm are equal)

Question 5.

Can a quadrilateral ABCD be a parallelogram, give reasons in support of your answer.

(i) ∠A + ∠C= 180°?

(ii) AD = BC = 6 cm, AB = 5 cm, DC = 4.5 cm?

(iii) ∠B = 80°, ∠D = 70°?

(iv) ∠B + ∠C= 180°?

Solution:

Quadrilateral ABCD can be a parallelogram of opposite sides

are equal and opposite angles are equal.

∴ ∠A = ∠C and ∠B = ∠D

and AB = DC, AD = BC

(i) ∠A + ∠C = 180°

It may be a parallelogram and may not be.

(ii) ∵ AD = BC = 6 cm, AB = 5 cm, DC = 4.5 cm

∵ AB ≠ DC

(iii) ∠B = 80°, ∠D = 70°

But there are opposite angles and ∠B ≠ ∠D

(iv) ∴ ∠B + ∠C = 180°

It may be or it may not be.

Question 6.

In the following figures HOPE and ROPE are parallelograms. Find the measures of angles x, y and z. State the properties you use to find them.

Solution:

(i) In parallelogram HOPE, HO is produced to D

∠AOP + ∠POD = 180° (Linear pair)

∴ ∠AOP + 70°= 180°

∠AOP = 180°- 70°= 110°

But ∠AOP = ∠HEP (Opposite angles of a ||gm)

∠HEP = 110°

⇒ x = 110°

∠HPO = ∠EHP (Alternate angles)

∴ y = 40°

In ∆HOP, Ext. ∠POD = y + z

⇒ 70° = y + z

⇒ 70° = 40° + z

⇒ z = 70° – 40° = 30°

∴ x= 110°, y = 40°, z = 30°

(ii) In ||gm ROPE, RO is produced to D

∠POD = 80°, ∠EOP = 60°

∠P = ∠POD (Alternate angles)

∴ y = 80°

∠ROE + ∠EOP + ∠POD = 180° (Angles on one side of a line)

x + 60° + 80° = 180° ⇒ x + 140° = 180°

∴ x = 180°- 140° = 40°

z = x (Alternate angles)

∴ z = 40°

Hence, x = 40°, y = 80°, z = 40°

Question 7.

In the given figure TURN and BURN are parallelograms. Find the measures of x and y (lengths are in cm).

Solution:

(i) We know that opposite sides of a parallelogram are equal.

∴ TU = RN

4x + 2 = 28 ⇒ 4x = 28 – 2

⇒ 4x = 26

⇒ x = \(\frac{26}{4}\) = 6.5 cm

and 5y – 1 = 24

⇒ 5y = 24 + 1

⇒ 5y = 25

⇒ y = \(\frac{25}{5}\) = 5

∴ x = 6.5 cm, y = 5 cm

(ii) We know that the diagonal of a parallelogram bisect each other.

∴ BO = OR

⇒ x + y = 20 ………..(i)

and UO = ON

⇒ x + 3 = 18

⇒ x = 18 – 3 = 15 From (i)

15 + y = 20

⇒ y = 20 – 15 = 5

∴ x = 15, y = 5

Question 8.

In the following figure both ABCD and PQRS are parallelograms. Find the value of x.

Solution:

Two parallelograms ABCD and PQRS in which

∠A = 120° and ∠R = 50°

∠A + ∠B = 180° (Co-interior angles)

120° + ∠B = 180°

⇒ ∠B = 180°- 120° = 60°

∠P = ∠R (Opposite angles of a ||gm)

∠P = 50°

Now in ∆OPB,

∠POB + ∠P + ∠B = 180° (Angles of a triangle)

x + 50° + 60° = 180°

x + 110° = 180° ⇒ x = 180°- 110° = 70°

∴ x = 70°

Question 9.

In the given figure, ABCD, is a parallelogram and diagonals intersect at O. Find :

(i) ∠CAD

(ii) ∠ACD

(iii) ∠ADC

Solution:

(i) ∠DBC = ∠BDA = 46° (alternate angles)

In ∆ AOD,

46° + 68° + ∠CAD = 180° (∵ ∠CAD = ∠OAD)

∠CAD = 180°- 114° = 66°

(ii) ∠AOD + ∠COD = 180° (straight angle)

∴ ∠COD= 180°- 68°= 112°

In ∆COD, 112° + 30° + ∠ACD = 180° (∵ ∠ACD = ∠OCD)

∠ACD = 180° – 112° – 30° = 38°

(iii) ∠ADC = 30° + 46° = 76° (∵ ∠ADC = ∠ADO + ∠ODC)

Question 10.

In the given figure, ABCD is a parallelogram. Perpendiculars DN and BP are drawn on diagonal AC. Prove that:

(i) ∆DCN ≅ ∆BAP

(ii) AN = CP

Solution:

In the given figure,

ABCD is a parallelogram AC is it’s one diagonal.

BP and DN are perpendiculars on AC.

To prove :

(i) ∆DCN ≅ ∆BAP

(ii) AN = CP

Proof: In ∆DCN and ∆BAP

DC=AB (Opposite sides of a ||gm)

∠N = ∠P (Each 90°)

∠DCN = ∠PAB (Alternate angle)

∴ ∆DCN ≅ ∆BAP (AAS axiom)

∴ NC = AP (c.p.c.t.)

Subtracting NP from both sides.

NC – NP = AP – NP

∴ AN = CP

Question 11.

In the given figure, ABC is a triangle. Through A, B and C lines are drawn parallel to BC, CA and AB respectively, which forms a ∆PQR. Show that

2(AB + BC + CA) = PQ + QR + RP.

Solution:

In the given figure, ABC is a triangle.

Through A, B and C lines are drawn parallel to

BC, CA and AB respectively which forms ∆PQR.

To prove:

2(AB + BC + CA) = PQ + QR + RP

∵ BC || PR, AC || RQ

∴ ARBC is a ||gm

∴ AR = CB ….(i)

Similarly ABCP is a ||gm

∴ AP = BC …(ii)

From (i) and (ii),

AR = AP or PR = 2BC …(iii)

Similarly we can prove that

RQ = 2A and PQ = 2AB

Now perimeter of ∆PQR = PQ + QR + RP

= 2AB + 2AC + 2BC

= 2(AB + BC + CA)

Hence PQ + QR + RP = 2(AB + BC + CA)