**Chapter 5 Quadrilaterals Practice Set 5.1**

Chapter 5 Quadrilaterals Practice Set 5.1

**Question 1.Diagonals of a parallelogram WXYZ intersect each other at point O. If âˆ XYZâˆ = 135Â°, then measure of âˆ XWZ and âˆ YZW? If l(OY) = 5 cm, then l(WY) = ?Solution:**

i. âˆ XYZ = 135Â°

â–¡WXYZ is a parallelogram.

âˆ XWZ = âˆ XYZ

âˆ´ âˆ XWZ = 135Â° â€¦..(i)

ii. âˆ YZW + âˆ XYZ = 180Â° [Adjacent angles of a parallelogram are supplementary]

âˆ´ âˆ YZW + 135Â°= 180Â° [From (i)]

âˆ´ âˆ YZW = 180Â°- 135Â°

âˆ´ âˆ YZW = 45Â°

iii. l(OY) = 5 cm [Given]

âˆ´ I(WY) = 2 x l(OY)

= 2 x 5

âˆ´ I(WY) = 10 cm

âˆ´âˆ XWZ = 135Â°, âˆ YZW = 45Â°, l(WY) = 10 cm

**Question 2.In a parallelogram ABCD, if âˆ A = (3x + 12)Â°, âˆ B = (2x â€“ 32)Â°, then liptl the value of x and the measures of âˆ C and âˆ D.Solution:**

â–¡ABCD is a parallelogram. [Given]

âˆ´ âˆ A + âˆ B = 180Â° [Adjacent angles of a parallelogram are supplementary],

âˆ´ (3x + 12)Â° + (2x-32)Â° = 180Â°

âˆ´ 3x + 12 + 2x â€“ 32 = 180

âˆ´ 5x â€“ 20 = 180

âˆ´ 5x= 180 + 20

âˆ´ 5x = 200

âˆ´ x = 40

ii. âˆ A = (3x + 12)Â°

= [3(40) + 12]Â°

=(120 +12)Â°= 132Â°

âˆ B = (2x â€“ 32)Â°

= [2(40) â€“ 32]Â°

= (80 â€“ 32)Â° = 48Â°

âˆ´ âˆ C = âˆ A = 132Â°

âˆ D = âˆ B = 48Â° [Opposite angles of a parallelogram]

âˆ´ The value of x is 40, and the measures of âˆ C and âˆ D are 132Â° and 48Â° respectively.

**Question 3.Perimeter of a parallelogram is 150 cm. One of its sides is greater than the other side by 25 cm. Find the lengths of all sides.Solution:**

i. Let â–¡ABCD be the parallelogram and the length of AD be x cm.

One side is greater than the other by 25 cm.

âˆ´ AB = x + 25 cm

AD = BC = x cm

AB = DC = (x + 25) cm [Opposite angles of a parallelogram]

ii. Perimeter of â–¡ABCD = 150 cm [Given]

âˆ´ AB + BC + DC + AD = 150

âˆ´ (x + 25) +x + (x + 25) + x â€“ 150

âˆ´ 4x + 50 = 150

âˆ´ 4x = 150 â€“ 50

âˆ´ 4x = 100

âˆ´ x = 25

iii. AD = BC = x = 25 cm

AB = DC = x + 25 = 25 + 25 = 50 cm

âˆ´ The lengths of the sides of the parallelogram are 25 cm, 50 cm, 25 cm and 50 cm.

**Question 4.If the ratio of measures of two adjacent angles of a parallelogram is 1 : 2, find the measures of all angles of the parallelogram.Solution:**

i. Let â–¡ABCD be the parallelogram.

The ratio of measures of two adjacent angles of a parallelogram is 1 : 2.

Let the common multiple be x.

âˆ´ âˆ A = xÂ° and âˆ B = 2xÂ°

âˆ A + âˆ B = 180Â° [Adjacent angles of a parallelogram are supplementary]

âˆ´ x + 2x = 180

âˆ´ 3x = 180

âˆ´ x = 60

ii. âˆ A = xÂ° = 60Â°

âˆ B = 2xÂ° = 2 x 60Â° = 120Â°

âˆ A = âˆ C = 60Â°

âˆ B = âˆ D= 120Â° [Opposite angles of a parallelogram]

âˆ´ The measures of the angles of the parallelogram are 60Â°, 120Â°, 60Â° and 120Â°.

**Question 5.Diagonals of a parallelogram intersect each other at point O. If AO = 5, BO show that â–¡ABCD is a rhombus.Given: AO = 5, BO = 12 and AB = 13.To prove: â–¡ABCD is a rhombus.Solition:**

Proof:

AO = 5, BO = 12, AB = 13 [Given]

AO

^{2}+ BO

^{2}= 5

^{2}+ 12

^{2}

= 25 + 144

âˆ´ AO

^{2}+ BO

^{2}= 169 â€¦..(i)

AB

^{2}= 13

^{2}= 169 â€¦.(ii)

âˆ´ AB

^{2}= AO

^{2}+ BO

^{2}[From (i) and (ii)]

âˆ´ âˆ†AOB is a right-angled triangle. [Converse of Pythagoras theorem]

âˆ´ âˆ AOB = 90Â°

âˆ´ seg AC âŠ¥ seg BD â€¦..(iii) [A-O-C]

âˆ´ In parallelogram ABCD,

âˆ´ seg AC âŠ¥ seg BD [From (iii)]

âˆ´ â–¡ABCD is a rhombus. [A parallelogram is a rhombus perpendicular to each other]

**Question 6.In the adjoining figure, â–¡PQRS and â–¡ABCR are two parallelograms. If âˆ P = 110Â°, then find the measures of all the angles of â–¡ABCR.Solution:**

â–¡PQRS is a parallelogram. [Given]

âˆ´ âˆ R = âˆ P [Opposite angles of a parallelogram]

âˆ´ âˆ R = 110Â° â€¦..(iii)

â–¡ABCR is a parallelogram. [Given]

âˆ´ âˆ A + âˆ R= 180Â° [Adjacent angles of a parallelogram are supplementary]

âˆ´ âˆ A+ 110Â°= 180Â° [From (i)]

âˆ´ âˆ A= 180Â°- 110Â°

âˆ´ âˆ A = 70Â°

âˆ´ âˆ C = âˆ A = 70Â°

âˆ´ âˆ B = âˆ R= 110Â° [Opposite angles of a parallelogram]

âˆ´ âˆ A = 70Â°, âˆ B = 110Â°,

âˆ´ âˆ C = 70Â°, âˆ R = 110Â°

**Question 7.In the adjoining figure, â–¡ABCD is a parallelogram. Point E is on the ray AB such that BE = AB, then prove that line ED bisects seg BC at point F.Given: â–¡ABCD is a parallelogram.BE = ABTo prove: Line ED bisects seg BC at point F i.e. FC = FBSolution:**

Proof:

â–¡ABCD is a parallelogram. [Given]

âˆ´ seg AB â‰… seg DC â€¦â€¦.(i) [Opposite angles of a parallelogram]

seg AB â‰… seg BE â€¦â€¦..(ii) [Given]

seg DC â‰… seg BE â€¦â€¦..(iii) [From (i) and (ii)]

side DC || side AB [Opposite sides of a parallelogram]

i.e. side DC || seg AE and seg DE is their transversal. [A-B-E]

âˆ´ âˆ CDE â‰… âˆ AED

âˆ´ âˆ CDF â‰… âˆ BEF â€¦..(iv) [D-F-E, A-B-E]

In âˆ†DFC and âˆ†EFB,

seg DC = seg EB [From (iii)]

âˆ CDF â‰… âˆ BEF [From (iv)]

âˆ DFC â‰… âˆ EFB [Vertically opposite angles]

âˆ´ âˆ†DFC â‰… âˆ†EFB [SAA test]

âˆ´ FC â‰… FB [c.s.c.t]

âˆ´ Line ED bisects seg BC at point F.

**Intext Questions and Activities**

**Question 1.Write the following pairs considering â–¡ABCD. (Textbook pg. no 57)**

Pairs of adjacent sides:

i. AB, AD

ii. AD, DC

iii. DC, BC

iv. BC, AB

Pairs of adjacent angles:

i. âˆ A, âˆ B

ii. âˆ C, âˆ D

iii. âˆ B, âˆ C

iv. âˆ D, âˆ A

Pairs of opposite sides:

i. AB, DC

ii. AD, BC

Pairs of opposite angles:

i. âˆ A, âˆ C

ii. âˆ B, âˆ D

**Question 2.Complete the following tree diagram. (Textbook pg. no 57)**

**Question 3.In the above theorem, to prove âˆ DAB â‰… âˆ BCD, is any change in the construction needed? If so, how will you write the proof making the change? (Textbook pg. no. 60)Solution:**

Yes

Construction: Draw diagonal BD.

Proof:

side AB || side CD and diagonal BD is their transversal. [Given]

âˆ´ âˆ ABD â‰… âˆ CDB â€¦â€¦..(i) [Alternate angles]

side BC || side AD and diagonal BD is their transversal. [Given]

âˆ´ âˆ ADB â‰… âˆ CBD â€¦â€¦..(ii) [Alternate angles]

In âˆ†DAB and âˆ†BCD,

âˆ ABD â‰… âˆ CDB [From (i)]

seg BD â‰… seg DB [Common side]

âˆ´ âˆ ADB â‰… âˆ CBD [From (ii)]

âˆ´ âˆ†DAB â‰… âˆ†BCD [ASA test]

âˆ´ âˆ DAB â‰… âˆ BCD [c.a.c.t.]

Note: âˆ DAB s âˆ BCD can be proved using the same construction as in the above theorem.

âˆ BAC â‰… âˆ DCA â€¦..(i)

âˆ DAC â‰… âˆ BCA â€¦â€¦(ii)

âˆ´ âˆ BAC + âˆ DAC â‰… âˆ DCA + âˆ BCA [Adding (i) and (ii)]

âˆ´ âˆ DAB â‰… âˆ BCD [Angle addition property]