Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 5.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 5 Quadrilaterals.

## Practice Set 5.1 Geometry 9th Std Maths Part 2 Answers Chapter 5 Quadrilaterals

Question 1.

Diagonals of a parallelogram WXYZ intersect each other at point O. If ∠XYZ∠ = 135°, then measure of ∠XWZ and ∠YZW? If l(OY) = 5 cm, then l(WY) = ?

Solution:

i. ∠XYZ = 135°

□WXYZ is a parallelogram.

∠XWZ = ∠XYZ

∴ ∠XWZ = 135° …..(i)

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

∴ ∠YZW + 135°= 180° [From (i)]

∴ ∠YZW = 180°- 135°

∴ ∠YZW = 45°

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

l(OY) = \(\frac { 1 }{ 2 }\) l(WY) [Diagonals of a parallelogram bisect each other]

∴ l(WY) = 2 x l(OY)

= 2 x 5

∴ l(WY) = 10 cm

∴∠XWZ = 135°, ∠YZW = 45°, l(WY) = 10 cm

Question 2.

In a parallelogram ABCD, if ∠A = (3x + 12)°, ∠B = (2x – 32)°, then liptl the value of x and the measures of ∠C and ∠D.

Solution:

□ABCD is a parallelogram. [Given]

∴ ∠A + ∠B = 180° [Adjacent angles of a parallelogram are supplementary],

∴ (3x + 12)° + (2x-32)° = 180°

∴ 3x + 12 + 2x – 32 = 180

∴ 5x – 20 = 180

∴ 5x= 180 + 20

∴ 5x = 200

∴ x = \(\frac { 200 }{ 5 }\)

∴ x = 40

ii. ∠A = (3x + 12)°

= [3(40) + 12]°

=(120 +12)°= 132°

∠B = (2x – 32)°

= [2(40) – 32]°

= (80 – 32)° = 48°

∴ ∠C = ∠A = 132°

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

∴ The value of x is 40, and the measures of ∠C and ∠D are 132° and 48° 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 □ABCD be the parallelogram and the length of AD be x cm.

One side is greater than the other by 25 cm.

∴ AB = x + 25 cm

AD = BC = x cm

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

ii. Perimeter of □ABCD = 150 cm [Given]

∴ AB + BC + DC + AD = 150

∴ (x + 25) +x + (x + 25) + x – 150

∴ 4x + 50 = 150

∴ 4x = 150 – 50

∴ 4x = 100

∴ x = \(\frac { 100 }{ 4 }\)

∴ x = 25

iii. AD = BC = x = 25 cm

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

∴ 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 □ABCD be the parallelogram.

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

Let the common multiple be x.

∴ ∠A = x° and ∠B = 2x°

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

∴ x + 2x = 180

∴ 3x = 180

∴ x = \(\frac { 180 }{ 3 }\)

∴ x = 60

ii. ∠A = x° = 60°

∠B = 2x° = 2 x 60° = 120°

∠A = ∠C = 60°

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

∴ The measures of the angles of the parallelogram are 60°, 120°, 60° and 120°.

Question 5.

Diagonals of a parallelogram intersect each other at point O. If AO = 5, BO show that □ABCD is a rhombus.

Given: AO = 5, BO = 12 and AB = 13.

To prove: □ABCD is a rhombus.

Solition:

Proof:

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

AO^{2} + BO^{2} = 5^{2} + 12^{2}

= 25 + 144

∴ AO^{2} + BO^{2} = 169 …..(i)

AB^{2} = 13^{2} = 169 ….(ii)

∴ AB^{2} = AO^{2} + BO^{2} [From (i) and (ii)]

∴ ∆AOB is a right-angled triangle. [Converse of Pythagoras theorem]

∴ ∠AOB = 90°

∴ seg AC ⊥ seg BD …..(iii) [A-O-C]

∴ In parallelogram ABCD,

∴ seg AC ⊥ seg BD [From (iii)]

∴ □ABCD is a rhombus. [A parallelogram is a rhombus perpendicular to each other]

Question 6.

In the adjoining figure, □PQRS and □ABCR are two parallelograms. If ∠P = 110°, then find the measures of all the angles of □ABCR.

Solution:

□PQRS is a parallelogram. [Given]

∴ ∠R = ∠P [Opposite angles of a parallelogram]

∴ ∠R = 110° …..(iii)

□ABCR is a parallelogram. [Given]

∴ ∠A + ∠R= 180° [Adjacent angles of a parallelogram are supplementary]

∴ ∠A+ 110°= 180° [From (i)]

∴ ∠A= 180°- 110°

∴ ∠A = 70°

∴ ∠C = ∠A = 70°

∴ ∠B = ∠R= 110° [Opposite angles of a parallelogram]

∴ ∠A = 70°, ∠B = 110°,

∴ ∠C = 70°, ∠R = 110°

Question 7.

In the adjoining figure, □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: □ABCD is a parallelogram.

BE = AB

To prove: Line ED bisects seg BC at point F i.e. FC = FB

Solution:

Proof:

□ABCD is a parallelogram. [Given]

∴ seg AB ≅ seg DC …….(i) [Opposite angles of a parallelogram]

seg AB ≅ seg BE ……..(ii) [Given]

seg DC ≅ seg BE ……..(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]

∴ ∠CDE ≅ ∠AED

∴ ∠CDF ≅ ∠BEF …..(iv) [D-F-E, A-B-E]

In ∆DFC and ∆EFB,

seg DC = seg EB [From (iii)]

∠CDF ≅ ∠BEF [From (iv)]

∠DFC ≅ ∠EFB [Vertically opposite angles]

∴ ∆DFC ≅ ∆EFB [SAA test]

∴ FC ≅ FB [c.s.c.t]

∴ Line ED bisects seg BC at point F.

**Maharashtra Board Class 9 Maths Chapter 5 Quadrilaterals Practice Set 5.1 Intext Questions and Activities**

Question 1.

Write the following pairs considering □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. ∠A, ∠B

ii. ∠C, ∠D

iii. ∠B, ∠C

iv. ∠D, ∠A

Pairs of opposite sides:

i. AB, DC

ii. AD, BC

Pairs of opposite angles:

i. ∠A, ∠C

ii. ∠B, ∠D

Question 2.

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

Question 3.

In the above theorem, to prove ∠DAB ≅ ∠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]

∴ ∠ABD ≅ ∠CDB ……..(i) [Alternate angles]

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

∴ ∠ADB ≅ ∠CBD ……..(ii) [Alternate angles]

In ∆DAB and ∆BCD,

∠ABD ≅ ∠CDB [From (i)]

seg BD ≅ seg DB [Common side]

∴ ∠ADB ≅ ∠CBD [From (ii)]

∴ ∆DAB ≅ ∆BCD [ASA test]

∴ ∠DAB ≅ ∠BCD [c.a.c.t.]

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

∠BAC ≅ ∠DCA …..(i)

∠DAC ≅ ∠BCA ……(ii)

∴ ∠BAC + ∠DAC ≅ ∠DCA + ∠BCA [Adding (i) and (ii)]

∴ ∠DAB ≅ ∠BCD [Angle addition property]