## Maharashtra State Board Class 9 Maths Solutions Chapter 5 Quadrilaterals Practice Set 5.4

Question 1.

In □IJKL, side IJ || side KL, ∠I = 108° and ∠K = 53°, then find the measures of ∠J and ∠L.

Solution:

i. ∠I = 108° [Given]

side IJ || side KL and side IL is their transveral. [Given]

∴ ∠I + ∠L = 180° [Interior angles]

∴ 108° + ∠L = 180°

∴ ∠L = 180° – 108° = 72°

ii. ∠K = 53° [Given]

side IJ || side KL and side JK is their transveral. [Given]

∴ ∠J + ∠K = 180° [Interior angles]

∴ ∠J + 53° = 180°

∴ ∠J= 180°- 53° = 127°

∴ ∠L = 72°, ∠J = 127°

Question 2.

In □ABCD, side BC || side AD, side AB ≅ side DC. If ∠A = 72°, then find the measures of ∠B and ∠D.

Construction: Draw seg BP ⊥ side AD, A – P – D, seg CQ ⊥ side AD, A – Q – D.

Solution:

i. ∠A = 72° [Given]

In □ABCD, side BC || side AD and side AB is their transversal. [Given]

∴ ∠A + ∠B = 180° [Interior angles]

∴ 72° +∠B = 180°

∴ ∠B = 180° – 72° = 108°

ii. In ∆BPA and ∆CQD,

∠BPA ≅ ∠CQD [Each angle is of measure 90°]

Hypotenuse AB ≅ Hypotenuse DC [Given]

seg BP ≅ seg CQ [Perpendicular distance between two parallel lines]

∴ ∆BPA ≅ ∆CQD [Hypotenuse side test]

∴ ∠BAP ≅ ∠CDQ [c. a. c. t.]

∴ ∠A = ∠D

∴ ∠D = 72°

∴ ∠B = 108°, ∠D = 72°

Question 3.

In □ABCD, side BC < side AD, side BC || side AD and if side BA ≅ side CD, then prove that ∠ABC = ∠DCB.

Given: side BC < side AD, side BC || side AD, side BA = side CD

To prove: ∠ABC ≅ ∠DCB

Construction: Draw seg BP ⊥ side AD, A – P – D

seg CQ ⊥ side AD, A – Q – D

Solution:

Proof:

In ∆BPA and ∆CQD,

∠BPA ≅ ∠CQB [Each angle is of measure 90°]

Hypotenuse BA ≅ Hypotenuse CD [Given]

seg BP ≅ seg CQ [Perpendicular distance between two parallel lines]

∴ ∆BPA ≅ ∆CQD [Hypotenuse side test]

∴ ∠BAP ≅ ∠CDQ [c. a. c. t.]

∴ ∠A = ∠D ….(i)

Now, side BC || side AD and side AB is their transversal. [Given]

∴ ∠A + ∠B = 180°…..(ii) [Interior angles]

Also, side BC || side AD and side CD is their transversal. [Given]

∴ ∠C + ∠D = 180° …..(iii) [Interior angles]

∴ ∠A + ∠B = ∠C + ∠D [From (ii) and (iii)]

∴ ∠A + ∠B = ∠C + ∠A [From (i)]

∴ ∠B = ∠C

∴ ∠ABC ≅ ∠DCB