## Chapter 15 Area Set 15.3

Question 1.

In the given figure, ☐ABCD is a trapezium, side AB || side DC, l(AB) = 13 cm, l(DC) = 9 cm, l(AD) = 8 cm, find the area ☐ABCD.

Question 3.

☐PQRS is an isosceles trapezium. l(PQ) = 7 cm, seg PM ⊥ seg SR, l(SM) = 3 cm. Distance between two parallel sides is 4 cm, find the area of ☐PQRS.

Solution:

☐PQRS is an isosceles trapezium.

l(PQ) = 7 cm, seg PM ⊥ seg SR,

l(SM) = 3 cm, l(PM) = 4cm

Draw seg QN ⊥ seg SR.

In ☐PMNQ,

seg PQ || seg MN

∠PMN = ∠QNM = 90°

∴ ☐PMNQ is a rectangle.

Opposite sides of a rectangle are congruent.

∴ l(PM) = l(QN) = 4 cm and

l(PQ) = l(MN) = 7 cm

In ∆PMS, m∠PMS = 90°

∴ [l(PS)]² = [l(PM)]² + [l(SM)]² … [Pythagoras theorem]

∴ [l(PS)]² = (4)² + (3)²

∴ [l(PS)]² = 16 + 9 = 25

∴ l(PS) = √25 = 5 cm

…[Taking square root of both sides]

☐PQRS is an isosceles trapezium.

∴ l(PS) = l(QR) = 5 cm

In ∆QNR, m ∠QNR = 90°

∴ [l(QR)]² = [l(QN)]² + [l(NR)]²

… [Pythagoras theorem]

∴ (5)² = (4)² + [l(NR)]²

∴ 25 = 16 + [l(NR)]²

∴ [l(NR)]² = 25 – 16 = 9

∴ l(NR) = √9 = 3 cm

…[Taking square root of both sides]

l(SR) = l(SM) + l(MN) + l(NR)

= 3 + 7 + 3

= 13 cm