Chapter 15 Area Practice Set 15.3 Solutions

☐ABCD is a trapezium, side AB || side DC, l(AB) = 13 cm, l(DC) = 9 cm, l(AD) = 8 cm, Area of a trapezium = x sum of lengths of parallel sides x height ∴ A (☐ABCD) = x [l(AB) + l(DC)] x l(AD) = x (13 + 9) x 8 = x 22 x 8 = 11 x 8 = 88 sq.cm ∴ The area of ☐ABCD is 88 sq. cm. [Note: The question is modified.]

Length of the two parallel sides of a trapezium are 8.5 cm and 11.5 cm and its height is 4.2 cm. Area of a trapezium = x sum of lengths of parallel sides x height = x (8.5 + 11.5) x 4.2 = x 20 x 4.2 = 10 x 4.2 = 42 sq. cm ∴ The area of the trapezium is 42 sq. cm.

☐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 Area of a trapezium = x sum of lengths of parallel sides x height ∴ A(☐PQRS) = x [l(PQ) + l(SR)] x l(PM) = x (7+ 13) x 4 = x 20 x 4 = 40 sq.cm ∴ The area of ☐PQRS is 40 sq. cm.