The parallel sides of an isosceles trapezium are in the ratio 2 : 3. If its height is 4 cm and area is 60 cm², find the perimeter.
Mensuration
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The parallel sides of an isosceles trapezium are in the ratio 2 : 3. If its height is 4 cm and area is 60 cm², find the perimeter.

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Since, ABCD is an isosceles trapezium so, BC = AD. Since, parallel sides of an isosceles trapezium are in the ratio 2 : 3. ∴ CD = 2a and AB = 3a. Construct perpendicular DN from D to AB and perpendicular CM from C to AB. Given, Area = 60 cm² By formula, Area of trapezium = × sum of parallel sides × height ⇒ 60 = × (AB + DC) × DN ⇒ 60 = × (3a + 2a) × 4 ⇒ 60 = 2 × 5a ⇒ 10a = 60 ⇒ a = 6 cm. ⇒ AB = 3a = 3 × 6 = 18 cm and CD = 2a = 2 × 6 = 12 cm. In △ADN and △BCM, ⇒ ∠AND = ∠CMB = 90° ⇒ DN = CM = 4 cm ⇒ AD = CB = x cm (let) (As ABCD is an isosceles trapezium). ∴ △ADN ≅ △BCM by RHS axiom. ∴ AN = MB ........(1) Since, DNMC is a rectangle. ∴ NM = DC = 12 cm. (As opposite sides of a rectangle are equal.) From figure, ⇒ AN + NM + MB = 18 ⇒ AN + 12 + MB = 18 ⇒ AN + MB = 6 ⇒ 2AN = 6 (As AN = MB) ⇒ AN = = 3 cm. ⇒ MB = 3 cm. In right angle triangle AND, ⇒ AD² = AN² + DN² ⇒ x² = 4² + 3² ⇒ x² = 16 + 9 ⇒ x² = 25 ⇒ x = = 5 cm. From figure, Perimeter = AB + BC + CD + DA = 18 + 5 + 12 + 5 = 40 cm. Hence, perimeter of trapezium = 40 cm.

Mathematics

The parallel sides of an isosceles trapezium are in the ratio 2 : 3. If its height is 4 cm and area is 60 cm², find the perimeter.

Mensuration

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The parallel sides of an isosceles trapezium are in the ratio 2 : 3. If its height is 4 cm and area is 60 cm2, find the perimeter.

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