Question

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Answer 1

1. The figure is an isosceles trapezium: ABCD, having AB = 15 cm, CD = 25 cm and BC = AD = 12 cm.

The distance between AB and CD = [12^2–5^2]^0.5 = [144–25]^0.5 = 119^0.5 = 10.90871211 cm. The area of the isosceles trapezium = (15+25)*10.90871211/2 = 218.1742423 sq cm.

2. The figure is a trapezium: ABCD, having AB = 15 cm, CD = 25 cm, <A = <D = 90 deg and BC = 12 cm. The distance between AB and CD = [12^2–10^2]^0.5 = [144–100]^0.5 = 44^0.5 = 6.633249581 cm. The area of the trapezium = (15+25)*6.633249581/2 = 132.6649916 sq cm.

You have two choices for the area of the trapezium as 218.1742423 sq cm. or 132.6649916 sq cm.

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Answer 2

Answer:

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