What is the area (in cm2) of an isosceles trapezium in which the two parallel sides are 12 cm and 6 cm and the two non-parallel sides are 5 cm each?
Explanation:
In an isosceles trapezium ABCD. AD = BC = 5 and DE = FC = 3
In right triangle BCF, BC = 5, FC = 3 hence BF = 4 (pythagorean triplet)
∴ Area of the trapezium = ½ × (6 + 12) × 4 = 36 cm2.
Hence, 36.
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