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If the lengths of the parallel sides of an isosceles trapezium are 20 cm and 30 cm and the area is 100 cm2, then what is the length of the non-parallel sides ?

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Area of the trapezium = \(\frac{1}{2}(AB + EF)\) x height

⇒ 100 = \(\frac{1}{2}(20 + 30)\) x h ⇒ = 4 ⇒ AC = BD = 4 cm

∴ EA = BF = \(\sqrt{AC^2 + EC^2}\) = \(\sqrt{16 + 25}\) = \(\sqrt{41}\) cm.

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