1. ABCD is an isosceles trapezium such that AD||BC, AB = 5 cm, AD = 8 cm and BC = 14 cm. What is the area $$(in cm^2)$$ of trapezium?
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By: anil on 05 May 2019 01.47 am
Given : ABCD is an isosceles trapezium and AB = 5 cm, AD = 8 cm and BC = 14 cm To find : ar(ABCD) = ? Solution : In an isosceles trapezium AB = CD and BE = CF Also, AD = EF = 8 cm => BE + CF = 14 - 8 = 6 cm => BE = CF = $$frac{6}{2}=3$$ cm In $$ riangle$$ ABE, => $$(AE)^2=(AB)^2-(BE)^2$$ => $$(AE)^2=(5)^2-(3)^2$$ => $$(AE)^2=25-9=16$$ => $$AE=sqrt{16}=4$$ cm $$ herefore$$ ar(ABCD) = $$frac{1}{2} imes (AD+BC) imes(AE)$$ = $$frac{1}{2} imes(8+14) imes4$$ = $$22 imes2=44$$ $$cm^2$$ => Ans - (B)
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