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Question
In the below fig. ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm,
and distance between AB and DC is 4cm. Find the value of x and area of trapezium ABCD.
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Solution
Draw AL ⊥ DC, BM ⊥ DC Then ,
AL = BM = 4cm and LM = 7 cm
In ΔADL, we have
AD2 = AL2 + DL2 ⇒ 25 = 16 + DL2 ⇒ DL= 3 cm
`Similarly MC = sqrt(BC^2-BM^2)=sqrt(25-16)=3cm`
∴ x = CD = CM + ML + CD = 3 + 7 + 3 = 13cm
`ar (trap . ABCD ) = 1/2 ( AB + CD) × AL 1/2 (7 + 13) xx 4cm^2`
= 4cm2
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