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Q 6

Ans x=13 cm , 40 cm2

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Q.6. In the given figure, AD = BC = 5 cm, AB = 7 cm. The parallel sides AB, DC are 4 cm apart. DC = x cm. Find x and the area of the trapezium ABCD,

$IntherightangletriangleADE,\phantom{\rule{0ex}{0ex}}D{E}^{2}=A{D}^{2}-A{E}^{2}\phantom{\rule{0ex}{0ex}}={5}^{2}-{4}^{2}\phantom{\rule{0ex}{0ex}}=25-16\phantom{\rule{0ex}{0ex}}=9\phantom{\rule{0ex}{0ex}}DE=3cm\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}ThereforeDC=3+7+3=13cm\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}AreaoftrapeziumABCD=\frac{1}{2}\times AE\times \left(AB+DC\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 4\times \left(7+13\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 4\times 20\phantom{\rule{0ex}{0ex}}=40c{m}^{2}$

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