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Question

ABCD is a trapezium in which ABDC, DC=7cm distance between AB and DC is 4cm. Find AB.
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Solution

AED is a right angled triangle
So,
AD2=AE2+ED2
AE2=5242
AE=2516
AE=9
AE=3 cm
Similarly in BFC we have
BC2=BF2+FC2
BF2=5242
BF=2516
BF=9
BF=3 cm
Since EFCD is a rectangle and have one pair of parallel sides equal so the other pair of parallel sides are also equal i.e EF=7=CD
AB=3+7+3=13 cm

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