Question

$ABCD$ is a trapezium in which $AB\parallel DC$, $DC=7cm$ distance between $AB$ and $DC$ is $4cm$. Find $AB$.

Answer 1

$△AED$ is a right angled triangle

So,

$A{D}^2=A{E}^2+E{D}^2$

$A{E}^2=5^2-4^2$

$AE=\sqrt{25-16}$

$AE=\sqrt{9}$

$AE=3cm$

Similarly in $△BFC$ we have

$B{C}^2=B{F}^2+F{C}^2$

$B{F}^2=5^2-4^2$

$BF=\sqrt{25-16}$

$BF=\sqrt{9}$

$BF=3cm$

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 $EF=7=CD$

$\therefore AB=3+7+3=13cm$