Question

In the adjoining figure, $ABCD$ is an isosceles trapezium. If $\angle A={60}^o,DC=20cm$ and $AD=15cm$, find the length of $AB$.

Answer 1

In triangle $ADM.$

$\cos {60}^0=\frac{AM}{12}$

$AM=12\cos {60}^0=6cm$

So by symmetry $BN=6cm$

Hence $AB=AM+NM+NB=6+16+6=28cm$

Now$,$

$\sin {60}^0=\frac{DM}{12}$

$\frac{\sqrt{3}}{2}=\frac{DM}{12}$

$\Rightarrow DM=6\sqrt{3}$

So are of trapezium $=\frac{1}{2}\times \left(AB+CD\right)\times DM$

$=\frac{1}{2}\left(28+16\right)\times 6\sqrt{3}=132\sqrt{3}c{m}^2\approx 16cm$

We get,

length of $AB=16cm$