Question

Question 185
In the following figure, $AB\parallel DC$ and $AD=BC$. Find the value of x.

Answer 1

Given, an isosceles trapezium, where $AB\parallel DC$, AD = BC and $\angle A={60}^{\circ }$.
Then, $\angle B={60}^{\circ }$.
Draw a line parallel to BC through D which intersects the line AB at E (say).

Then, DEBC is a parallelogram, where
BE = CD = 20 cm and DE = BC = 10 cm
Now, $\angle DEB+\angle CBE={180}^{\circ }$
[adjacent angles are supplementary in parallelogram]
$\Rightarrow \angle DEB={180}^{\circ }-{60}^{\circ }={120}^{\circ }$
$\therefore In\Delta ADE,\angle ADE={60}^{\circ }$ [exterior angle]
Also, $\angle DEA={60}^{\circ }$ [$\because$ AD = DE = 10cm and $\angle DAE={60}^{\circ }$]
Then, $\Delta ADE$ is an equilateral triangle.
$\therefore AE=10cm$
$\Rightarrow AB=AE+EB=10+20=30cm$
Hence, x = 30 cm.