Question

In trapezium ABCD, AB is parallel to DC and given that AB = AD = BC = 13 cm and CD = 23 cm. Find the area of the trapezium.

Answer 1

From B draw BE || AD, and BF $\perp$ DC
Since ABED is a parallelogram, DE = 13 cm.
$\therefore$ EC = 23 cm - 13cm = 10 cm
Also BE = 13 cm
Therefore BEC is an isosceles triangle.



Since BF $\perp$ EC, therefore F is the midpoint of EC
$\therefore FC=\frac{1}{2}\times 10cm=5cm$
In the right triangle BFC
$B{F}^2=B{C}^2-F{C}^2={13}^2-5^2=144$
$\therefore$ BF = 12 cm
Area of trapezium = $\frac{1}{2}$ sum of parallel sides × height
= $\frac{1}{2}(13+23)\times 12c{m}^2$
= 216 $c{m}^2$