Question

In the figure, ABCD is a parallelogram and AB = 6 cm. If the area of triangle ADE is $\frac{3}{4}$ that of parallelogram ABCD, find the length of BE. [1 MARK]

Answer 1

Concept: 1 Mark

Let the length of BE be y cm. Let the height from D to the base AB be h cm.
This will also be the height for $\Delta ADE$

Area of $\Delta$ADE
$=\frac{3}{4}\times areaofABCD$

Therefore
$\frac{1}{2}\times (6+y)\times h=\frac{3}{4}\times 6\times h$

$6+y=2\times \frac{3}{4}\times 6$

$6+y=9$

$y=3$

The length of BE is 3 cm.