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, then find the length of BE.

Answer 1

Let the length of BE be y cm. Let the height from D to the base AB be h cm.

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

AE = AB+BE = (6 + y) cm.

Area of a triangle is half the product of base and height = $\frac{1}{2}AE\times height$ = $\frac{1}{2}\times (6+y)\times h$

Area of parallelogram is product of base and height = $AB\times height$ = 6 x h
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.