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. [4 Marks]

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 $=\frac{3}{4}\times areaofABCD$

$AE=AB+BE$
$AE=(6+y)cm$ [1 Mark]

Area of a triangle is half the product of base and height
$=\frac{1}{2}AE\times height$

$=\frac{1}{2}\times (6+y)\times h$ [1 Mark]

Area of parallelogram is product of base and height
$=AB\times height$

$=6\times h$ [1 Mark]

Therefore,
Area of triangle ADE $=\frac{3}{4}\times$ area of parallelogram

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

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

$\Rightarrow 6+y=9$

$\Rightarrow y=3$ [1 Mark]

The length of BE is 3 cm.