Question

ABCD is a rectangle such that

CD = 30 cm

BC = 40 cm

F is a point on the interior of the rectangle and the perpendicular from F meets the side AB at its midpoint i.e., E.

If EF = 32 cms.

Find the area of the region AFBCD [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 3 Marks

We know that the area of a rectangle

$=length\times breadth$

$=(30\times 40)c{m}^2$

$=1200c{m}^2$

Now the area of the $\Delta AFB=\frac{1}{2}\times Base\times Height$

$\Rightarrow \frac{1}{2}\times AB\times EF$

$\Rightarrow \frac{1}{2}\times 30\times 32$

$\Rightarrow (15\times 32)$

$\Rightarrow 480c{m}^2$

Hence Area of the shaded region.

$AFBCD=Area\left(ABCD\right)-Areaof\Delta AFB$.

$\Rightarrow 1200c{m}^2-480c{m}^2$

$\Rightarrow 720c{m}^2$