Question

In quadrilateral $\text{ABCD}$ shown in figure, $\text{AB}\parallel \text{DC}\text{and}\text{AD}\perp \text{AB}$, Also $\text{AB = 8m,}\text{DC = BC = 5m}\text{.}$ Find the area of the quadrilateral.

Answer 1

In quadrilateral ABCD:

Assume CE is a perpendicular dropped to AB,

Now we have a right angled $△$ CEB and a rectangle AECD

$\therefore$ Area of quadrilateral ABCD = Area of right angled $△$ CEB + Area of rectangle AECD

AB=AE+EB

EB=AB-AE = 8-5 = 3m ($\because$ AE=DC , opposite sides of a rectangle)

In $△$ CEB:

CB${}^2$ = CE${}^2$ + EB${}^2$

CE${}^2$ = CB${}^2$ - EB${}^2$

CE${}^2$ = 5${}^2$ - 3${}^2$= 25-9=16

CE = $\sqrt{1}6$ = 4m

Now, Area of $△$ CEB = $\frac{1}{2}\times 3\times 4$ = 6m${}^2$

Area of rectangle AECD = DC $\times$ CE = 5 $\times$ 4 =20m${}^2$

Area of quadrilateral ABCD = 6 m${}^2$ + 20m${}^2$ = 26m${}^2$