Question

Some measures are given in the adjacent figure, find the area of ABCD

Answer 1

Given:

In the fig. 1, m$\angle$ BAD=90${}^{\circ }$, AB=40 m, BT=13 m, DC=60 m, AD=9 m

Area of $\square$ ABCD= area of $△$ BAD+ area of $△$ BDC

area of $△$ BAD $=\frac{1}{2}\times$ base$\times$ height

$=\frac{1}{2}\times 40\times 9$

$=20\times 9=180m^2$

First we need to calculate the area of $△$ BDC

Base DC =60 m and Height BT =13 m

Area ($△$ BDC)$=\frac{1}{2}$ base $\times$ height

Area⁡($△$ BDC) $=\frac{1}{2}\times 60\times 13$

Area⁡($△$ BDC)=$30\times 13$

Area⁡($△$ BDC)=390 sq.m.

Area of $\square$ABCD= area of $△$ BAD + area of BDC

Area of $\square$ABCD=180+390

Area of $\square$ABCD=570 sq.m.

Hence, the area of the quadrilateral is 570 sq. m.