Question

In fig. ABCD is a field in the form of a quadrilateral whose sides are indicated in the figure. if $\angle DAB={90}^o$ , find the area of the field.

Answer: $306$ $m^2$

State whether answer given is true or not

• A. True
• B. False

Answer 1

The correct option is A True

$=ar△ABD+ar△BDC$

Now, $ar△ABD=\frac{1}{2}\times$ base$\times$ height$=\frac{1}{2}\times AB\times AD=\frac{1}{2}\times 40\times 9$

$=180㎡$

Again in $△ABD,BD=\sqrt{{\left(AB\right)}^2+{\left(AD\right)}^2}$ (Using Pythagoras theorem)

$=\sqrt{{\left(40\right)}^2+{\left(9\right)}^2}=\sqrt{1681}=41㎝$

Now, $ar△BDC=\sqrt{s(s-a)(s-b)(s-c)},s=\frac{a+b+c}{2}=\frac{41+15+28}{2}=42㎝$

$\therefore ar△BDC=\sqrt{42(42-4)(42-15)(42-28)}=\sqrt{42\times 1\times 27\times 14}$

$=\sqrt{14\times 3\times 9\times 3\times 14}=3\times 3\times 14=126㎡$

$\therefore$Area of quadrilateral ABCD$=180+126=306㎡$