Question

Find the area of the quadrilateral whose sides are $9m,40m,28m,15m$. The angle between first two sides is ${90}^o$.

• A. $348$ sq. m
• B. $315$ sq. m
• C. $306$ sq. m
• D. None of these

Answer 1

The correct option is C $306$ sq. m

Area of quadrilateral $ABCD$ = ar $\left[△ABC\right]$ + ar $\left[△ACD\right]$

In $△ABC$ :

$ar\left[△ABC\right]=\frac{1}{2}\times BC\times AB=\frac{1}{2}\times 40m\times 9m=180m^2$

Using pythagoras theorem,

$AC=\sqrt{{AB}^2+{BC}^2}=\sqrt{{\left(9\right)}^2+{\left(40\right)}^2}=\sqrt{1681}=41m$

Using Heron's formula,

$ar\left[△ACD\right]=\sqrt{s(s-a)(s-b)(s-c)}s=\frac{a+b+c}{2}=\frac{41+28+15}{2}=\frac{84}{2}=42m$

$ar\left[△ACD\right]=\sqrt{42(42-41)(42-28)(42-15)}=\sqrt{42\times 1\times 14\times 27}=126m^2$

Therefore, area of quadrilateral $ABCD$

$=180+126$

$=306m^2$