Question

Find the area of a Quadrilateral ABCD whose sides are 9m, 40m, 28m, 15m, respectively and the angle between the first two sides is a right angle.

• A. 180 sqm
• B. 306 sqm
• C. 360 sqm
• D. 108 sqm

Answer 1

The correct option is B 306 sqm

Let ABCD be a quadrilateral, then

$AB=9m,BC=40m,CD=28m,DA=15m$ & $\angle ABC={90}^{\circ }$

$ar\square ABCD=ar△ABC+ar△ACD$

In $△ABC$, using Pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2\Rightarrow AC=\sqrt{9^2+{40}^2}=\sqrt{1681}=41m$

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

$ar△ACD=\sqrt{s(s-a)(s-b)(s-c)},S=\frac{a+b+c}{2}=\frac{15+28+41}{2}=42m$

$=\sqrt{42(42-15)(42-28)(42-41)}$

$=\sqrt{42\times 27\times 14\times 1}=126m^2$

$\therefore ar\square ABCD=180+126=306m^2$