Question

In a quadrilateral the sides are $9,40,28,15$ units and the angle between first two sides is a right angle. The area of quadrilateral is

• A. $106$ sq units
• B. $206$ sq units
• C. $306$ sq units
• D. $406$ sq units

Answer 1

The correct option is C $306$ sq units

Given, $\angle DAB$ is a right-angled triangle. Therefore, $D{B}^2=D{A}^2+A{B}^2$ [Using Pythagoras Theorem]
$\Rightarrow D{B}^2=9^2+{40}^2$
$\Rightarrow DB=\sqrt{81+1600}=\sqrt{1681}=41$ unit
For $\Delta DAB$, we have $s=\frac{9+40+41}{2}=\frac{90}{2}=45$ unit
Therefore, Area of $\Delta DAB=A_1=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{45(45-9)(45-40)(45-41)}=\sqrt{45\times 36\times 5\times 4}=180$ sq. unit
For $\Delta DCB$, we have $s=\frac{28+15+41}{2}=\frac{84}{2}=42$ unit
$\Rightarrow A_2=$ Area of $\Delta DCB$
$=\sqrt{42\times (42-28)\times (42-15)\times (42-41)}$ sq. units
$=\sqrt{42\times 14\times 27\times 1}$
$=126$
Hence, area of the Field $=A_1+A_2=(180+126)=306$ sq. units