Question

A quadrilateral ABCD has $\angle C=90,$ AB=11 cm,CD=8cm, BD=10 cm and AD=9 cm. Find its area.

• A. $30\sqrt{2}+24$
• B. $60\sqrt{2}$
• C. $48$
• D. none of these

Answer 1

The correct option is A $30\sqrt{2}+24$

It is given that $\angle C=90º,AB=11$ cm, $BD=10$ cm, $CD=8$ cm and $AD=9$ cm

In $\Delta BCD$,

By applying Pythagoras theorem,

$B{C}^2=B{D}^2+-C{D}^2$

$\Rightarrow B{C}^2={10}^2-8^2$

$\Rightarrow B{C}^2=100-64=36$

$BC=6$ cm

Area of $\Delta BCD$$=\frac{1}{2}\times 6\times 8=24$ cm${}^2$

Now, semi perimeter of $\Delta ABD$$=\frac{1}{2}(9+11+10)=\frac{30}{2}=15$ cm${}^2$

Using heron's formula,

Area of $\Delta ABD$ is:

$A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{15(15-10)(15-11)(15-9)}=\sqrt{15\times 5\times 4\times 6}=\sqrt{1800}=30\sqrt{2}$cm${}^2$

Area of quadrilateral $ABCD$ = Area of $\Delta BCD+$ Area of $\Delta ABD$ that is $30\sqrt{2}+24$ cm${}^2$.