Question

A quadrilateral $ABCD$ has $\angle C=90,AB=9cm,BC=8cm,CD=6cm$ and $AD=7cm.$ How much area does it occupy?

• A. $6\sqrt{26}+24$
• B. $48$
• C. $12\sqrt{26}$
• D. none of these

Answer 1

The correct option is A $6\sqrt{26}+24$

It is given that $\angle C=90º,AB=9$ cm, $BC=8$ cm, $CD=6$ cm and $AD=7$ cm

$BD$ is joined.

In $\Delta BCD$,

By applying Pythagoras theorem,

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

$\Rightarrow B{D}^2=8^2+4^2$

$\Rightarrow B{D}^2=64+16=80$

$BD=4\sqrt{5}$ cm

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

Now, semi perimeter of $\Delta ABD$$=\frac{1}{2}(7+9+4\sqrt{5})=\frac{16+4\sqrt{5}}{2}=8+2\sqrt{5}$ cm${}^2$

Using heron's formula,

Area of $\Delta ABD$ is:

$A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{8+2\sqrt{5}(8+2\sqrt{5}-4\sqrt{5})(8+2\sqrt{5}-9)(8+2\sqrt{5}-7)}$

$=\sqrt{(8+2\sqrt{5})(8-2\sqrt{5})(2\sqrt{5}-1)(2\sqrt{5}+1)}=\sqrt{[{\left(8\right)}^2-{\left(2\sqrt{5}\right)}^2][{\left(2\sqrt{5}\right)}^2-{\left(1\right)}^2]}=\sqrt{[64-20][20-1]}=\sqrt{44\times 19}$

$=\sqrt{936}=6\sqrt{26}$ cm${}^2$

Area of quadrilateral $ABCD$ = Area of $\Delta BCD+$ Area of $\Delta ABD$

= $6\sqrt{26}+24$ cm${}^2$.