Question

In a quadrilateral $ABCD,AB=7cm,BC=6cm,CD=12cm,DA=15cm,AC=9cm$. Its area is

• A. $(\sqrt{440}+54)$ $c{m}^2$
• B. $(\sqrt{440}+44)$ $c{m}^2$
• C. $(\sqrt{110}+44)$ $c{m}^2$
• D. $(\sqrt{340}+64)$ $c{m}^2$

Answer 1

The correct option is A $(\sqrt{440}+54)$ $c{m}^2$

By applying heron's formula in triangle ABC having the sides $7cm,6cm$ and $9cm$, we have
$s=\frac{7+6+9}{2}=11cm$
Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{11(11-7)(11-6)(11-9)}$
$=\sqrt{440}c{m}^2$
Now, in triangle $ADC$ having sides $15cm,12cm$ and $9cm$, we have
$s=\frac{15+12+9}{2}=18cm$
$=\sqrt{18(18-15)(18-12)(18-9)}$
$=54c{m}^2$
There, the area of quadrilateral $ABCD=(\sqrt{440}+54){cm}^2$