Question

In quadrilateral ABCD, given that $AB=7cm,BC=12cm,CD=12cm,DA=9cm$ and diagonal $AC=15cm$. It's area is $54c{m}^2$.

• A. True
• B. False

Answer 1

The correct option is B False

In $△ABC$, we have
$a=7cm,b=12cm$ and $c=15cm$
$\therefore s=\frac{a+b+c}{2}=\frac{7cm+12cm+15cm}{2}$
$=\frac{34cm}{2}=17cm$
Area of $\Delta ABC=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{17(17-7)(17-12)(17-15)}$
$=\sqrt{17\times 10\times 5\times 2}$
$=10\sqrt{17}c{m}^2$
Similarly, in $△ADC$, we have
$a=9cm,b=12cm$ and $c=15cm$
$\therefore s=\frac{a+b+c}{2}=\frac{9cm+12cm+15cm}{2}$
$=\frac{36cm}{2}=18cm$
Area of $\Delta ADC=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{18(18-9)(18-12)(18-15)}$
$=\sqrt{18\times 9\times 6\times 3}$
$=54c{m}^2$
Area of quadrilateral ABCD $=$ Area $\Delta ABC$ $+$ Area $\Delta ADC$ = $10\sqrt{17}c{m}^2+54c{m}^2$