Question

Find the area of a quadrilateral whose sides are $12,5,6$ and $15$. The angle between the first two sides is $90$. (Use Heron's formula)

• A. $20+$ $2\sqrt{2805}$
• B. $30+$ $2\sqrt{805}$
• C. $30+$ $2\sqrt{374}$
• D. $30+$ $2\sqrt{1805}$

Answer 1

Answer: (C)

$30+$ $2\sqrt{374}$

Consider $ABCD$ is a quadrilateral where,

$AB=12,BC=5,CD=6,DA=15$ and $\angle ABC={90}^o$

Area of $ABCD=$ Area of $\Delta ABC+$Area of $\Delta ACD$
In $\Delta$ ABC, $\angle B={90}^o$
Apply Pythagoras theorem in $\Delta ABC$

Therefore, $A{C}^2=A{B}^2+B{C}^2={12}^2+5^2$
So, $AC=13$

Area of $\Delta ABC=\frac{1}{2}\times AB\times BC=\frac{1}{2}\times 12\times 5=30m^2$

In $\Delta ACD$, let $s$ be the semiperimeter,

$S=\frac{6+15+13}{2}=17$m
Applying Heron's formula,

Area of $\Delta ACD$ = $\sqrt{S(S-a)(S-b)(S-c)}$ = $\sqrt{17(17-13)(17-15)(17-6)}$

= $\sqrt{17\left(4\right)\left(2\right)\left(11\right)}=2\sqrt{374}$

Hence, Area of quadrilateral $ABCD=30+2\sqrt{374}$

So, option C is correct.