Question

Find the area of a quadrilateral whose sides are $3cm,4cm,2cm$, and $5cm$. The angle between the first two sides is ${90}^o$. (Use Heron's formula)

• A. $14c{m}^2$
• B. $16c{m}^2$
• C. $(6+\sqrt{24})c{m}^2$
• D. $20c{m}^2$

Answer 1

Answer: (C)

$(6+\sqrt{24})c{m}^2$

In quadrilateral $ABCD$,

$AB=3cm,BC=4cm,CD=2cm,DA=5cm$ and $\angle ABC={90}^0$

Area of $ABCD$ = Area of $\Delta ABC+$ Area of $\Delta ACD$

In $\Delta ABC,\angle B={90}^0$

$\therefore$ Area of $\Delta ABC$ = $\frac{1}{2}\times AB\times BC=\frac{1}{2}\times 3\times 4c{m}^2=6c{m}^2$

In $\Delta ABC$, applying pythagorous theorem we get,

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

$\Rightarrow A{C}^2=3^2+4^2$

$\Rightarrow A{C}^2=9+16$

$\Rightarrow A{C}^2=25$

$\Rightarrow AC=5$

$\therefore AC=5cm$

In $\Delta ACD$, semi-perimeter $s=\frac{2+5+5}{2}cm=6$ $cm$

Applying Heron's formula,
Area of $\Delta ACD$ = $\sqrt{S(S-a)(S-b)(S-c)}$ = $\sqrt{6(6-5)(6-5)(6-2)}$
= $\sqrt{6\left(1\right)\left(1\right)\left(4\right)}=\sqrt{24}$

Hence, Area of $\square ABCD=(6+\sqrt{24})c{m}^2$

So, option $C$ is correct.