Question

Find the area of the quarilateral ABCD if AB = 5 m, BC = 5 m, CD = 6 m and AD = 6 m.

• A. $3\sqrt{3}+4{\text{m}}^2$
• B. $3(\sqrt{3}+4){\text{m}}^2$
• C. $3(3\sqrt{3}+4){\text{m}}^2$
• D. $3\sqrt{3}{\text{m}}^2$

Answer 1

The correct option is C $3(3\sqrt{3}+4){\text{m}}^2$

Diagonal AC divides the quadrilateral ABCD into two traingles: $△ABC$ and $△ACD$.
For $△ACD$, sides are 6 m, 6 m and 6 m.


Then, $△ACD$ is an equilateral triangle.
We know, area of an equilateral triangle with side 'a' is $\frac{\sqrt{3}}{4}\times a^2$.
So, $ar\left(△ACD\right)=\frac{\sqrt{3}}{4}\times 6^2=9\sqrt{3}{\text{m}}^2$

For $△ABC$, sides are 5 m, 5 m and 6 m.
The area of a triangle with sides a, b and c can be calculated using Heron's formula, given by:
Area $=\sqrt{s(s-a)(s-b)(s-c)}$,
where 's' is the semi-perimeter.
[$s=\frac{a+b+c}{2}$]
Then, $s=\frac{5+5+6}{2}=8\text{m}$
So,
$ar\left(△ABC\right)=\sqrt{8(8-5)(8-5)(8-6)}=\sqrt{8\times 3\times 3\times 2}=12{\text{m}}^2$

$\therefore$ Area of quarilateral ABCD$=9\sqrt{3}+12=3(3\sqrt{3}+4){\text{m}}^2$