Question

The area of the given polygon is ${\text{cm}}^2$.

• A. $6+2\sqrt{21}$
• B. $6-2\sqrt{21}$
• C. $6\sqrt{21}$

Answer 1

The correct option is A $6+2\sqrt{21}$

Quadrilateral ABCD can be divided into two triangles ABC and ADC.

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.
[$s=\frac{a+b+c}{2}$]

$\text{For}△ABC,\text{semi-perimeter}=\frac{3+4+5}{2}=6\text{cm}$
Then, $ar\left(△ABC\right)=\sqrt{6\times (6-3)\times (6-4)\times (6-5)}=6{\text{cm}}^2$

Similarly, for $△ADC$,
$\text{semi-perimeter}=\frac{5+5+4}{2}=7\text{cm}$
$ar\left(ADC\right)=\sqrt{7\times (7-5)\times (7-5)\times (7-4)}=2\sqrt{21}{\text{cm}}^2$
So, area of quadrilateral ABCD
= ar(ABC) + ar(ADC)
= $6+2\sqrt{21}{\text{cm}}^2$