Question

The area of triangle (without using Heron's Formula) having side lengths,
$a=2\sqrt{13}b=6c=8$
is sq. units.

• A. 12$\sqrt{3}$
• B. 4$\sqrt{39}$
• C. 3$\sqrt{39}$

Answer 1

The correct option is A 12$\sqrt{3}$

Calculating angle A using cosine rule -

$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{6^2+8^2-(2\sqrt{13}{)}^2}{2\times 6\times 8}=\frac{100-52}{2\times 48}=\frac{1}{2}\Rightarrow A={60}^o$

Now the area can be calculated as
$Area=\frac{1}{2}\times bc\sin A=\frac{1}{2}\times 6\times 8\times \frac{\sqrt{3}}{2}=12\sqrt{3}sq.units.$