Assume the area of the triangle be A.
A = \(\frac{1}{2}\) × b × c × sin A (b, c are both the sides which have included angle A)
both the sides b, c are doubled and the side A remains same
A' (new area) = \(\frac{1}{2}\) × 2b × 2c × sinA
A' = 2bcsinA
= 4A
So the New Area becomes 4 times of previous area.
The correct option is (C)