SSC CGL 2022 PYQ — Question 29

The correct answer is option 4 i.e. $\frac{3}{2}$ ab

Here we don't know anything about the values of a and b so let's suppose,

a = b then the respective sides become,

$\sqrt{a^2 + b^2} = \sqrt{a^2 + a^2} = \sqrt2a$

$\sqrt{(2a)^2 + b^2} = \sqrt{4a^2 + a^2} = \sqrt5a$

$\sqrt{a^2 + (2b)^2} = \sqrt{a^2 + 4a^2} = \sqrt5a$

Now one can say that the triangle must be an isosceles triangle.

Area of an isosceles triangle = $\frac{b}{4} \times \sqrt{4a^2 - b^2}$

$\frac{\sqrt2 a}{4} \times \sqrt{4 \times (\sqrt5a)^2 - (\sqrt2a)^2}$

= 3/2 × a² = 3/2 ab.