Three sides of a triangle are a^2 + b^2 ,\, ( 2a )^2 + b^2 and a^2 + ( 2b )^2 units. What is the area (in unit squares) of the triangle?

Answer: Option B

Solution

Answer: Option B

Solution:

$$\begin{array}{rl}& \sqrt{a^2+b^2},\sqrt{{\left(2a\right)}^2+b^2},\sqrt{a^2+{\left(2b\right)}^2}\\ & \text{Let }a=b\\ & \sqrt{2}a,\sqrt{5}a,\sqrt{5}a\end{array}$$

$$\begin{array}{rl}& \text{Height}\left(h\right)=\sqrt{l^2-b^2}\\ & =\sqrt{{\left(\sqrt{5}a\right)}^2-{\left(\frac{\sqrt{2}}{2}a\right)}^2}\\ & =\sqrt{\frac{9a^2}{2}}\\ & =\frac{3a}{\sqrt{2}}\\ & \text{Area}=\frac{1}{2}\times \frac{3a}{\sqrt{2}}\times \sqrt{2}a\\ & =\frac{3}{2}{a}^2\\ & =\frac{3}{2}ab\end{array}$$