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

Solution (By Examveda Team)

$$\eqalign{ & \sqrt {{a^2} + {b^2}} ,\,\sqrt {{{\left( {2a} \right)}^2} + {b^2}} ,\,\sqrt {{a^2} + {{\left( {2b} \right)}^2}} \cr & {\text{Let }}a = b \cr & \sqrt 2 a,\,\sqrt 5 a,\,\sqrt 5 a \cr} $$

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