question_answer

The area of a right angled isosceles triangle having hypotenuse \[16\sqrt{2}\,cm\] is                 [SSC (CGL) 2010]

A) \[144\,\,c{{m}^{2}}\]              

B) \[112\,\,c{{m}^{2}}\]

C) \[128\,\,c{{m}^{2}}\]

D) \[110\,\,c{{m}^{2}}\]

Correct Answer: C

Solution :

\[AB=BC,\]\[\angle ABC=90{}^\circ \]

Let \[AB=BC=x\]

Then, by Pythagoras theorem,

\[A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}\]

\[\Rightarrow \]   \[{{x}^{2}}+{{x}^{2}}={{(16\sqrt{2})}^{2}}\]

\[\Rightarrow \]   \[2{{x}^{2}}={{16}^{2}}\times 2\]

\[\Rightarrow \]   \[{{x}^{2}}={{16}^{2}}\]\[\Rightarrow \]\[x=16\]

\[\therefore \]Area of triangle \[=\frac{1}{2}\times AB\times BC\]

\[=\frac{1}{2}\times 16\times 16=8\times 16=128\,\,c{{m}^{2}}\]