question_answer

The area of a right-angled isosceles triangle having hypotenuse \[16\sqrt{2}\,\,\text{cm}\] is

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

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

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

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

Correct Answer: B

Solution :

\[AB=BC=x\] \[AC=16\sqrt{2}\] \[\therefore \]      \[{{x}^{2}}+{{x}^{2}}={{(16\sqrt{2})}^{2}}\] \[\Rightarrow \]   \[2{{x}^{2}}=16\times 16\times 2\] \[\Rightarrow \]   \[{{x}^{2}}=16\times 16\] \[\Rightarrow \]   \[x=16\] Area of triangle \[=\frac{1}{2}\times \text{base}\times \text{height}\] \[=\frac{1}{2}\times 16\times 16=128\,\,\text{c}{{\text{m}}^{2}}\]