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