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 \[=\frac{1}{2}\times \text{base}\times \text{height}\] \[=\frac{1}{2}\times 16\times 16=128\,\,\text{c}{{\text{m}}^{2}}\]You need to login to perform this action.
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