Quantitative Aptitude Ques 760
Question: The area of a right angled isosceles triangle having hypotenuse $ 16\sqrt{2},cm $ is
Options:
A) $ 144cm^{2} $
B) $ 112cm^{2} $
C) $ 128cm^{2} $
D) $ 110cm^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
- $ AB=BC, $ $ \angle ABC=90{}^\circ $ Let $ AB=BC=x $ Then, by Pythagoras theorem, $ AB^{2}+BC^{2}=AC^{2} $
$ \Rightarrow $ $ x^{2}+x^{2}={{(16\sqrt{2})}^{2}} $
$ \Rightarrow $ $ 2x^{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=128cm^{2} $
BETA
