question_answer

The altitude drawn to the based of isosceles triangle is \[8\,\,cm\] and the perimeter is\[32\,\,cm\]. The area of the triangle is:

A) \[24c{{m}^{2}}\]                                

B) \[32c{{m}^{2}}\]

C) \[48c{{m}^{2}}\]          

D)         None of these

Correct Answer: C

Solution :

 Let base \[=x\] cm \[\therefore \]Side\[=\sqrt{64+\frac{{{x}^{2}}}{4}}\] \[\therefore \]\[x+2\sqrt{64+\frac{{{x}^{2}}}{4}}=32\] i.e.\[x+\sqrt{256+{{x}^{2}}}=32\]                              ? (i) Also\[(256+{{x}^{2}})-{{x}^{2}}=256\]                     ? (ii) Dividing (ii) by (i) Subtracting (i) and (ii)             \[2x=24\] \[\Rightarrow \]   \[x=12\] \[\therefore \]Area of triangle             \[=\frac{1}{2}\times 12\times 8=48\,\,sq.\,\,cm\]