question_answer

The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. The area of the triangle is

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

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

C) \[48\,\text{c}{{\text{m}}^{\text{2}}}\] 

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 Eq. (ii) by Eq. (i), Subtracting Eq. (i) and Eq. (ii), \[2x=24\] \[\Rightarrow \]   \[x=12\] \[\therefore \] Area of triangle \[=\frac{1}{2}\times 12\times 8=48\,\,sq\,\,cm\]