Question

The altitude drawn to the base of a isosceles triangle is of length $8$ cm and the perimeter is $32$ cm. The area of the triangle is ___________.

• A. $32{\text{cm}}^2$
• B. $40{\text{cm}}^2$
• C. $48{\text{cm}}^2$
• D. $56{\text{cm}}^2$

Answer 1

The correct option is C $48{\text{cm}}^2$

Let $ABC$ be the isosceles triangle, the $AD$ be the altitude

Let $AB=AC=x$, then $BC=32-2x$ [because parameter $=2$ (side) $+$ Base]

Since in an issoceles triangle the altitude bisects the base, so

$BD=DC=16-x$

In a triangle $ADC$, we get

$A{C}^2=A{D}^2+C{D}^2$

$\Rightarrow x^2=8^2+(16-x{)}^2$

$\Rightarrow x=10$ cm

$BC=32-2x=32-20=12$ cm

Hence, required area $=\frac{1}{2}AD\times BC=\frac{1}{2}\times 8\times 12=48{\text{cm}}^2$