Question

The base of an isosceles triangle is $16$ cm and its perimeter is $36$ cm. The area of the triangle is ........

• A. $72c{m}^2$
• B. $36c{m}^2$
• C. $24c{m}^2$
• D. $48c{m}^2$

Answer 1

Answer: (D)

$48c{m}^2$

In isosceles $△ABC$,

perimeter $=AB+BC+CA=36$

Let $AB=AC=x$
$x+x+16=36$

$x=\frac{1}{2}(36-16)cm=10cm$

In an isosceles triangle, median is the perpendicular bisector.

$\angle AEB={90}^o$
Now, $△ABE$ is a right angle triangle.

Then by pythagoras theorem, we have

$A{E}^2+B{E}^2=A{B}^2$

$h^2+8^2={10}^2$

$h^2=100-64$

$\Rightarrow h=6cm$

$ar\left[\Delta ABC\right]=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 16cm\times 6cm=48c{m}^2$