Question

The base of an isosceles triangle measures $24cm$ and its area is $192c{m}^2$. Find its perimeter.

• A. $64cm$
• B. $46cm$
• C. $84cm$
• D. $54cm$

Answer 1

Answer: (A)

$64cm$

Given $b=24cm,Area=192c{m}^2$

Let $△ABC$ is the given triangle with $BC=b=24cm,AD=h=16cm$

$Area=\frac{1}{2}\left(b\right)\left(h\right)=192$

$\Rightarrow h=2\times \frac{192}{24}=16$

we know that altitude of isosceles triangle bisects base

In $△ADC$

$A{C}^2=A{D}^2+C{D}^2=A{D}^2+{\left(\frac{BC}{2}\right)}^2$

${\left(AC\right)}^2={16}^2+{12}^2=400$

$AC=20=AB$

$Perimeter=AB+AC+BC=2\left(20\right)+24=64$