Question

The area of an isosceles triangle is $60{cm}^2$ and the length of each one of its equal sides is $13cm$. Find its base.

Answer 1

Let $ABC$ be the given isosceles triangle in which $AB=AC=13cm$. Draw $AD$ perpendicular from $A$ on $BC$

Let $BC=2x$ cm. Then $BD=DC=x$ cm

In $△ABD$, we have

${AB}^2+{AD}^2+{BD}^2$(By pythagoras theorem)

$\Rightarrow {13}^2={AD}^2+x^2\Rightarrow AD=\sqrt{{13}^2-x^2}=\sqrt{169-x^2}$

Now, Area $=60{cm}^2$

$\Rightarrow \frac{1}{2}(BC\times AD)=60\Rightarrow \frac{1}{2}\{2x\times \sqrt{169-x^2}\}=60$

$\Rightarrow x\sqrt{169-x^2}=60\Rightarrow x^2(169-x^2)=3600$

$\Rightarrow (x^2-144)(x^2-25)=0\Rightarrow x^2=144or{x}^2=25$

$\Rightarrow x=12orx=5$

Hence, Base $2x=24cm$ or $10cm$