Question

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

• A. 24 cm, or 20 cm
• B. 14 cm, or 10 cm
• C. 24 cm, or 10 cm
• D. 44 cm, or 10 cm

Answer 1

The correct option is B. $24cm$ or $10cm$

Let $ABC$ be an isosceles triangle such that $AB=AC=13cm$

Let $BC=2x$

Now, drop $AD\perp BC$

$\therefore BD=DC$ [Perpendicular of an isosceles triangle bisects the third side.]

$\Rightarrow BD=DC=\frac{2x}{2}=x$

Consider, right $\Delta ABD$, by Pythagoras Theorem,

$\Rightarrow A{D}^2+B{D}^2=A{B}^2$

$\Rightarrow A{D}^2+x^2={13}^2$

$\therefore A{D}^2=169-x^2$...(i)

Also, area of $\Delta ABC=60c{m}^2$

$\Rightarrow \frac{1}{2}\times BC\times AD=60$

$\Rightarrow \frac{1}{2}\times 2x\times AD=60$

$\Rightarrow AD=\frac{60}{x}$...(ii)

From (i) and (ii), we get,

$\Rightarrow \frac{3600}{x^2}=169-x^2$

$\Rightarrow 3600=169x^2-x^4$

$\Rightarrow x^4-169x^2+3600=0$

$\Rightarrow x^4-144x^2-25x^2+3600=0$

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

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

$\Rightarrow x=12,5$

Therefore, $BC=24cm$ or $10cm$.