Question

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

Answer 1

Framing of equation: 1 Mark
Steps: 2 Marks
Answer: 1 Mark



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

Let $BC=2xcm$. Then, $BD=DC=xcm$

In $\Delta ABD$, we have

$A{B}^2=A{D}^2+B{D}^2$ [By Pythagoras Theorem]

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

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

Now, Area $=60c{m}^2$

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

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

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

$\Rightarrow x^2(169-x^2)=3600$

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

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

$\Rightarrow x^2=144or,x^2=25\Rightarrow x=12or,x=5$

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