Question

The area of an isosceles triangle is 12 sq.m. If one of its equal sides is 5m, then find its base.

Answer 1

$\begin{array}{c}\text{Area of triangle}=12cm2\\ \text{Given,}a=b=5cm\\ s=\frac{a+b+c}{2}=5+\frac{c}{2}\\ A=\sqrt{s(s-a)(s-b)(s-c)}\\ \Rightarrow 12=\sqrt{(5+\frac{c}{2})\times \frac{c^2}{4}\times (5-\frac{c}{2})}\\ \Rightarrow 12=\sqrt{(5+\frac{c}{2})\times \frac{c^2}{4}\times (5-\frac{c}{2})}\end{array}$

$\begin{array}{c}\Rightarrow 144=(25-\frac{c^2}{4})\times \frac{c^2}{4}\\ \Rightarrow 144=(\frac{25c^2}{4}-\frac{c^4}{16})\\ \Rightarrow c^4-100c^2+2304=0\\ \Rightarrow c^4-36c^2-64c^2+2304=0\end{array}$

$\begin{array}{c}\Rightarrow (c^2-64)(c^2-36)=0\\ \Rightarrow c=8\text{or}6\\ \text{Thus, the other side of the triangle is}8cm\text{or}6cm\text{.}\end{array}$