Question

The base of an isosceles triangle is 24 cm and its area is 192 cm². Find its perimeter.

Answer 1

Dear student,
Here is the solution of your asked query:
$$\begin{gathered} Letequalsidesofanisoscelestrianglebea. \\ Baseofanisoscelestriangle=24cm \\ So,perimeteroftriangle=24+a+a=24+2a \\ So,semiperimeteroftriangle;s=\frac{24+2a}{2}=12+a \\ Nowareaoftriangleisgivenby; \\ Area=\sqrt{s\left(s-a\right)\left(s-a\right)\left(s-24\right)}=192c{m}^2 \\ \Rightarrow \sqrt{\left(12+a\right)\left(12+a-a\right)\left(12+a-a\right)\left(12+a-24\right)}=192 \\ \Rightarrow \sqrt{\left(12+a\right)\times 12\times 12\left(a-12\right)}=192 \\ \Rightarrow 12\sqrt{a^2-144}=192 \\ \Rightarrow \sqrt{a^2-144}=16 \\ \Rightarrow a^2-144={\left(16\right)}^2 \\ \Rightarrow a^2-144=256 \\ \Rightarrow a^2=400 \\ \Rightarrow a=\pm \sqrt{400}=\pm 20 \\ Sincelengthcan\text{'}tbenegative,soneglectinga=-20 \\ So,a=20 \\ soperimeteroftriangle=24+2\times 20=64cm \end{gathered}$$
Regards