Question

$\text{The base of an isosceles triangle is 16 cm and its area is 48}c{m}^2.\text{The perimeter of the triangle is}\left(a\right)41cm\left(b\right)36cm\left(c\right)48cm\left(d\right)324cm$

Answer 1

let h be the perpendicular height from vertex to the base of length 16cm in given isosceles triangle then the area of isosceles triangle

$\frac{1}{2}\left(16\right)\left(h\right)=4812\left(16\right)\left(h\right)=48$

h=6 cm

Now, using the Pythagorean theorem, each of the equal sides of the isosceles triangle is =$\sqrt{8^2+6^2}=10cm$

hence the perimeter of the isosceles triangle=sum of all the sides=10+10+16=36cm