Question

The base of an isosceles triangle measures 80 cm and its area is
360 cm². Find the perimeter of the triangle. ​

Answer 1

Given :

• Base of an isosceles triangle is 80 cm
• Area of the triangle is 360 cm²

To find :

• the perimeter of the triangle

Solution :

As we know,

$\purple \star\red{\boxed{ \sf{ \green{area \: of \: a \: triangle = \frac{1}{2} \times base \times height}}}}$

$\sf \to360 = \frac{1}{2} \times 80 \times height$

$\sf \to \: height = \frac{ {360} \times 2}{{80}}$

$\sf \to \: height = 9 \: cm$

Since the height of the triangle, divides it into two right angled triangles,

$\purple \star \color{gold} {\boxed{ \sf{ \blue{side(a) \: of \: a \: triangle \implies {a}^{2} = {h}^{2} + ( \frac{b}{2} ){}^{2} }}}}$

$\sf \to {a}^{2} = {9}^{2} + ( \cancel \frac{80}{2} {)}^{2}$

$\sf \to {a}^{2} = 81 + 4 {0}^{2}$

$\sf \to {a}^{2} = 81 + 1600$

$\sf \to {a}^{2} = 1681$

$\sf \to \: a = \sqrt{1681}$

$\sf \to \: a = 41$

Hence,

Side of the triangle is 41 cm

$\red{ \star} \color{cyan}\boxed{ \sf{ \color{magenta}{perimeter \: of \: the \: triangle = sum \: of \: all \: sides}}}$

$\sf \to \: a + a + b$

$\sf \to41 + 41 + 80$

$\sf \to162 \: cm$

Hence, the perimeter of the triangle is 162 cm.