Question

the base of an isosceles triangle is 16 cm and its area is 48 sqcm then perimeter of triangle is

Answer 1

Pic have the solution

Answer 2

Step-by-step explanation:

Let ∆ABC be an isosceles triangle and Let AF⊥BC

Area(∆ABC) = $\frac{1}{2}$×BC×AF

∴ 48 = $\frac{1}{2}$×16×h

h = $\frac{48}{8}$

h = 8 cm

and AF = 6 cm

Let 'a' be the two equal sides of an isosceles triangle

∴ a = $\sqrt{BF²+AF²}$

= $\sqrt{(8)²+(6)²}$

= $\sqrt{100}$

= 10 cm

Hence, Perimeter = 10+10+16

= 36 cm