Question

find the area of a triangle whose two sides are 24 cm and 10 cm and the perimeter of the triangle is 62 cm . (answer should be step-by-step and the answer should come 21 sq.)

Answer 1

Answer:

Step-by-step explanation:

let a, b and c are sides of triangle

Given :

a = 24 cm

b = 10 cm

a + b + c = 62 cm (perimeter)

c = 62 - (24 + 10)

c = 28

s = perimeter / 2

 = 62 / 2

 = 31 cm

according to Heron's Formula area of triangle is

A = $\sqrt{s(s-a)(s-b)(s-c)}$

   = $\sqrt{31(31-24)(31-10)(31-28)\\}$

    = $\sqrt{31 * 7 * 21 * 3}$

    = $\sqrt{31 * 7 * 7 * 3 * 3}$

    = 21 $\sqrt{31}$ sq cm

Answer mentioned 21 Sq cm seems wrong