Question

The perimeter of a triangle is 240 dm. If two of its sides are 78 dm and 50 dm, find the length of perpendicular on side of length 50 dm from opposite vertex.

Answer 1

P = 240
S = 240/2 = 120

By Heron's Formula
Area = $\sqrt{s(s-a)(s-b)(s-c)}$
Area = $\sqrt{120*8*70*42}$
Area = 1680

Area of Triangle = $\frac{1}{2}$ × B × H
2 × 1680 = 50 × H
H = $\frac{2*1680}{50}$
H = 67.2dm

Answer 2

Solution :-

Perimeter of the triangle = 240 dm

Two sides (given) = 78 dm and 50 dm

Third side of the triangle = 240 - (78 + 50)

= 240 dm - 128 dm

= 112 dm

s = (a + b + c)/2

= (78 + 50 + 112)/2

= 240/2

s = 120 dm

Area of the triangle = √s(s - a)(s - b)(s - c)

⇒ √120 (120 - 50)(120 - 78)(120 - 112)

⇒ √120*70*42*8

⇒ √2822400

Area of the triangle = 1680 sq dm

Now, 

Area of the triangle = 1/2*base*height

Height = 2*area/base

 ⇒ (2*1680)/50 

⇒ 336/5

height = 67.2 dm

Answer.