The sides of a triangle are 24 cm, 32 cm and 40 cm. What is the height of the perpendicular drawn on the longest side?

Given:

The sides of a triangle are 24 cm, 32 cm, and 40 cm.

Formula used:

Area of triangle(A) = \(\sqrt{s(s-a)(s-b)(s-c)}\)

s = \(\dfrac{a + b + c}{2}\) (where, a, b, c are the sides of the triangle and s is the semi-perimeter)

Calculations:

Let a = 40, b = 32, c = 24

s = \(\dfrac{40 + 32 + 24}{2}\)

= \(\dfrac{96}{2}\)

= 48

A = \(\sqrt{48(48-40)(48-32)(48-24)}\)

= \(\sqrt{48(8)(16)(24)}\)

= \(\sqrt{(147456)}\)

= 384 sq. cm.

Let the height corresponding to the longest side (40 cm) h

Area(384) = \(\dfrac{1}{2}\) × 40 × h

h = \(\dfrac{384 × 2}{40}\)

= 19.2 cm

∴ The answer is 19.2 cmShortcut Trick 

Since the given triangle is right angle triangle, the length of the height of the perpendicular drawn on the longest side is given by:

(P × B)/H

(24 × 32)/40 = 19.2 cm

Alternate Method 

(1/2) × 32 × 24 = (1/2) × m × 40

32 × 24 = m × 40

m × 40 = 32 × 24 

m  = (32 × 24)/40

m = 19.2 cm