A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

Solution:

Given that, A triangle and a parallelogram have the same base and the same area

We know that, Heron's formula

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

where, s is the semi-perimeter = half of the perimeter

and  a, b and c are the sides of the triangle 

Let ABCD is a parallelogram and ABE is a triangle having a common base with parallelogram ABCD.

For ∆ABE, a = 30 cm, b = 26 cm, c = 28 cm

Semi Perimeter = s  = Perimeter/2

s = (a $+$ b $+$ c)/2

= (30 $+$ 26 $+$ 28)/2

= 84/2

= 42 cm

Thus, we get

Area of a ΔABE = $\sqrt{s (s - a) (s - b) (s -c)}$

= $\sqrt{42(42 - 30)(42 - 28)(42 - 26)}$

= $\sqrt{42 \times 12 \times 14 \times 16}$

= 336 cm²

Hence we get,

Area of parallelogram ABCD = Area of ΔABE 

Base × Height = 336 cm²

28  × Height = 336 cm²

On rearranging, we get

Height = 336/28 cm = 12 cm

Thus, height of the parallelogram is 12 cm.