# A triangle and a parallelogram have the same base and the same area.

Question:

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of a parallelogram.

Solution:

The sides of the triangle DCE are

DC = 15 cm,

CE = 13 cm,

ED = 14 cm

Let the h be the height of parallelogram ABCD

Now, for the area of triangle DCE

Perimeter = DC + CE + ED

2s = 15 cm + 13 cm + 14 cm

s = 21 cm

By using Heron's Formula,

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

$=\sqrt{21 \times(7) \times(8) \times(6)}$

$=84 \mathrm{~cm}^{2}$

Also, area of triangle DCE $=$ Area of parallelogram $\mathrm{ABCD} \Rightarrow 84 \mathrm{~cm}^{2}$

$24 \times h=84 \mathrm{~cm}^{2}$

h = 6 cm.