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

Given:

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

The sides of a triangle are $26\ cm, 28\ cm$ and $30\ cm$, and the parallelogram stands on the base $28\ cm$.

To do:

We have to find the height of the parallelogram.

Solution:

The sides of the triangle are $a=26\ cm, b=28\ cm$ and $c=30\ cm$.

The semi-perimeter of the triangle $s=\frac{a+b+c}{2}$

$=\frac{28+26+30}{2}$

$=42\ cm$
Therefore, by Heron's formula,
$A=\sqrt{s(s-a)(s-b)(s-c)}$

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

$=\sqrt{42(14)(16)(12)}$

$=\sqrt{112896}$

$=336\ cm^2$ 

The area of the parallelogram is equal to the area of the triangle.

Area of parallelogram $=$ Area of triangle
Base $\times$ corresponding height $=336\ cm^2$
$\Rightarrow 28 \times$ corresponding height $=336\ cm^2$

$\Rightarrow\ Height =\frac{336}{28}$

$=12\ cm$
Hence, the height of the parallelogram is $12\ cm$.