Question

The area of a parallelogram with a base $24\text{cm}$ is equal to the area of a triangle of sides $10\text{cm},24\text{cm}$ and $26\text{cm}$. Find the altitude of the parallelogram.

Answer 1

Let the height of the parallelogram be $h.$

Area of a parallelogram $=base\times height$

$=24\times h$

$=24h$

Semi-Perimeter of triangle with sides $10\text{cm}$, $24\text{cm}$ and $26\text{cm}$ is,

$\begin{array}{cc}s& =\frac{a+b+c}{2}\\ & =\frac{10+24+26}{2}\\ & =\frac{60}{2}\\ & =30\text{cm}\end{array}$

Now by using heron's formula,

Area of triangle $=\sqrt{30(30-10)(30-24)(30-26)}$

$=\sqrt{30\times 20\times 6\times 4}$

$=\sqrt{6\times 5\times 5\times 4\times 4\times 6}$

$=6\times 5\times 4$

$=120\text{sq. cm}$

But it is given that area of the parallelogram is equal to the area to triangle,

$24h=120$

$\Rightarrow h=5\text{cm}$

Hence, the height of the parallelogram is equal to $5\text{cm}.$