Question

# An isosceles triangle has perimeter $30cm$ and each of the equal sides is $12cm$. Find the area of the triangle.

Hint: Determine the sides of the triangle first. And then use Heron’s formula for the area of a triangle.

According to the question, the perimeter of the triangle is $30cm$. Then the semi-perimeter will be:
$\Rightarrow s = \dfrac{{30}}{2} = 15cm$.
Length of two equal sides of the triangle is $12cm$ (given in the question).
Let the length of the third side is $x$. Then we can use perimeter to find the value of $x$. We’ll get:
$\Rightarrow 12 + 12 + x = 30, \\ \Rightarrow x = 30 - 24, \\ \Rightarrow x = 6 \\$
So, we have $12cm, 12cm$ and $6cm$ as the length of three sides of a triangle whose semi-perimeter is $s = 15cm$.
Now, we can use Heron’s Formula to determine its area. We have:
$\Rightarrow A = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$, where $a,b,c$ are length of sides of triangle in any order.
So, putting values from above, we’ll get:
$\Rightarrow A = \sqrt {15\left( {15 - 12} \right)\left( {15 - 12} \right)\left( {15 - 6} \right)} , \\ \Rightarrow A = \sqrt {15 \times 3 \times 3 \times 9} , \\ \Rightarrow A = 3 \times 3 \times \sqrt {15} , \\ \Rightarrow A = 9\sqrt {15} \\$
Thus the area of the triangle is $9\sqrt {15} c{m^2}$.
Note:
We can also use $Area = \dfrac{1}{2} \times b \times h$ to find out its area.
For an isosceles triangle, base is always the unequal side. So, in this case $b = 6cm$.
And we can easily find out the height of an isosceles using Pythagoras Theorem.
$\Rightarrow h = \sqrt {{{12}^2} - {{\left( {\dfrac{b}{2}} \right)}^2}} , \\ \Rightarrow h = \sqrt {144 - 9} , \\ \Rightarrow h = \sqrt {135} , \\ \Rightarrow h = 3\sqrt {15} \\$
Now, we can use $\dfrac{1}{2} \times b \times h$. We’ll get the same result.