Question

Find the area of an isosceles triangle having base $acm$ and one side $bcm$.

Answer 1

Step 1: Find the semi-perimeter of the triangle

Since it is known that the triangle is isosceles so its two sides will be equal.

Hence,

$$\begin{gathered} x=acm \\ y=bcm \\ z=bcm \end{gathered}$$

It is known that the semi-perimeter of the triangle can be given as: $s=\frac{x+y+z}{2}$

So the semi-perimeter of the triangle will be,

$\begin{array}{rcl}s& =& \frac{acm+bcm+bcm}{2}\\ & =& \frac{a+2b}{2}\\ & =& (\frac{a}{2}+b)cm\end{array}$

Step 2: Find the area of the triangle

To find the area of the triangle we will use Heron's formula.

Heron's formula can be given as: $A=\sqrt{s(s-x)(s-y)(s-z)}$

Where $s$ is the semi-perimeter of the triangle and $x,y,z$ are the sides of the triangle.

So, the area of the triangle will be,

$\begin{array}{rcl}A& =& \sqrt{(\frac{a}{2}+b)(\frac{a}{2}+b-a)(\frac{a}{2}+b-b)(\frac{a}{2}+b-b)}\\ & =& \sqrt{\frac{a}{2}+b(b-\frac{a}{2})\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)}\\ & =& \frac{a}{2}\sqrt{b^2-\frac{a^2}{4}}c{m}^2\end{array}$

Hence, the area of the triangle is $\frac{a}{2}\sqrt{b^2-\frac{a^2}{4}}c{m}^2$