Question

Using Heron’s formula, find the area of an isosceles triangle, the measure of one of its equal sides being $a$ units and the third $2b$ units.
(A) $b\sqrt{a^2-b^2}$ sq.units.
(B) $a\sqrt{a^2+b^2}$ sq.units.
(C) $a\sqrt{a^2-b^2}$ sq.units.
(D) $b\sqrt{a^2+b^2}$ sq.units.

Answer 1

Hint: Heron’s formula to determine the area of a triangle is $A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$, where $a,b\text{ and }c$ are the sides of the triangle and $s$ is its semi-perimeter. Use this formula for the given isosceles triangle to determine the answer.

Complete step-by-step solution:
According to the question, an isosceles triangle is given such that the length of its equal sides is $a$ units and the length of the third side is $2b$ units. This is shown in the below figure:

We know that Heron’s formula to determine the area of a triangle is $A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$, where $a,b\text{ and }c$ are the sides of the triangle and $s$ is its semi-perimeter. The value of the semi-perimeter of the above triangle is:
$$\begin{gathered} \Rightarrow s={\displaystyle \frac{a+a+2b}{2}} \\ \Rightarrow s={\displaystyle \frac{2\left(a+b\right)}{2}} \\ \Rightarrow s=a+b \end{gathered}$$
Putting the values of semi-perimeter and the lengths of sides of the triangle in Heron’s formula, we’ll get:
$$\begin{gathered} \Rightarrow A=\sqrt{\left(a+b\right)\left(a+b-2b\right)\left(a+b-a\right)\left(a+b-a\right)} \\ \Rightarrow A=\sqrt{\left(a+b\right)\left(a-b\right)\cdot b\cdot b} \end{gathered}$$
We know that $\left(a+b\right)\left(a-b\right)=a^2-b^2$. Using this formula, we’ll get:
$$\begin{gathered} \Rightarrow A=\sqrt{b^2\left(a^2-b^2\right)} \\ \Rightarrow A=b\sqrt{\left(a^2-b^2\right)} \end{gathered}$$
Thus the area of the given isosceles triangle is $b\sqrt{\left(a^2-b^2\right)}$ sq.units.
(A) is the correct option.

Note: The general formula for finding the area of a triangle is given as:
$\Rightarrow A={\displaystyle \frac{1}{2}}\times \text{base}\times \text{height}$
If $a$ and $b$ are the lengths of two sides of a triangle and $\theta$ is the angle between them as shown in the below figure, the area of the triangle is:

$\Rightarrow A={\displaystyle \frac{1}{2}}\times a\times b\sin \theta$