Question

The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is

(a) $\sqrt{15}{\mathrm{cm}}^2$

(b) $\frac{\sqrt{15}}{2}{\mathrm{cm}}^2$

(c) $2\sqrt{15}{\mathrm{cm}}^2$

(d) $4\sqrt{15}{\mathrm{cm}}^2$

Answer 1

Given:
The base of an isosceles triangle is 2 cm.
The length of one of the equal sides 4 cm.

Using Heron's formula:
If a, b and c are three sides of a triangle, then
area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.


Here, $s=\frac{2+4+4}{2}=\frac{10}{2}=5$

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{5\left(5-2\right)\left(5-4\right)\left(5-4\right)} \\ =\sqrt{5\left(3\right)\left(1\right)\left(1\right)} \\ =\sqrt{15}{\mathrm{cm}}^2 \end{gathered}$$

Hence, the correct option is (a).