Question

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

• A. $\sqrt{15}c{m}^2$
• B. $\sqrt{\frac{15}{2}c{m}^2}$
• C. $2\sqrt{15}c{m}^2$
• D. $4\sqrt{15}c{m}^2$

Answer 1

The correct option is A $\sqrt{15}c{m}^2$

Let ABC be an isosceles triangle in which AB = AC = 4cm and BC = 2cm.

In right angled $\Delta ADB$,
$A{B}^2=A{D}^2+B{D}^2$ [by Pythagoras theorem]
$\Rightarrow (4{)}^2=A{D}^2+1$
$\Rightarrow A{D}^2=16-1$
$\Rightarrow A{D}^2=15$
$\therefore AD=\sqrt{15}cm$
[taking positive square root because length is always positive]
$\therefore Areaof\Delta ABC=\frac{1}{2}\times BC\times AD[\because areaoftriangle=\frac{1}{2}(base\times height\left)\right]$
$=\frac{1}{2}\times 2\times \sqrt{15}=\sqrt{15}c{m}^2$
Alternate Method
We know that,
Area of an isosceles triangle $=\frac{a}{4}\sqrt{4b^2-a^2}$
Where, b is the length of equal sides and a is the length of the base.
Here, the length of side be b = 4cm and a = 2cm
$\therefore$ Area of an isosceles triangle $=\frac{2\sqrt{4}(4{)}^2-4}{4}=\frac{\sqrt{64-4}}{2}$
$=\frac{\sqrt{60}}{2}=\frac{2\sqrt{15}}{2}=\sqrt{15}c{m}^2$