Question

Find the area of the equilateral triangle given below:

Given: AD=10 cm.

• A. $100c{m}^2$
• B. $10\sqrt{3}c{m}^2$
• C. $25\sqrt{3}c{m}^2$
• D. $\frac{100}{3}\sqrt{3}c{m}^2$

Answer 1

The correct option is D

$\frac{100}{3}\sqrt{3}c{m}^2$

Let us assume AB=BC =AC =2a cm

Then BD =DC = a cm

In $\Delta ABD,$

$A{D}^2+B{D}^2=A{B}^2$

${10}^2+a^2=(2a{)}^2$ [Since AD = 10 cm]

$\Rightarrow 4a^2-a^2=100$

$\Rightarrow 3a^2=100$

$\Rightarrow a^2=\frac{100}{3}---\left(1\right)$

Area of equilateral triangle $=\frac{\sqrt{3}}{4}(side{)}^2$

$=\frac{\sqrt{3}}{4}(2a{)}^2$

$=\frac{\sqrt{3}}{4}.4a^2$

$=\sqrt{3}{a}^2$

$=\sqrt{3}.\frac{100}{3}$ [From (1)]

$=\frac{100}{3}\sqrt{3}c{m}^2$

Alternate Solution:

Area of equilateral triangle with height${}^{\prime }{h}^{\prime }=\frac{h^2}{\sqrt{3}}$

$\therefore$ Area $=\frac{{10}^2}{\sqrt{3}}=\frac{100}{\sqrt{3}}=\frac{100}{3}\sqrt{3}c{m}^2$