Question

The area of an equilateral triangle whose height is 3 m is

• A. $2\sqrt{3}{m}^2$
• B. $3\sqrt{3}{m}^2$
• C. $3\sqrt{2}{m}^2$
• D. None of the above

Answer 1

The correct option is B

$3\sqrt{3}{m}^2$

Let $△ABC$ be an equilateral with side 'a' m and AD is perpendicular to BC as shown in the figure below.

Using Pythagoras Theorem for $△ADB$, we have,
$A{B}^2=A{D}^2+B{D}^2$
$i.e.a^2=3^2+(\frac{a}{2}{)}^2$
$\Rightarrow a^2-\frac{a^2}{4}=9$
$\Rightarrow \frac{3}{4}{a}^2=9$
$\Rightarrow a^2=12\Rightarrow a=\sqrt{12}i.e.a=2\sqrt{3}m$

We know that Area of a trianlge = $\frac{1}{2}\times base\times height$

$\therefore Areaof△ABC=\frac{1}{2}\times 2\sqrt{3}\times 3$
$=3\sqrt{3}{m}^2$