Question

Altitude of an equilateral triangle is $h$ unit. Find its area.

• A. $\frac{1}{\sqrt{3}}{h}^2$ unit${}^2$
• B. $h^2$ unit${}^2$
• C. $\frac{2}{\sqrt{3}}{h}^2$ unit${}^2$

Answer 1

The correct option is A $\frac{1}{\sqrt{3}}{h}^2$ unit${}^2$

Let each side of the equilateral triangle is $a$ unit.
The altitude of a triangle will touch the opposite side of the vertex at the midpoint.


Hence, $QS=SR=\frac{a}{2}$

Now, consider the right-angled triangle $△PQS.$

PQ is the hypoteneuse of the triangle.
$\therefore a^2=h^2+(\frac{a}{2}{)}^2$
$\Rightarrow a^2-\frac{a^2}{4}=h^2$
$\Rightarrow \frac{3}{4}{a}^2=h^2$
$\Rightarrow a^2=\frac{4}{3}{h}^2$
(takeing square root both side)
$\Rightarrow a=\frac{2}{\sqrt{3}}h$

$\therefore$ Dimension of each side of the equilateral triangle is $\frac{2}{\sqrt{3}}h$ unit.

Now, area of the triangle $△PQS$ is half the area of the rectangle TPSQ with sides $h$ unit and $\frac{a}{2}$ unit.

Area of rectangle TPSQ $=h\times \frac{a}{2}$ unit${}^2$

$\therefore$ Area of the triangle $△PQS=\frac{1}{2}\times h\times \frac{a}{2}$ unit${}^2$

Also, the altitude divides the $△PQR$ into two halves.

$\therefore$ Area of the triangle $△PQR$ is double the area of $△PQS.$

$\Rightarrow$ Area of the $△PQR$ $=2\times \frac{1}{2}\times h\times \frac{a}{2}$ unit${}^2$
$=\frac{1}{2}\times a\times h$ unit${}^2$
(substituting the expression of $a$)
$=\frac{1}{2}\left(\frac{2}{\sqrt{3}}h\right)\left(h\right){\text{unit}}^2=\frac{1}{\sqrt{3}}\times h^2{\text{unit}}^2=\frac{1}{\sqrt{3}}{h}^2{\text{unit}}^2$