# How do you find the perimeter of a triangle if the altitude of an equilateral triangle is 32cm?

Apr 9, 2016

$64 \sqrt{3}$ cm..

#### Explanation:

Ia a is a side length, altitude = $a \left(\cos {30}^{o}\right) = a \frac{\sqrt{3}}{2}$ = 32 cm.
$a = \frac{64}{\sqrt{3}}$cm.
So, the perimeter = 3a = $64 \sqrt{3}$cm.

Apr 9, 2016

$\textcolor{red}{\text{Solving without using Trig}}$

$\text{perimeter "=64sqrt(3)color(white)(.) cm" }$ (exact value)

$\text{perimeter } \approx 110.85 \textcolor{w h i t e}{.} c m$ to 2 decimal places

#### Explanation:

This is solvable by comparing to a standardised triangle and then using ratios.

The perimeter of the standardised equilateral triangle is $3 \times 2 = 6 \text{ units}$

The 'altitude' of the standardised triangle is $\sqrt{3}$

The given 'altitude' in the question is 32 cm

Let the length of 1 side be $L$
Let the perimeter be $P$

Then by ratio of $\left(\text{triangle in question")/("standardised triangle}\right)$

$\implies \frac{L}{2} = \frac{32}{\sqrt{3}}$

$\implies L = \frac{2 \times 32}{\sqrt{3}} = \frac{64}{\sqrt{3}}$

Then $P = 3 \times \frac{64}{\sqrt{3}}$

$P = \frac{192}{\sqrt{3}}$

Multiply by 1 but in the form of $1 = \frac{\sqrt{3}}{\sqrt{3}}$

$P = \frac{192 \sqrt{3}}{3} = 64 \sqrt{3}$

As an approximate figure $P \approx 110.85$ to 2 decimal places