How do you find the area of a 30, 60, 90, triangle?

Jun 16, 2015

1) ${L}^{2} / 8 \sqrt{3}$
2) ${b}^{2} / 2 \sqrt{3}$
3) ${h}^{2} \frac{\sqrt{3}}{6}$

Explanation:

You know a 30-60-90 triangle is half an equilateral triangle, so the base is half of the hypotenuse L

$b = \frac{L}{2}$

You can use now Pythagoras theorem and calculate the height:

$h = \sqrt{{b}^{2} + {L}^{2}} = \sqrt{{L}^{2} / 4 + {L}^{2}} = \sqrt{\frac{3}{4} {L}^{2}} = \frac{L}{2} \sqrt{3} = b \sqrt{3}$

So we can determine the area only knowing one of the edges:

1) If we know $L$:

$A = \frac{b h}{2} = \frac{1}{2} \left(\frac{L}{2}\right) \left(\frac{L}{2} \sqrt{3}\right) = {L}^{2} / 8 \sqrt{3}$

2) If we know $b$:

$A = \frac{b h}{2} = {b}^{2} / 2 \sqrt{3}$

3) If we know $h$:

$A = \frac{b h}{2} = \frac{1}{2} {h}^{2} / \sqrt{3} = \frac{1}{2} {h}^{2} \frac{\sqrt{3}}{3} = \frac{{h}^{2} \sqrt{3}}{6}$