# Pythagorean Theorem (Right Triangle) Calculator

## Solve right triangles using the Pythagorean theorem

The calculator will try to find all sides of the right-angled triangle (the legs and the hypotenuse) using the Pythagorean theorem. It will also find all angles, as well as perimeter and area. The solution steps will be shown.

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.

Solve the triangle, if $a = 6$, $b = 6 \sqrt{3}$, $C = 90^0$.

### Solution

According to the Pythagorean theorem: $c^{2} = a^{2} + b^{2}$.

In our case, $c^{2} = 6^{2} + \left(6 \sqrt{3}\right)^{2} = 144$.

Thus, $c = 12$.

According to the definition of the sine: $\sin{\left(A \right)} = \frac{a}{c}$.

Thus, $\sin{\left(A \right)} = \frac{1}{2}$.

There are two possible cases:

1. $A = 30^0$

The third angle is $B = 180^0 - \left(A + C\right)$.

In our case, $B = 180^0 - \left(30^0 + 90^0\right) = 60^0$.

The area is $S = \frac{1}{2} a b = \left(\frac{1}{2}\right)\cdot \left(6\right)\cdot \left(6 \sqrt{3}\right) = 18 \sqrt{3}$.

The perimeter is $P = a + b + c = 6 + 6 \sqrt{3} + 12 = 6 \left(\sqrt{3} + 3\right)$.

2. $A = 150^0$

The third angle is $B = 180^0 - \left(A + C\right)$.

In our case, $B = 180^0 - \left(150^0 + 90^0\right) = -60^0$.

This case is impossible, since the angle is nonpositive.

$a = 6$A

$b = 6 \sqrt{3}\approx 10.392304845413264$A

$c = 12$A

$A = 30^0$A

$B = 60^0$A

$C = 90^0$A

Area: $S = 18 \sqrt{3}\approx 31.176914536239791$A.

Perimeter: $P = 6 \left(\sqrt{3} + 3\right)\approx 28.392304845413264$A.