What is the sum of the lengths, in centimeters, of the two legs of a 30-60-90 right triangle, if the length of the hypotenuse is \(2\sqrt{6}\) centimeters?

Guest May 14, 2018

#1**+2 **

Hi Guest!

There is a rule about 30-60-90 triangles.

As shown by the image, each of the sides of the triangle has a relationship.

So if the length of the hypotenuse is \(2\sqrt6\), we can write an equation to solve for the legs.

We call the legs x and √3x

Using the pythagorean theorem, we get:

\(x^2+(\sqrt3x)^2=(2\sqrt6)^2\\ x^2+3x^2=2\cdot6\\ \)

You can solve for x, then plug in x to find the value of the two legs.

You then add the values to find your answer!

I hope this helped,

Gavin

GYanggg
May 14, 2018

#3**+1 **

wait. in your answer, you used \(2x^2\) as a leg. shouldn't it be \(x^2+(2\sqrt{6})^2=(2x)^2\)?

edit: nvmd read my own question wrong :P

Guest May 14, 2018

edited by
Guest
May 15, 2018

#4**+2 **

Gavin, I believe you made another error here. For your application of the Pythagorean's theorem, you did the following:

\(x^2+(2x)^2=(2\sqrt6)^2\\ x^2+4x^2=4\cdot6\\ 5x^2=24\\\)

However, I do not understand why *2x *is the representation of a leg in this equation, but you show in your diagram that it is clearly the representation of the hypotenuse's length. If you wanted to use this method to get the answer, then *x *and* x√3 *should be the lengths. The work would look something like this:

\(x^2+\left(\textcolor{red}{x\sqrt{3}}\right)^2=\left(2\sqrt{6}\right)^2\\ x^2+\textcolor{red}{3}x^2=2^2*\sqrt{6}^2\\ \textcolor{red}{4}x^2=4*6\\ \textcolor{red}{4}x^2=24\)

That small error from your work makes the discrepancies highlighted in red. You can, then, solve for x and plug them in for the legs, but you can also use the method below.

The ratio of the sidelengths of the 30-60-90 special right triangle is \(1:\sqrt{3}:2\), so Pythagorean's theorem is not necessary. Of course, you can still use it if you wish. However, it is also possible to generate this answer with your knowledge of proportions.

\(\frac{2\sqrt{6}}{2}=\frac{\text{leg}_1}{1}\\ \text{leg}_1=\sqrt{6}\)

\(\frac{2\sqrt{6}}{2}=\frac{\text{leg}_2}{\sqrt{3}}\\ \text{leg}_2=\sqrt{6}*\sqrt{3}=\sqrt{18}=3\sqrt{2} \)

Now, add them together.

\(\left(3\sqrt{2}+\sqrt{6}\right)\text{cm}\approx 6.6921\text{cm}\) is, then, the sum of the length of the sides.

TheXSquaredFactor
May 15, 2018