Equilateral Triangles

May 2018
73
13
Idaho, USA
Hello,

So, I was puzzling over a math problem, when I noticed something interesting I wanted to share.

Imagine an equilateral triangle, the sides are all the same, as are the angles.

Let's say the side lengths are 1, and the angles are 60. The way people would normally find the height is to do the side squared minus the side divided by 2 squared, to equal the height squared.

In other words, The square root of (1^2 - 0.5^2)

This is where it got interesting:

The answer, or height, is 0.866025404

This got me thinking, and what if instead of doing The square root of (1^2 - 0.5^2) to find the height, what about taking the side and just multiplying it by 0.866025404?

I tried this out, this time with side lengths of 1.5. Both ways, with my method, and with the other method, I got the same answer for the height, which is 1.299038106.

It worked when I did The square root of (1.5^2 - 0.75^2), and when I did the side times 0.866025404.

In a nutshell, the side multiplied by 0.866025404. It's quicker, and just as accurate as the other method.

So, while trying to learn one thing, I learned something else!

HOWEVER, so far, it only works with triangles that have the same side length and the same angles. I am still experimenting with it, though.

That's what I learned today! I love geometry!

Jared
 
Last edited by a moderator:

greg1313

Forum Staff
Oct 2008
8,008
1,174
London, Ontario, Canada - The Forest City
\(\displaystyle \sin(60^\circ)=\frac{\sqrt{3}}{2}\)
 

skipjack

Forum Staff
Dec 2006
21,478
2,470
The number 0.866025404 was found by using the first method, so one is relying on the first method even when using the second method. The number 0.866025404 is an approximation to the square root of 0.75. Hence the second method gives an approximate answer, and one can't improve the accuracy of the approximation without using the first method again (but finding the square root to greater accuracy).

The second method can be generalized to deal with triangles that aren't equilateral. However, the value of the multiplier may be different. When the triangle isn't equilateral, one needs to specify which height is to be found. If the original equilateral triangle is ABC, with the height drawn from A to BC, this height remains the same if A and B are kept fixed, but BC is extended or shortened by changing the position of C without changing the angle B. In this case, the multiplier is still the same as for the equilateral triangle, provided that the multiplicand is the length of the side AB.
 
May 2018
73
13
Idaho, USA
@skipjack, thanks for helping me out. You are right, the answer would be an approximation. The calculator I have at home only goes to the amount of digits in the number I mentioned, 0.866025404. So, both ways, I got the same answer. I can see where you're coming from, though. If you can get the square root of 0.75 more accurately, the answer would be more accurate. Correct?

By the way, thanks for always being willing to help me out with learning. I appreciate it.

Jared
 

skipjack

Forum Staff
Dec 2006
21,478
2,470
Correct. You could use this online calculator to get the answer as accurately as you want (within reason),
or use \(\displaystyle \frac{\sqrt3}{2}\) (which is exact) if you don't need the answer in decimal notation.

If a triangle has sides of length a, b, and c, its corresponding heights are of length bc/d, ac/d and ab/d, where d is the diameter of the circle that passes through the vertices of the triangle.
 
May 2018
73
13
Idaho, USA
@skipjack, I found what you said BC/D and the circle diameter that passes through the vertices of the triangle interesting. I think I will try that out! Thanks!

Jared