Finding the height of an equilateral triangle

mathisfun_user · 2016-05-16 18:19:58

Each and every side of an equilateral triangle (60-60-60 triangle) is equal to (10/ square root 2). When finding the height of the equilateral triangle, I assume that it's suffice we divide the equilateral triangle into two halves. Both halves are congruent to each other and we find that they are both 30-60-90 triangles. What I am trying to figure out is how exactly would I be able to find the length opposite of the 60 degree angle and the length opposite of the 30 degree angle given that the side opposite of the 90 degree angle is (10/ square root 2).

All help is appreciated!

bobbym · 2016-05-16 18:24:39

Why not try SOHCAHTOA or Pythagoras for the length opposite the 60 degree angle which is the height of the triangles?

The length opposite the 30 degree angle is defined in your problem.

The side opposite the 90 degree angle is called the hypotenuse

Why did you delete your post? Please do not do that.

mathisfun_user · 2016-05-16 18:53:51

If the original sides of an equilateral triangle is (10/ square root 2) and I divide the equilateral triangle into halves, would I divide
(10/ square root 2) / 2 to obtain the new length for one of the sides in the 30-60-90 triangle (the side opposite of the 30 degree angle)? Then continue on with the Pythagorean theorem?

bobbym · 2016-05-16 18:55:58

That is how you would get the base of one of your two congruent triangles, You already know what the hypotenuse is so Pythagoras can be used.

mathisfun_user · 2016-05-16 19:03:39

I know that the answers to this question are: a height of (5 square root 6 / 2). However, I don't quite understand how this answer was obtained. There isn't much of an explanation in the book I have.

Also, I am in a section where everything is based solely on geometry, so I find that it wouldn't be quite necessary to use socahtoah as it would be more suitable for a trig problem. I know that the Pythagorean theorem is one of the options I can choose in solving this problem, however, I only have difficulties with solving it because of having a square root involved in the problem.

Last edited by mathisfun_user (2016-05-16 19:28:38)

bobbym · 2016-05-16 19:09:07

What do you not understand?

mathisfun_user · 2016-05-16 19:10:32

How the height and (new) base were obtained step by step.

(Primarily the height)

Last edited by mathisfun_user (2016-05-16 21:55:48)

bobbym · 2016-05-16 19:14:03

Do you know what the Pythagorean theorem says:

a^2 + b^2 = c^2 where a can be the height, b can be the base and c is the hypotenuse.

Are you clear up to here?

mathisfun_user · 2016-05-16 19:22:52

How did you get the 5 / square root 2? Did you just divide 10 / square root 2 by 2?

Last edited by mathisfun_user (2016-05-16 19:27:24)

bobbym · 2016-05-16 19:28:06

I got it from right here in your OP.

mathisfun_user wrote:

I assume that it's suffice we divide the equilateral triangle into two halves.

I did not question it back then to prevent more confusion.

When you divide

by 2 you get

It just so happens that is correct but it is a bit loose to say that. You would need a proof that when you drop a perpendicular line through the apex of your equilateral triangle it divides the base in half.

mathisfun_user · 2016-05-16 19:39:33

Well that explains quite a bit! Would it be correct to subtract (5 / square root 2)^2 from both sides of your previous equation, and come up with h^2 = (10 / square root 2)^2 - (5 / square root 2)^2 to obtain the height of the triangle? And if so, how would I continue with solving it? I have difficulties at this point.

Also, if the sides of an equilateral triangle were 20 / square root 2 and I halved it, would the new base of the 30-60-90 triangles be 10 / square root 2?
(Just checking to see if I have a full understanding of this)

Last edited by mathisfun_user (2016-05-16 19:46:44)

bobbym · 2016-05-16 19:46:28

You would need a proof that when you drop a perpendicular line through the apex of your equilateral triangle it divides the base in half.

Let us assume that this is true for the moment...

You want to solve this:

Clean up the squares:

Do you understand up to here?

mathisfun_user · 2016-05-16 19:57:14

Yes! I follow through by canceling out the square roots of the two's and then multiplying the numerators by themselves. Thus, we obtain the new numbers of 25/2 and 100/2. 100/2 = 50. Subtract both sides by 25/2. h^2 = 50 - (25/2). h^2 = 75/2. h = square root (75 / 2) or square root 75 / square root 2. Am I correct so far?

(I should have left the 100/2 as it was)

Last edited by mathisfun_user (2016-05-16 19:58:58)

bobbym · 2016-05-16 20:03:47

Clean up the squares:

Multiply everything by 2:

Which can be simplified a bit more.

mathisfun_user · 2016-05-16 20:11:57

Awesome! So now that we have found h = square root (75/2). How would the process of simplifying work out? I am a bit confused when approaching this step.

I continued forth by simplifying the square root 75 into square root 25 and square root 3. The square root 25 became 5 and so I came up with an answer of (5 square root 3 / square root two). However, the final answer is supposed to be 5 square root 6 / 2.

Last edited by mathisfun_user (2016-05-16 20:34:30)

bobbym · 2016-05-16 20:14:00

One idea is:

Which can be further simplified...

mathisfun_user · 2016-05-16 20:23:19

So . . .

5 square root 3 times square root 2 / square root 2 times square root 2

5 square root 6 / square root 4

5 square root 6 / 2!

Thanks for helping me solve this lengthy problem bobbym!

Last edited by mathisfun_user (2016-05-16 20:25:39)

bobbym · 2016-05-16 20:28:20

Now we have to deal with the only assumption made here:

I assume that it's suffice we divide the equilateral triangle into two halves.

One of the properties of an equilateral triangle:

In an equilateral triangle, the median, angle bisector, and altitude are equal.

We are done.

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#1 2016-05-16 18:19:58

Finding the height of an equilateral triangle

#2 2016-05-16 18:24:39

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