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The Height Of An 45-45-90 Isoceles Triangle Is Always Half It's Base

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mstrhz
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The Height Of An 45-45-90 Isoceles Triangle Is Always Half It's Base [#permalink] New post   Updated on: 18 Sep 2015, 17:18
Hello Everyone!

I tried looking for a post similar to this but couldn't find one. Regardless, I spent a long time proving this little geometric rule, one which I learnt a long time ago. So, as a refresher for everyone preparing for the GMAT, here is a simple, time saving method for calculating the height of a 45-45-90 isosceles triangle.

Height = 1/2 * base

As simple as it gets! I am sure a lot of you know this but I will show the proof below:


Figure 1: Isosceles Triangle taken from https://mathworld.wolfram.com/images/eps-gif/IsoscelesTriangle_800.gif

1. In the triangle above, the base has been bisected by the line h, essentially dividing the original triangle into two smaller ones. The base of the new triangles are both 1/2 a.
2. The angles marked are both 45 each, as is the case with all isosceles triangles. Therefore, [tan 45 = h / (1/2 * a))]
3. Since tan 45 degrees = 1, the relationship is now [1 = h / (1/2 * a)]
4. If you balance the equations, you get [1/2 *a =h]

a is your base and h is the height.

There are even easier ways to find the height, once you understand the concept. Take the two smaller triangles that you divided the main triangle into. You know that one side is 45 and the other 90. This means that those triangles are isosceles too and you can just eye the missing side without calculating anything!

There you have it, a small proof that hopefully goes a long way!

Edit: only works for 45-45-90

Originally posted by mstrhz on 17 Sep 2015, 18:54.
Last edited by mstrhz on 18 Sep 2015, 17:18, edited 1 time in total.
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Re: The Height Of An 45-45-90 Isoceles Triangle Is Always Half It's Base [#permalink] New post   17 Sep 2015, 21:01
mstrhz wrote:
Hello Everyone!

I tried looking for a post similar to this but couldn't find one. Regardless, I spent a long time proving this little geometric rule, one which I learnt a long time ago. So, as a refresher for everyone preparing for the GMAT, here is a simple, time saving method for calculating the height of an isosceles triangle.

Height = 1/2 * base

As simple as it gets! I am sure a lot of you know this but I will show the proof below:


Figure 1: Isosceles Triangle taken from https://mathworld.wolfram.com/images/eps-gif/IsoscelesTriangle_800.gif

1. In the triangle above, the base has been bisected by the line h, essentially dividing the original triangle into two smaller ones. The base of the new triangles are both 1/2 a.
2. The angles marked are both 45 each, as is the case with all isosceles triangles. Therefore, [tan 45 = h / (1/2 * a))]
3. Since tan 45 degrees = 1, the relationship is now [1 = h / (1/2 * a)]
4. If you balance the equations, you get [1/2 *a =h]

a is your base and h is the height.

There are even easier ways to find the height, once you understand the concept. Take the two smaller triangles that you divided the main triangle into. You know that one side is 45 and the other 90. This means that those triangles are isosceles too and you can just eye the missing side without calculating anything!

There you have it, a small proof that hopefully goes a long way!


your method only works for right-anged isosceles triangles, NOT isosceles triangles in general.
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OptimusPrepJanielle
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Re: The Height Of An 45-45-90 Isoceles Triangle Is Always Half It's Base [#permalink] New post   17 Sep 2015, 21:39
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mstrhz wrote:
Hello Everyone!

I tried looking for a post similar to this but couldn't find one. Regardless, I spent a long time proving this little geometric rule, one which I learnt a long time ago. So, as a refresher for everyone preparing for the GMAT, here is a simple, time saving method for calculating the height of an isosceles triangle.

Height = 1/2 * base

As simple as it gets! I am sure a lot of you know this but I will show the proof below:


Figure 1: Isosceles Triangle taken from https://mathworld.wolfram.com/images/eps-gif/IsoscelesTriangle_800.gif

1. In the triangle above, the base has been bisected by the line h, essentially dividing the original triangle into two smaller ones. The base of the new triangles are both 1/2 a.
2. The angles marked are both 45 each, as is the case with all isosceles triangles. Therefore, [tan 45 = h / (1/2 * a))]
3. Since tan 45 degrees = 1, the relationship is now [1 = h / (1/2 * a)]
4. If you balance the equations, you get [1/2 *a =h]

a is your base and h is the height.

There are even easier ways to find the height, once you understand the concept. Take the two smaller triangles that you divided the main triangle into. You know that one side is 45 and the other 90. This means that those triangles are isosceles too and you can just eye the missing side without calculating anything!

There you have it, a small proof that hopefully goes a long way!


I am afraid but this method works only for a "special" case for isosceles triangles. One in which the angles are 45, 45, 90
There can be innumerable number of isosceles triangles such as 30, 30, 120 and 20, 20, 140 etc.
In all such cases, this formula will not hold true.
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Re: The Height Of An 45-45-90 Isoceles Triangle Is Always Half It's Base [#permalink] New post   17 Sep 2015, 22:22
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mstrhz

This result is too narrow to be of any value for the GMAT. Instead, the height of an isosceles triangle can be related to its base and the two slant sides that are of equal lengths. If we consider the figure you posted, one can apply the Pythagorean theorem to the right triangle and obtain: \(h^2 + (\frac{a}{2})^2 = b^2\)

This will give you the height for any isosceles triangle. Such questions are tested on the GMAT. For example, what is the area of a triangle with sides 16, 16, and 17?

Cheers,
Dabral
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Re: The Height Of An 45-45-90 Isoceles Triangle Is Always Half It's Base [#permalink] New post   18 Sep 2015, 17:16
dabral wrote:
mstrhz

This result is too narrow to be of any value for the GMAT. Instead, the height of an isosceles triangle can be related to its base and the two slant sides that are of equal lengths. If we consider the figure you posted, one can apply the Pythagorean theorem to the right triangle and obtain: \(h^2 + (\frac{a}{2})^2 = b^2\)

This will give you the height for any isosceles triangle. Such questions are tested on the GMAT. For example, what is the area of a triangle with sides 16, 16, and 17?

Cheers,
Dabral


I don't really understand what you mean by 'it's too narrow'. Also, the whole reason for the explanation is to avoid using Pythagorean theorem on 45-45-90 triangles.
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dabral
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Re: The Height Of An 45-45-90 Isoceles Triangle Is Always Half It's Base [#permalink] New post   19 Sep 2015, 12:33
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mstrhz

By narrow I meant that the result you mentioned applies to only right isosceles triangles, for which one is expected to know the ratios of sides in a 45-45-90 triangle.

Dabral
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Re: The Height Of An 45-45-90 Isoceles Triangle Is Always Half It's Base [#permalink]
19 Sep 2015, 12:33

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