Author Topic: Geometry Question  (Read 14412 times)

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Offline Munchor

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Geometry Question
« on: May 05, 2011, 03:05:53 pm »
I have this image and I need to find out the height of the triangle.

What I know:
  • The triangle is rectangular so that angle above is 90º
  • The base of the triangle is 50 ( sqroot(40²+30²) ).

I need to find out height (please without really advanced stuff, if I were in USA, I'd be highschool freshman, so something about that level).



The triangle can also be divided in two rectangular triangles.

EDIT: Also, no trigonometry. I know how to use it, but can't use it this year.
« Last Edit: May 05, 2011, 03:26:03 pm by Scout »

Offline Spyro543

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Re: Geometry Question
« Reply #1 on: May 05, 2011, 03:30:25 pm »
Seems like you can split that triangle in half and use the Pythagorean Theorem.

Offline Stefan Bauwens

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Re: Geometry Question
« Reply #2 on: May 05, 2011, 03:31:03 pm »
Seems like you can split that triangle in half and use the Pythagorean Theorem.
No, that can't because he doesnt know the hight.


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Offline Munchor

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Re: Geometry Question
« Reply #3 on: May 05, 2011, 03:31:39 pm »
Seems like you can split that triangle in half and use the Pythagorean Theorem.

Only seems like cos I really can't. I only have 1 side of each triangle (both splitted).

Offline Spyro543

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Re: Geometry Question
« Reply #4 on: May 05, 2011, 03:35:52 pm »
Ok if you split the triangle the hypotenuse is 30 and the bottom is 25. You can use the Pythagorean Theorem to find missing legs, too.

Offline Builderboy

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Re: Geometry Question
« Reply #5 on: May 05, 2011, 03:44:10 pm »
Why would the bottom be 25?  The triangle's base is not going to be split evenly in half.  Before I solve this Scout, you are familiar with what Sin/Cos and the like do?

Offline Munchor

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Re: Geometry Question
« Reply #6 on: May 05, 2011, 03:49:31 pm »
Code: [Select]
30²=x²+y²
40²=(50-x)²+y²

Got it, y is height, thanks much everyone though!

Offline Horrowind

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Re: Geometry Question
« Reply #7 on: May 05, 2011, 03:49:40 pm »
well, were is no need for advanced sin/cos-stuff... one can simply use the area of the triangle:
A = (50 * h)/2 = 30*40/2
which leeds to h = 24, if i'm not mistaken

Offline Munchor

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Re: Geometry Question
« Reply #8 on: May 05, 2011, 03:52:14 pm »
well, were is no need for advanced sin/cos-stuff... one can simply use the area of the triangle:
A = (50 * h)/2 = 30*40/2
which leeds to h = 24, if i'm not mistaken

I have no idea of what you did Horrowind :S

Solving my system of equations:

Code: [Select]
30²=x²+y²
40²=(50-x)²+y²

I get x=18 and y=24, which is exactly the answer (24).

Offline Builderboy

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Re: Geometry Question
« Reply #9 on: May 05, 2011, 03:59:44 pm »
well, were is no need for advanced sin/cos-stuff... one can simply use the area of the triangle:
A = (50 * h)/2 = 30*40/2
which leeds to h = 24, if i'm not mistaken

That is an elegant solution!  Here I am in my corner stuck on using Sin and Cos :P

Offline Munchor

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Re: Geometry Question
« Reply #10 on: May 05, 2011, 04:10:21 pm »
@Builderbot: I tried trigonometry but I couldn't do it because I don't know how to find out the two lower angles. Did you make it? I'd like to know a sin/cos solution too.

Offline Jim Bauwens

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Re: Geometry Question
« Reply #11 on: May 05, 2011, 04:12:33 pm »
well, were is no need for advanced sin/cos-stuff... one can simply use the area of the triangle:
A = (50 * h)/2 = 30*40/2
which leeds to h = 24, if i'm not mistaken
Mind explaining? I'm curious to know why the area is the same as the height :D

Edit: Nevermind, got it!
« Last Edit: May 05, 2011, 04:14:06 pm by jimbauwens »

Offline Munchor

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Re: Geometry Question
« Reply #12 on: May 05, 2011, 04:13:56 pm »
well, were is no need for advanced sin/cos-stuff... one can simply use the area of the triangle:
A = (50 * h)/2 = 30*40/2
which leeds to h = 24, if i'm not mistaken
Mind explaining? I'm curious to know why the area is the same as the height :D

It isn't, the area can be calculated by:

Code: [Select]
(b*h)/2    ;b=50 and h=24
(50*24)/2
A = 600

Offline Jim Bauwens

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Re: Geometry Question
« Reply #13 on: May 05, 2011, 04:14:43 pm »
Yes I just noticed, I read it wrong. Nice solution!

Offline Builderboy

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Re: Geometry Question
« Reply #14 on: May 05, 2011, 04:18:16 pm »
So i made a new picture to name the angles, there is angle A,B,C and D.  We first need to find angle B.  Sin = opposite over the hypotenuse.  In triangle ABC, the hypotenuse is 50, and the opposite of angle B is 40.  So Sin(B)=40/50.  

Now we also have triangle ABD, where 30 is the hypotenuse, and side Y is the opposite to angle B.  So in this case, Sin(B) = Y/30.  

Now we substitute both equations for Sin(B) and get 40/50 = Y/30.  Solving yields 24.