﻿ Geometry Question
26 May, 2013, 08:14:42
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Munchor

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 « on: 05 May, 2011, 21:05:53 » 0

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: 05 May, 2011, 21:26:03 by Scout » Logged
Spyro543

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 « Reply #1 on: 05 May, 2011, 21:30:25 » 0

Seems like you can split that triangle in half and use the Pythagorean Theorem.
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Stefan Bauwens
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 « Reply #2 on: 05 May, 2011, 21:31:03 » 0

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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Munchor

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 « Reply #3 on: 05 May, 2011, 21:31:39 » 0

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).
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Spyro543

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 « Reply #4 on: 05 May, 2011, 21:35:52 » 0

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.
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Builderboy
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 « Reply #5 on: 05 May, 2011, 21:44:10 » 0

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?
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Munchor

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 « Reply #6 on: 05 May, 2011, 21:49:31 » 0

 123 30²=x²+y²40²=(50-x)²+y²

Got it, y is height, thanks much everyone though!
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Horrowind
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 « Reply #7 on: 05 May, 2011, 21:49:40 » +3

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
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Munchor

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 « Reply #8 on: 05 May, 2011, 21:52:14 » 0

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:

 12 30²=x²+y²40²=(50-x)²+y²

I get x=18 and y=24, which is exactly the answer (24).
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Builderboy
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 « Reply #9 on: 05 May, 2011, 21:59:44 » 0

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
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Munchor

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 « Reply #10 on: 05 May, 2011, 22:10:21 » 0

@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.
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Jim Bauwens
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 « Reply #11 on: 05 May, 2011, 22:12:33 » 0

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

Edit: Nevermind, got it!
 « Last Edit: 05 May, 2011, 22:14:06 by jimbauwens » Logged

Munchor

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 « Reply #12 on: 05 May, 2011, 22:13:56 » 0

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

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

 1234 (b*h)/2    ;b=50 and h=24(50*24)/2A = 600
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Jim Bauwens
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 « Reply #13 on: 05 May, 2011, 22:14:43 » 0

Yes I just noticed, I read it wrong. Nice solution!
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Builderboy
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 « Reply #14 on: 05 May, 2011, 22:18:16 » 0

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.
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