Geometry Question

[Munchor]

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.

[Spyro543]

Seems like you can split that triangle in half and use the Pythagorean Theorem.

[Stefan Bauwens]

Seems like you can split that triangle in half and use the Pythagorean Theorem.

No, that can't because he doesnt know the hight.

[Munchor]

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).

[Spyro543]

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.

[Builderboy]

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?

[Munchor]

Code: Select All | Copy To Clipboard

1 2 3

30²=x²+y² 40²=(50-x)²+y²

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

[Horrowind]

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

[Munchor]

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 All | Copy To Clipboard

1 2	30²=x²+y² 40²=(50-x)²+y²

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

[Builderboy]

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

[Munchor]

@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.

[Jim Bauwens]

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!

[Munchor]

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:

Code: Select All | Copy To Clipboard

1 2 3 4

(b*h)/2    ;b=50 and h=24 (50*24)/2 A = 600

[Jim Bauwens]

Yes I just noticed, I read it wrong. Nice solution!

[Builderboy]

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.