# How to find the largest square inside a triangle?

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• April 11th 2011, 10:12 PM
Medusa
How to find the largest square inside a triangle?
I am wondering how you would go about to find the largest square that would fit inside a triangle?

An example of what i am asking is:

What is the side length of the largest square that would fit inside a right angled triangle with the sides 5,12, and 13?

Any help would be appreciated

Medusa
• April 11th 2011, 10:26 PM
johnny
The largest rectangle that would fit inside a right triangle with the sides 5, 12, 13 is a square. Let the square have side x. By Pythagorean theorem, $(12\,-\,x)^2\,+\,x^2\,=\,(13\,-\,\sqrt{2x^2\,-\,10x\,+\,25})^2.$ Solving the equation gives x = 60/17.
• April 11th 2011, 10:33 PM
Medusa
Thank you for the help. I trying to use this method but i kept stuffing up the second part. You have made it a lot clearer now.

Thanks

Medusa
• April 24th 2011, 05:54 AM
Medusa
After doing further research into this question and many discussions with my math genius friends i was wondering if anyone could explain how the formula: bh/b+h works in this situation? b = base, h = height. I am unsure how this works and any help explaining it would be appreciated.
• April 24th 2011, 08:19 AM
Soroban
Hello, Medusa!

Quote:

Can anyone explain how the formula: bh/(b+h) works in this situation?
b = base, h = height.

Code:

```-  - Ao
:  :  | *
: h-x |   *
:  :  |     *
:  - Do-------oE
h  :  |   x   | *
:  :  |       |   *
:  x  |      x|     *
:  :  |       |       *
:  :  |       |          *
-  -  o-------o-------------o
      B - x - F - - b-x - - C
      : - - - -  b  - - - - :```

We have right triangle ABC with AB = h, BC = b.

We have square BDEF with sides x.

Since ∆ABC ~ ∆ADE, we have:

. . .h. . . .h - x
. . --- .= .----- . . → . . hx .= .b(h - x) . . → . . hx .= .bh - bx
. . .b. . . . . x

. . bx + hx .= .bh . . → . . (b + h)x .= .bh

. . . . . . . . . . . . . . .bh
Therefore: . x .= .-------
. . . . . . . . . . . . . .b + h

• April 24th 2011, 08:33 AM
Medusa
So the formula would work for a right angled triangle.. What about if the triangle was isoceles? Would the formula still work if the triangle isn't right angled is pretty much what i am asking.

Medusa
• April 24th 2011, 10:16 AM
earboth
1 Attachment(s)
Quote:

Originally Posted by Medusa
So the formula would work for a right angled triangle.. What about if the triangle was isoceles? Would the formula still work if the triangle isn't right angled is pretty much what i am asking.

Medusa

1. Draw a sketch. see attachment

2. Use similar triangles. You'll get the proportions:

http://latex.codecogs.com/png.latex?...dfrac{bh}{b+h}
• April 26th 2011, 03:29 AM
Medusa
Thanks for the help :) I sat there and worked it out using your advice and it makes sense after completely working it out by hand :) Thanks for help

Medusa
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