# Thread: Square Inscirbed in a Triangle

1. ## Square Inscirbed in a Triangle

There is a square inscribed in a 3 - 4 -5 right angled triangle. That is, the opposite is 3cm, the adjacent is 4cm and the hypotenuse is 5cm.

There is a diagram, here is an attempt to describe it: The square is situated such that two of it's sides are parts of adjacent and opposite, and the other two sides go from the opposite to meet the hypotenuse, and then down from that point to meet the adjacent.

I'll attempt to draw it...

Code:
|  -
|      -
|          -
|              -
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|               |    -
|               |         -
|               |             -
|               |                  -
|               |                       -
|_______________|___________________
What fraction of the triangle does the square occupy?

My uncle says that all of the triangles are scalar multiples of each other, so they all have the same tangents (same opp/adj), then he worked it out from there, but I can't see how he got this?

2. Originally Posted by Bruce
There is a square inscribed in a 3 - 4 -5 right angled triangle. That is, the opposite is 3cm, the adjacent is 4cm and the hypotenuse is 5cm.

There is a diagram, here is an attempt to describe it: The square is situated such that two of it's sides are parts of adjacent and opposite, and the other two sides go from the opposite to meet the hypotenuse, and then down from that point to meet the adjacent.

I'll attempt to draw it...

Code:
|  -
|      -
|          -
|              -
||||||||||||||     -
|               |    -
|               |         -
|               |             -
|               |                  -
|               |                       -
|_______________|___________________
What fraction of the triangle does the square occupy?

My uncle says that all of the triangles are scalar multiples of each other, so they all have the same tangents (same opp/adj), then he worked it out from there, but I can't see how he got this?
The question boils down to finding the sidelength of the square. You can do this using similar triangles. Let the sidelength be x. Then:

$\frac{3-x}{x} = \frac{x}{4 - x} \Rightarrow x = \, ....$

3. But why does $\frac {3 - x}{x} = \frac {x}{4 -x}$

I don't understand why they are equal, or what those fractions signify. I realise that the opp and adjacent are 3 - x and x, and x and 4 - x, but what does that signify, and why are they equal?

4. Originally Posted by Bruce
But why does $\frac {3 - x}{x} = \frac {x}{4 -x}$

I don't understand why they are equal, or what those fractions signify. I realise that the opp and adjacent are 3 - x and x, and x and 4 - x, but what does that signify, and why are they equal?
The top triangle has sides of length 3 - x and x. The bottom triangle has sides of length x and 4 - x. The triangles are similar because the angles of each are congruent. Hence the ratio of sides is the same: Similar Triangles

5. Ah, I get it now. Thanks, thanked.