# Rectangle in an isosceles triangle

• Oct 30th 2010, 01:17 AM
Educated
Rectangle in an isosceles triangle
There is a rectangle inside an isosceles triangle. One side of the rectangle is on the base of the isosceles triangle. The 2 corners on the other side of the rectangle are touching the equal lengths of the 2 isosceles triangle.

Image:
http://www.mathhelpforum.com/math-he...e-triangle.png

Attachment 19526

"Write down, in terms of x the height, h, of the rectangle."

Here's what I came up with:

One of the longer lengths of the triangle is $\displaystyle a + b$ which equals $\displaystyle \sqrt{10^2 + 2.5^2} = \sqrt{106.25}$

Length a + length b = sqrt(106.25)

The equation becomes:

$\displaystyle \underbrace{\sqrt{ \bigg(\frac{1}{2}x \bigg)^2 + (10-h)^2}}_{\text{length a}} + \underbrace{\sqrt{h^2 + \bigg(\frac{5-x}{2} \bigg)^2}}_{\text{length b}} = \sqrt{106.25}$

That doesn't seem correct and seems way too complex. I cannot simplify it down to just h=...

Can anyone help me?
• Oct 30th 2010, 01:37 AM
Wilmer
If the question is: what is h in terms of x,
then there is no need to calculate hypotenuses; use similar triangles:

10 / 2.5 = h / [(5 - x) / 2]
4 = 2h / (5 - x)
Wrap it up...
• Oct 30th 2010, 01:37 AM
Sudharaka
Quote:

Originally Posted by Educated
There is a rectangle inside an isosceles triangle. One side of the rectangle is on the base of the isosceles triangle. The 2 corners on the other side of the rectangle are touching the equal lengths of the isosceles triangle.

Image:
http://www.mathhelpforum.com/math-he...e-triangle.png

Attachment 19524

"Write down, in terms of x the height, h, of the rectangle."

Here's what I came up with:

The longer length of the triangle is $\displaystyle \sqrt{10^2 + 2.5^2} = \sqrt{106.25}$

The hypotenuse of the smaller bottom left/right right angle triangle + the hypotenuse of the smaller top left/right right angle triangle = sqrt(106.25)
The equation becomes:

$\displaystyle \sqrt{h^2 + (\frac{5-x}{2})^2} + \sqrt{(\frac{1}{2}x)^2 + (10-h)^2} = \sqrt{106.25}$

That doesn't seem correct and seems way too complex. I cannot simplify it down to just h=...

Can anyone help me?

Dear Educated,

You can consider two similar traingles and their corresponding sides would be in the same ratio. For example cosidering the upper right angled traingle and the lower right angled traingle you can come up with, $\displaystyle \dfrac{10-h}{h}=\dfrac{\dfrac{x}{2}}{\dfrac{5-x}{2}}$

Hope you can continue.
• Oct 30th 2010, 01:50 AM
Educated
Oh, that makes things so much simpler! Can you believe this was the hardest question for a test paper for Year 12 (about 11th the Grade) two years ago for algebra?

h = 10 - 2x

I can't believe I was overthinking it way too much, using pythagorus theorum twice... etc.

Thanks guys.