# Problem Solving #2

Whilst eating some toast I felt the need to have a go at another geometry problem this morning:

(not drawn to scale)

“If the radius of both circles is 2, and the area of both the equilateral triangle and rectangle are the same, find the dimensions of the rectangle”

I liked the diagram and the idea so I thought I’d have a go.

This question felt easier than the last one in that i have a direction to go in right from the start. I’m told the areas are the same, and I can use the diagonal of the rectangle to get a pythagorean formula, and a second formula from the triangle, then solve. So there’s less pondering at the initial stage:

Using the red triangle, we get

$x^2+y^2=4^2$

The second formula looked a bit trickier to determine. I decided to go for this triangle:

Which is a 30,60,90 triangle, and so has side ratios $1:2:\sqrt{3}$

I mixed up which side matched which part of the ratio initially, but it didnt look right so I went back and corrected it fairly quickly.

$\frac{a}{2}=\sqrt{3}\ so\ a=2\sqrt{3}$

Now I have the side length, I can derive the area:

$2\sqrt{3}*0.5*3=3\sqrt{3}$

And so I have my second formula for the rectangle:

$xy=3\sqrt{3}$

squaring this gives :

$x^2y^2=27, so\ x^2=\frac{27}{y^2}$

Substitute into the other equation:

$y^2+\frac{27}{y^2}=16$

$y^4-16y^2+27=0$

Now I have a quartic, which I’m not happy about. Fortunately as I only have a power of 4 and 2 to deal with, I can substitute with say, z:

$z = y^2$

$z^2-16z+27=0$

$z=8\pm\sqrt{37}$

$y^2=8\pm\sqrt{37}$

This is getting yucky. At this point I square rooted everything and started messing with

$xy=3\sqrt{3}$

Which got even messier to the point that I stopped and had a rethink. I went back to

$y^2=8\pm\sqrt{37}$

And it dawned on me that if I use

$x^2=16-(8\pm\sqrt{37})$

then I can see by observation that when:

$y^2=8+\sqrt{37}$

$x^2=8-\sqrt{37}$

and when

$y^2=8-\sqrt{37}$

$x^2=8+\sqrt{37}$

Hence the rectangle has dimensions $\sqrt{8+\sqrt{37}} ,\ \sqrt{8-\sqrt{37}}$