Here are a couple of problems I saved a while back from the Drexel University Math Forum:

1. A square with side length s is inscribed in an equilateral triangle of side length t. t = ks. Find k, expressed as a fraction in simplest radical form.

2. A regular hexagon and an equilateral triangle are both inscribed in the same circle so that the hexagon and the triangle share three vertices. The radius of the circle is 8 units. What is the area of the region between the two polygons?

1. February 17, 2007 12:42 pm

For the first, I started with an equilateral triangle with side S, and dropped a vertical of length S to make a 30-60-90 triangle. Finding the hypotenuse was just algebra. And then taking care to keep the parts straight and offer the correct ratio.

Jonathan

2. February 17, 2007 12:58 pm

I’m not following — you dropped a vertical in the equilateral triangle with side S? Then it wouldn’t have length S. I used the little 30-60-90’s already present on the sides at the bottom, with legs S and 1/2(T-S) and hypotenuse (T-S) and then, as you said, it’s all algebra.

3. February 17, 2007 1:46 pm

I started with a 60-60-60 with side S (in other words, the little triangle on top)

If we labeled the endpoints of the various segments A,B,C,D,E,F,G so that the A is on top and the whole thing is ACF:

I started with ABG (all sides are S) and added BD (length S), BC and CD (lengths derived from 30-60-90 ratios).

Jonathan

4. February 17, 2007 1:53 pm

Oh, I think that’s the same as what I did. It’s just that BD is already there, so your description confused me.

February 27, 2013 12:30 pm

Difference equals = 3Tan(30)r^2/cos(30)-1.5cos(30)r^2 or about 45 square units in this case.