Originally posted by wittywonka
Alrighty, here's yet another geometry problem, much harder. What is the circumference of a circle inscribed in an equilateral triangle with a height of (12 * square root of 3)?
That's not hard.
The inscribed circle of an equilateral triangle is centred on the triangle's centre. The triangle's centre is situated 1/3rd of the way along each of the triangle's three height lines. The radius of the circle is equal to the distance of the centre, along the height line, to the nearest side of the triangle. All these are easily seen when you draw a diagram, and flow elementarily from the equilaterality of the triangle.
The radius of the circle is, then, a third of 12*sqrt(3), is 4*sqrt(3). It circumference is therefore 2 pi r, equals 2 pi 4 sqrt(3), equals 8 pi sqrt(3). Its area is more interesting; that's pi r**2, which is pi 4**2 sqrt(3)**2, equals pi * 16 * 3, equals 48 pi, eliminating the square root.