Math Problem

A rectangle and an isosceles triangle are inscribed in a circle with a radius of 1. The rectangle has height, h, and base, b. For what value of h are the areas of the rectangle and triangle the same?

Missing information. Is the base of the isosceles triangle also b, or what?

yes, they both have base b

h = 2/3

The heights of the triangles (for there are two: http://i34.tinypic.com/2rcsprd.png) are {h \over 2}+r and r - {h \over 2}, where r is the radius of the circle.
b is fixed based on h, according to the formula b = sqrt{-{h^2 \over 4} + r^2}*2 (this follows from the formula for the circle).

The formula for the surface of the rectangle is sqrt{-{h^2 \over 4} + r^2}*2*h, and for the triangles {({{h \over 2}+r}) * sqrt{-{h^2 \over 4} + r^2}*2} \over 2 and {(r - {h \over 2}) * sqrt{-{h^2 \over 4} + r^2}*2} \over 2.
The problem therefore reduces to finding the values of h for sqrt{-{h^2 \over 4} + r^2}*2*h = {{({{h \over 2}+r}) * sqrt{-{h^2 \over 4} + r^2}*2} \over 2} on the one hand, and sqrt{-{h^2 \over 4} + r^2}*2*h = {{(r - {h \over 2}) * sqrt{-{h^2 \over 4} + r^2}*2} \over 2} on the other.
(Hint: they will be the same.)

Since this is your homework, I suggest you solve those.

>>5
Balls. That'll teach me to preview equations in Mimetex.
The relevant equations are:



(Fingers crossed.)

Bump for pretty TeX.

Can't be done by me, because I'm black.

>>5
I think there's a much easier way to do this problem.
h = rectangle height, H = triangle height, b = shared base

area of rectangle = hb = Hb/2 = area of triangle
h = H/2
2h = H

Draw the perpendicular bisector of the rectangle's height.
It is clear that H = r + h/2

Substituting, 2h = H = r + h/2
3h/2 = r
h = 2r/3

>>9
Don't forget H is also r - h/2, for the other triangle, so h is also 5r/2.

>>10
For h = 5r/2 > 4r/2 = 2r, the rectangle is no longer inscribed in the circle.