# What is the area of the largest trapezoid that can fit into a semicircle with radius $r$?

What is the area of the largest trapezoid that can fit into a semicircle with radius $$r$$ and whose bases are parallel to the diameter of the semi-circle?

I know this is a variant of some previously posted problems (e.g. this one and that one), but I can't seem to get the general answer.

First of all, to prove that the trapezoid must be inscribed in the circle, will it suffice to say that if the trapezoid is not inscribed in the circle, then one of its corners can be extended to be on the circle, thereby increasing its area?

Also, I can't seem to eliminate enough variables to only get one variable. I know the maximum area is achieved when the top base is equal to the radius, but I don't know how to show this. If I do manage to prove this, then I don't even think I'll need calculus anymore since I'll have already gotten a formula for the area of the trapezoid in terms of $$r.$$ As well, I don't know how to use calculus to approach this problem when there are two variables. Obviously, the derivative with respect to the radius should be $$0,$$ but the derivative with respect to the top base $$a$$ should be ...?

Showing that the trapezoid must be inscribed in the circle is very simple.

It is easy to see that the height of the trapezoid is given by $$\sqrt{r^2-\dfrac{a^2}{4}},$$ where $$a$$ is the shorter base of the trapezoid. As well, for the area to be maximal, we must have that the longer base is equal to $$2r.$$

Now, the area function is $$A=\left(r+\dfrac{a}{2}\right)\sqrt{r^2-\dfrac{a^2}{4}}.$$ Differentiating wrt $$a$$ gives $$A' = \dfrac{1}{2}\sqrt{r^2-\dfrac{a^2}{4}}-\dfrac{a\left(r+\dfrac{a}{2}\right)}{2\sqrt{r^2-\dfrac{a^2}{4}}}.$$ Solve for $$A'=0$$ and show that $$A'>0$$ when $$a and $$A'<0$$ when $$a>r$$ to solve your problem. To simplify things, note that the denominator is always positive and that you should get a quadratic that is easy to factor. In terms of radius $$r$$ and the angle $$a$$, The area of the trapezoid is,

$$A(a) = \frac12\cdot r\sin a \cdot(2r\cos a+ 2r) = r^2\sin a(\cos a +1)$$

Take the derivative with respect to $$a$$ and set it to zero,

$$A'(a) = r^2[\cos a ( \cos a +1) -\sin^2 a]=0$$

or

$$2\cos^2 a +\cos a -1=0$$

which yields $$\cos a = \frac12$$, or $$a = 60^\circ$$. Thus, the largest area is,

$$A(60^\circ) = r^2\sin 60^\circ(\cos 60^\circ +1) = \frac{3\sqrt3}{4}r^2$$

Since the bases are parallel to the diameter, it suffices to consider the case where one base is the diameter (otherwise you could move one base closer to the diameter to increase the area of the trapezium). Then there are two relevant variables left, the distance $$d$$ between the two bases and the length of the shorter base. The length of the shorter base can easily be found in terms of $$d$$ using the Pythagorean theorem, and then it should be a quick matter to maximize the area of the trapezium with respect to $$a$$.