# What is the value of $a$ where $√a$ is area of a trapezoid which touches the circle with center $O$ (diameter is 2)?

The sides $$AB, BC, CD$$ of trapezoid $$ABCD$$ touches the circle with center $$O$$ and they are equal. $$AD$$, goes through the point O. If diameter is 2, then the area of the trapezoid is $$√a$$ . What is the value of a?

I tried but I could not find any possible solution. How am I supposed to get $$BC + AD$$?

• Is it not just half of the regular circumscribed hexagon? – dfnu Jan 21 at 17:02
• Out of curiosity, do you have the actual answer? Maybe not the solution, but an answer at least? I ask because I think I might have a solution but I'd like to double-check because it feels a little convoluted. – Eevee Trainer Jan 21 at 17:02
• First show that $ABO$, $BOC$, and $OCD$ are equilateral triangles... – dfnu Jan 21 at 17:21

Warning: might be a little convoluted, uses a bit of assumed knowledge, and is probably not the most elegant/efficient solution. I've triple-checked the method and arithmetic though so unless there's a fundamental flaw with my solution, I think this should be correct.

We make a duplicate trapezoid of $$ABCD$$, with diameter $$AD$$, which basically inscribes the circle in a regular hexagon:

Since the duplicate trapezoid has the same area, the area of the hexagon is $$2\sqrt a$$.

The area of a regular hexagon can be shown to be given by $$\frac{3 \sqrt 3}{2} s^2$$, where $$s$$ is the length of a side of the hexagon. Thus,

$$2 \sqrt a = \frac{3 \sqrt 3}{2} s^2 \implies a = \frac{27}{16} s^4$$

The area of a regular polygon can also be shown (link above) to be given by $$xp/2$$, for perimeter $$p$$ and apothem $$x$$. Here, $$p = 6s$$, and $$x = 1$$, the radius of the circle, clear by the construction above and that the diameter is $$2$$. Thus, the area is given by $$(6s)(1)/2 = 3s$$.

Thus,

$$2 \sqrt a = 3s \implies a = \frac{9}{4}s^2$$

Thus,

$$a = \frac{27}{16}s^4 = \frac{9}{4}s^2$$

We subtract the right fraction from the left, and multiply through by $$16$$ as we begin solving for $$s$$. We get the equation

$$27s^4 - 36s^2 = 0$$

Let $$u = s^2$$ for ease of use, then the above equation becomes

$$27u^2 - 36u = 0 \implies 9u(3u - 4) = 0 \implies u = 0, u = \frac{4}{3}$$

$$u=0$$ is obviously not what we want, but $$u=4/3$$ is fine. Then, going back through our substitutions,

$$u=\frac{4}{3} \implies s^2 = \frac{4}{3} \implies s = \frac{2}{\sqrt 3} \implies 2 \sqrt a = 3 \cdot \frac{2}{\sqrt 3} \implies a = 3$$

• Just add up three areas of equilateral triangles with altitude equal to the circumference radius and get the same result much quicklier. – dfnu Jan 21 at 17:42

Hint.

Once you showed that $$ABO$$, $$OBC$$ and $$COD$$ are equilater triangles, note that the altitude of such triangles is equal to $$r=1$$. By Pythagorean theorem on half of the triangle determine the side, i.e. $$\frac{2\sqrt{3}}{3}$$. The total area of the trapezoid is equal to three times the area of the triangle, i.e. $$\sqrt{3}$$