Area of a sector

Geometry Level pending

$\Gamma$ is a circle with center $O$ . $A$ and $B$ are points on $\Gamma$ such that the sector $AOB$ has a perimeter of $40$ . Amongst all circular sectors with a perimeter of $40$ , what is the central measure of $\angle AOB$ (in radians) of the sector with the largest area?

Details and assumptions

A central angle is an angle whose vertex is the center of a circle, and whose sides pass through a pair of points on the circle.

The answer is 2.

1 solution

Arron Kau Staff
May 13, 2014

Let $r$ be the radius of the circle, let $\theta = \angle AOB$ and let $[AOB]$ denote the area of sector $AOB$ . From the question, we have that $\widehat{AB} = 40 - 2r$ . So, we have $\begin{aligned} [AOB] &= \pi r^2 \cdot \left(\frac{\theta}{2\pi}\right) \\ &= \frac{r (r \cdot \theta)}{2} \\ &= \frac{r \cdot \widehat{AB}}{2} \\ &= \frac{r \cdot (40 - 2r)}{2} \\ &= -r^2 + 20r \\ &= -(r - 10)^2 + 100 \\ \end{aligned}$

Hence the maximal area of sector $AOB$ is $100$ which occurs at $r = 10$ . Thus $\theta = \frac{2[AOB]}{r^2} = \frac{2(100)}{10^2} = 2$ .

Area of a sector

The answer is 2.

1 solution

0 pending reports