1952 AMC 12 Problem 37

Question

Two equal parallel chords are drawn

8

inches apart in a circle of radius

8

inches. The area of that part of the circle that lies between the chords is:

21\dfrac13\pi-32\sqrt3

32\sqrt3+21\dfrac13\pi

32\sqrt3+42\dfrac23\pi

16\sqrt3+42\dfrac23\pi

42\dfrac23\pi

Equal parallel chords lie the same distance from the center, so each is

4

inches from it

Subtract the two congruent outer circular segments from the whole circle

The equal chords are symmetrically

4

inches from the center. For either chord, the half-angle at the center satisfies

\cos\theta=\frac{4}{8}=\frac{1}{2},

so

\theta=60^\circ.

Each outer segment is a

120^\circ

sector minus the isosceles triangle:

\begin{aligned} &\frac{120}{360}\pi(8)^2 -\frac12(8)^2\sin120^\circ\\ &\qquad=\frac{64\pi}{3}-16\sqrt3. \end{aligned}

The area between the chords is therefore

\begin{aligned} &64\pi-2\left(\frac{64\pi}{3}-16\sqrt3\right)\\ &\qquad=\frac{64\pi}{3}+32\sqrt3\\ &\qquad=21\frac13\pi+32\sqrt3. \end{aligned}

Thus, the correct answer is B.

Problem 37 in other years