1979 AMC 12 Problem 8

Attempt Problem 8 of the 1979 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1979 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

Find the area of the smallest region bounded by the graphs of

y=|x|

and

x^2+y^2=4.

\frac\pi4

\frac{3\pi}4

\pi

\frac{3\pi}2

2\pi

The two rays of

y=|x|

meet at the center of the circle

Find the smaller central angle between the rays

y=x

and

y=-x

The two rays meet the circle at angles

45^\circ

and

135^\circ,

so the smallest bounded region is a

90^\circ

sector of the radius-

2

circle. Its area is

\frac{90^\circ}{360^\circ}\pi(2)^2=\pi.

Therefore, the correct answer is C.

Problem 8 in other years