1979 AMC 12 第 8 题

Question

求曲线

y=|x|

与

x^2+y^2=4

所围成的最小区域的面积。

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

y=|x|

的两条射线交于圆心

The two rays of

y=|x|

meet at the center of the circle

求射线

y=x

与

y=-x

之间较小的圆心角

Find the smaller central angle between the rays

y=x

and

y=-x

两条射线与圆的交点所对应的方向角分别为

45^\circ

和

135^\circ

，所以最小的有界区域是半径为

2

的圆的一个

90^\circ

扇形。其面积为

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

因此，正确答案是 C。

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.

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