2012 AMC 12B 第 15 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

15.

Jesse 沿两条半径剪开一个半径为 1212 的圆形纸片,形成两个扇形,其中较小扇形的圆心角为 120120 度。他用每个扇形作为一个圆锥的侧面,做成两个圆锥。较小圆锥的体积与较大圆锥的体积之比是多少?

Jesse cuts a circular paper disk of radius 1212 along two radii to form two sectors, the smaller having a central angle of 120120 degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger?

18\dfrac{1}{8}

14\dfrac{1}{4}

1010\dfrac{\sqrt{10}}{10}

56\dfrac{\sqrt{5}}{6}

105\dfrac{\sqrt{10}}{5}

答案:C
知识点:圆锥体积勾股定理
难度评级:1800
解答:

每个扇形形成的圆锥母线长为 1212。 较小扇形的弧长为 1203602π12=8π\tfrac{120}{360}\cdot2\pi\cdot12=8\pi, 因此底面半径为 44,高为 12242=82\sqrt{12^2-4^2}=8\sqrt2

较大扇形(圆心角 240240^\circ)的弧长为 16π16\pi, 底面半径为 88, 高为 12282=45\sqrt{12^2-8^2}=4\sqrt5

体积之比为 13π428213π8245=1010.\frac{\tfrac13\pi\cdot4^2\cdot8\sqrt2}{\tfrac13\pi\cdot8^2\cdot4\sqrt5} =\frac{\sqrt{10}}{10}.

因此正确答案是 C

Each sector forms a cone with slant height 12.12. The smaller sector's arc length is 1203602π12=8π,\tfrac{120}{360}\cdot2\pi\cdot12=8\pi, so its base radius is 44 and its height is 12242=82.\sqrt{12^2-4^2}=8\sqrt2.

The larger sector (central angle 240240^\circ) has arc length 16π,16\pi, base radius 8,8, and height 12282=45.\sqrt{12^2-8^2}=4\sqrt5.

The ratio of volumes is 13π428213π8245=1010.\frac{\tfrac13\pi\cdot4^2\cdot8\sqrt2}{\tfrac13\pi\cdot8^2\cdot4\sqrt5} =\frac{\sqrt{10}}{10}.

Thus, the correct answer is C.

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