AMC10 2012 B

AMC10 2012 B · Q17

AMC10 2012 B · Q17. It mainly tests Circle theorems, 3D geometry (volume).

Jesse cuts a circular paper disk of radius 12 along two radii to form two sectors, the smaller having a central angle of 120 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?

杰西将半径为12的圆纸盘沿两条半径切开，形成两个扇形，其中较小的中心角为120度。他用每个扇形制作一个圆锥，作为圆锥的侧面。较小圆锥体积与较大圆锥体积的比率为多少？

(A) 1/8 1/8

(B) 1/4 1/4

(D) $\sqrt{5}/6$ $\sqrt{5}/6$

(E) $\sqrt{10}/5$ $\sqrt{10}/5$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Each sector forms a cone with slant height 12. The circumference of the base of the smaller cone is \$\frac{120}{360}\cdot 2\cdot 12\cdot \pi=8\pi\$. Hence the radius of the base of the smaller cone is 4 and its height is \$\sqrt{12^2-4^2}=8\sqrt{2}\$. Similarly, the circumference of the base of the larger cone is \$16\pi\$. Hence the radius of the base of the larger cone is 8 and its height is \$4\sqrt{5}\$. The ratio of the volume of the smaller cone to the volume of larger cone is \[ \frac{\frac{1}{3}\pi\cdot 4^2\cdot 8\sqrt{2}}{\frac{1}{3}\pi\cdot 8^2\cdot 4\sqrt{5}}=\frac{\sqrt{10}}{10}. \]

答案（C）：每个扇形形成一个母线长为 12 的圆锥。较小圆锥底面圆周长为 \$\frac{120}{360}\cdot 2\cdot 12\cdot \pi=8\pi\$。因此较小圆锥的底面半径为 4，高为 \$\sqrt{12^2-4^2}=8\sqrt{2}\$。同理，较大圆锥的底面圆周长为 \$16\pi\$。因此较大圆锥的底面半径为 8，高为 \$4\sqrt{5}\$。较小圆锥体积与较大圆锥体积之比为 \[ \frac{\frac{1}{3}\pi\cdot 4^2\cdot 8\sqrt{2}}{\frac{1}{3}\pi\cdot 8^2\cdot 4\sqrt{5}}=\frac{\sqrt{10}}{10}. \]

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