AMC12 2017 B

AMC12 2017 B · Q14

AMC12 2017 B · Q14. It mainly tests 3D geometry (volume), 3D geometry (surface area).

An ice-cream novelty item consists of a cup in the shape of a 4-inch-tall frustum of a right circular cone, with a 2-inch-diameter base at the bottom and a 4-inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height 4 inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches?

一个冰激凌新奇物品由一个高 4 英寸的圆锥台形状杯子组成，底部基底直径 2 英寸，顶部基底直径 4 英寸，里面塞满冰激凌，再加上一个高 4 英寸的实心冰激凌圆锥，其底部基底即圆锥台的顶部基底。冰激凌总体积是多少立方英寸？

(A) $8\pi$ $8\pi$

(B) $\frac{28\pi}{3}$ $\frac{28\pi}{3}$

(D) $14\pi$ $14\pi$

(E) $\frac{44\pi}{3}$ $\frac{44\pi}{3}$

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): A frustum is constructed by removing a right circular cone from a larger right circular cone. The volume of the given frustum is the volume of a right circular cone with a 4-inch-diameter base and a height of 8 inches, minus the volume of a right circular cone with a 2-inch-diameter base and a height of 4 inches. (The stated heights come from considering similar right triangles.) Because the volume of a right circular cone is $\frac{1}{3}\pi r^2 h$, the volume of the frustum is $\frac{1}{3}\pi\cdot 2^2\cdot 8-\frac{1}{3}\pi\cdot 1^2\cdot 4=\frac{28}{3}\pi.$ The volume of the top cone of the novelty is $\frac{1}{3}\pi\cdot 2^2\cdot 4=\frac{16}{3}\pi.$ The requested volume of ice cream is the sum of the volume of each part of the novelty, namely $\frac{28}{3}\pi+\frac{16}{3}\pi=\frac{44}{3}\pi.$

答案（E）：圆台是通过从一个较大的直圆锥中去掉一个较小的直圆锥得到的。所给圆台的体积等于底面直径为 4 英寸、高为 8 英寸的直圆锥体积，减去底面直径为 2 英寸、高为 4 英寸的直圆锥体积。（这些高度来自相似直角三角形的考虑。）由于直圆锥体积为$\frac{1}{3}\pi r^2 h$，圆台的体积为 $\frac{1}{3}\pi\cdot 2^2\cdot 8-\frac{1}{3}\pi\cdot 1^2\cdot 4=\frac{28}{3}\pi.$ 该新奇物顶部圆锥的体积为$\frac{1}{3}\pi\cdot 2^2\cdot 4=\frac{16}{3}\pi.$ 所求冰淇淋的体积为该物体各部分体积之和，即$\frac{28}{3}\pi+\frac{16}{3}\pi=\frac{44}{3}\pi.$

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