AMC12 2001 A

AMC12 2001 A · Q8

AMC12 2001 A · Q8. It mainly tests Fractions, Circle theorems.

Which of the cones listed below can be formed from a $252^\circ$ sector of a circle of radius $10$ by aligning the two straight sides?

下面哪个圆锥可以由半径为 $10$ 的圆的 $252^\circ$ 扇形通过对齐两条直边形成？

(A)

(B)

(C)

(D)

(E)

Answer

Correct choice: (C)

正确答案：（C）

Solution

The blue lines will be joined together to form a single blue line on the surface of the cone, so $10$ will be the slant height of the cone. The red line will form the circumference of the base. We can compute its length and use it to determine the radius. The length of the red line is $\dfrac{252^{\circ}}{360^{\circ}}\cdot 2\pi \cdot 10 = \frac{7}{10} \cdot 2\pi \cdot 10 = 14\pi$. This is the circumference of a circle with radius $\frac{14\pi}{2\pi} = 7$. Therefore the correct answer is $\boxed{\textbf{(C)} \text{ A cone with slant height of } 10 \text{ and radius } 7}$.

蓝色线段将被拼合在一起，在圆锥侧面形成一条蓝色线段，因此 $10$ 将是圆锥的斜高。红色弧将形成底面的周长。我们可以计算其长度并据此确定底面半径。红色弧长为 $\dfrac{252^{\circ}}{360^{\circ}}\cdot 2\pi \cdot 10 = \frac{7}{10} \cdot 2\pi \cdot 10 = 14\pi$。这就是一个半径为 $\frac{14\pi}{2\pi}=7$ 的圆的周长。因此正确答案是 $\boxed{\textbf{(C)} \text{ A cone with slant height of } 10 \text{ and radius } 7}$。

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