AMC10 2007 B

AMC10 2007 B · Q23

AMC10 2007 B · Q23. It mainly tests Similarity, 3D geometry (volume).

A pyramid with a square base is cut by a plane that is parallel to its base and is 2 units from the base. The surface area of the smaller pyramid that is cut from the top is half the surface area of the original pyramid. What is the altitude of the original pyramid?

一个底面为正方形的金字塔被一个与底面平行且距底面 2 单位的平面切割。从顶部切下的小金字塔的表面积是原金字塔表面积的一半。原金字塔的高度是多少？

(A) 2 2

(B) $2 + \sqrt{2}$ $2 + \sqrt{2}$

(D) 4 4

(E) $4 + 2\sqrt{2}$ $4 + 2\sqrt{2}$

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Let $h$ be the altitude of the original pyramid. Then the altitude of the smaller pyramid is $h-2$. Because the two pyramids are similar, the ratio of their altitudes is the square root of the ratio of their surface areas. Thus $h/(h-2)=\sqrt{2}$, so $$ h=\frac{2\sqrt{2}}{\sqrt{2}-1}=4+2\sqrt{2}. $$

答案（E）：设 $h$ 为原棱锥的高，则小棱锥的高为 $h-2$。由于两个棱锥相似，它们的高之比等于表面积之比的平方根。因此 $\frac{h}{h-2}=\sqrt{2}$，所以 $$ h=\frac{2\sqrt{2}}{\sqrt{2}-1}=4+2\sqrt{2}. $$

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