Question

一个正方形底面的金字塔被一个平行于底面、且距底面 $2$ 个单位的平面切开。切下来的上方小金字塔的表面积是原金字塔表面积的一半。原金字塔的高是多少？

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?

$2$

$2+\sqrt2$

$1+2\sqrt2$

$4$

$4+2\sqrt2$

答案：E

知识点：长度、面积与体积的缩放关系棱锥表面积

难度评级：1880

解答：

设原金字塔高为 $h$ ，小金字塔高为 $h-2$ 。

两个金字塔相似，所以表面积之比等于高之比的平方。小金字塔表面积为原来的一半，因此 $\left(\dfrac{h-2}{h}\right)^2=\dfrac12$ ，从而 $\dfrac{h}{h-2}=\sqrt2$ 。

于是 $h=\sqrt2(h-2)$ ，所以 $h(\sqrt2-1)=2\sqrt2$ ， $h=\dfrac{2\sqrt2}{\sqrt2-1}=4+2\sqrt2$ 。

所以正确答案是 E。

Let $h$ be the altitude of the original pyramid; the smaller pyramid has altitude $h-2.$ The two pyramids are similar, so the ratio of their surface areas is the square of the ratio of their altitudes.

The smaller surface area is half the original, so $\left(\dfrac{h-2}{h}\right)^2=\dfrac12,$ giving $\dfrac{h}{h-2}=\sqrt2.$

Then $h=\sqrt2(h-2),$ so $h(\sqrt2-1)=2\sqrt2$ and $h=\dfrac{2\sqrt2}{\sqrt2-1}=4+2\sqrt2.$

Thus, the correct answer is E.

2007 AMC 10B 第 23 题

答案：E

难度评级：1880

解答：

其他年份的第 23 题