Question

一个正方形底面的封闭盒子要用一张正方形包装纸包裹。如左图所示，盒子放在包装纸中央，底面顶点位于正方形纸张的中线上。包装纸的四个角将沿盒子侧面向上折，并在盒子顶面中心、即右图中的点 $A$ 处相会。盒子底边长为 $w$ ，高为 $h.$ 包装纸的面积是多少？

A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point $A$ in the figure on the right. The box has base length $w$ and height $h.$ What is the area of the sheet of wrapping paper?

$2(w + h)^2$

$\dfrac{(w + h)^2}{2}$

$2w^2 + 4wh$

$2w^2$

$w^2 h$

答案：A

知识点：折纸勾股定理正方形（几何）

难度评级：1730

解答：

设包装纸边长为 $s$ ，盒子底面边长为 $w$ 。包装纸中心到一个角的距离为 $\tfrac{s}{\sqrt2}$ 。盒子底面相对于包装纸旋转了 $45^\circ$ ，所以从底面中心到一条边的距离为 $\tfrac{w}{2}$ 。一个角折到盒子顶部中心时，依次经过 $\tfrac{w}{2}$ 、高度 $h$ 和另一个 $\tfrac{w}{2}$ ，因此 $\tfrac{s}{\sqrt2} = \tfrac{w}{2} + h + \tfrac{w}{2} = w + h$ 。所以 $s = \sqrt2\,(w + h)$ ，包装纸面积为 $s^2 = 2(w + h)^2$ 。正确答案是 A。

Let the sheet have side $s.$ The base sits as a square of side $w$ turned $45^\circ,$ so the center is $\tfrac{w}{2}$ from each base edge. A corner of the sheet lies $\tfrac{s}{\sqrt2}$ from the center. Folding that corner up to the top center traces a straight line: $\tfrac{w}{2}$ out to the base edge, then $h$ up the side, then $\tfrac{w}{2}$ across the top. So $\tfrac{s}{\sqrt2} = \tfrac{w}{2} + h + \tfrac{w}{2} = w + h.$ Then $s = \sqrt2\,(w + h),$ and the area is $s^2 = 2(w + h)^2.$ Thus, A is the correct answer.

2018 AMC 10B 第 15 题

答案：A

难度评级：1730

解答：

其他年份的第 15 题