AMC10 2018 B

AMC10 2018 B · Q15

AMC10 2018 B · Q15. It mainly tests Area & perimeter, Ratios in geometry.

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?

一个底部为正方形的闭合盒子要用一张正方形包装纸包裹。盒子置于包装纸中央，底面的顶点位于包装纸正方形边中点处（如左图所示）。包装纸的四个角要向上折叠覆盖盒子侧面，并在盒子顶面中心点 A（右图）相遇。盒子底边长 $w$，高 $h$。包装纸的面积是多少？

(A) $2(w+h)^2$ $2(w+h)^2$

(B) $\frac{(w+h)^2}{2}$ $\frac{(w+h)^2}{2}$

(D) $2w^2$ $2w^2$

(E) $w^2 h$ $w^2 h$

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): The figure shows that the distance $AO$ from a corner of the wrapping paper to the center is $$ \frac{w}{2}+h+\frac{w}{2}=w+h. $$ The side of the wrapping paper, $AB$ in the figure, is the hypotenuse of a $45-45-90^\circ$ right triangle, so its length is $\sqrt{2}\cdot AO=\sqrt{2}(w+h)$. Therefore the area of the wrapping paper is $$ \left(\sqrt{2}(w+h)\right)^2=2(w+h)^2. $$

答案（A）：图中显示，从包装纸的一个角到中心的距离 $AO$ 为 $$ \frac{w}{2}+h+\frac{w}{2}=w+h. $$ 图中包装纸的一边 $AB$ 是一个 $45-45-90^\circ$ 直角三角形的斜边，因此其长度为 $\sqrt{2}\cdot AO=\sqrt{2}(w+h)$。因此，包装纸的面积为 $$ \left(\sqrt{2}(w+h)\right)^2=2(w+h)^2. $$

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