Question

下图中所有矩形都与外接矩形相似，且图按比例绘制。每个数字表示对应矩形的面积。求长度 $AB$ ？

All of the rectangles in the figure below, which is drawn to scale, are similar to the enclosing rectangle. Each number represents the area of the rectangle. What is length $AB?$

$4 + 4\sqrt5$

$10\sqrt2$

$5 + 5\sqrt5$

$10\sqrt[4]{8}$

$20$

答案：D

知识点：相似面积比矩形

难度评级：1730

小提示：

这些拼块的总面积为 $200$ ；设外接长方形的高为 $h$ ，宽 $AB=w$

The pieces have total area $200;$ let the enclosing rectangle have height $h$ and width $AB=w$

大提示：

沿左边，面积为 $32$ 和 $36$ 的拼块的竖边长度分别为 $h\sqrt{\frac{32}{200}}$ 和 $w\sqrt{\frac{36}{200}}$

Along the left edge, the vertical sides of the area- $32$ and area- $36$ pieces have lengths $h\sqrt{\frac{32}{200}}$ and $w\sqrt{\frac{36}{200}}$

解答：

十一个拼块的面积之和为 $200$ 。设外接矩形的高为 $h$ ，宽为 $w=AB$ 。相似图形的对应边长之比等于它们面积之比的平方根。由图可知，面积为 $32$ 的拼块把它的短边贡献给左边，长度为 $h\sqrt{\frac{32}{200}}=\tfrac25h$ 。在它上方，面积为 $36$ 的拼块贡献的是它的长边，长度为 $w\sqrt{\frac{36}{200}}=\tfrac{3\sqrt2}{10}w$ 。这两段合起来构成整个高，所以 $h=\tfrac25h+\tfrac{3\sqrt2}{10}w$ ，化简得 $h=\frac{w}{\sqrt2}$ 。由于 $wh=200$ ，我们得到 $\frac{w^2}{\sqrt2}=200$ ，从而 $w=\sqrt{200\sqrt2}=10\sqrt[4]{8}$ 。因此正确答案是 D。

The areas of the eleven pieces sum to $200.$ Let the enclosing rectangle have height $h$ and width $w=AB.$ Similar figures have corresponding side lengths in the square-root ratio of their areas. From the diagram, the area- $32$ piece contributes its short side to the left edge, of length $h\sqrt{\frac{32}{200}}=\tfrac25h.$ Above it, the area- $36$ piece contributes its long side, of length $w\sqrt{\frac{36}{200}}=\tfrac{3\sqrt2}{10}w.$ These two segments make the full height, so $h=\tfrac25h+\tfrac{3\sqrt2}{10}w,$ which simplifies to $h=\frac{w}{\sqrt2}.$ Since $wh=200,$ we get $\frac{w^2}{\sqrt2}=200$ and $w=\sqrt{200\sqrt2}=10\sqrt[4]{8}.$ Therefore, the answer is D.

2024 AMC 10A 第 16 题

答案：D

难度评级：1730

小提示：

大提示：

解答：

其他年份的第 16 题