Question

一个边长为 $3$ 的正方形内接于一个等腰三角形，其中正方形的一条边在三角形的底边上。另一个边长为 $2$ 的正方形有两个顶点在第一个正方形上，另外两个顶点在三角形的边上，如图所示。三角形的面积是多少？

A square with side length $3$ is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length $2$ has two vertices on the other square and the other two on sides of the triangle, as shown. What is the area of the triangle?

$19\frac14$

$20\frac14$

$21 \frac34$

$22\frac12$

$23\frac34$

答案：B

知识点：相似三角形面积正方形（几何）

难度评级：1660

解答：

设等腰三角形高为 $H$ ，底为 $B$ 。

大正方形的上边长为 $3$ ，距顶点 $H-3$ 。小正方形的上边长为 $2$ ，距顶点 $H-5$ 。因此 $\frac{3}{H-3}=\frac{2}{H-5}.$

解得 $H=9$ 。又 $\frac{B}{H}=\frac{3}{H-3}$ ，所以 $B=\frac{9}{2}$ 。面积为 $\frac12\cdot\frac92\cdot9=\frac{81}{4}=20\frac14$ 。

所以答案是 B。

Let the isosceles triangle have height $H$ and base $B.$ By similarity, horizontal widths in the triangle are proportional to distance from the top vertex.

The top side of the larger square has width $3$ and is $H-3$ units from the top. The top side of the smaller square has width $2$ and is $H-5$ units from the top. Hence $\frac{3}{H-3}=\frac{2}{H-5}.$

Solving gives $H=9.$ Also $\frac{B}{H}=\frac{3}{H-3},$ so $B=\frac{9}{2}.$ The area is $\frac12\cdot\frac92\cdot9=\frac{81}{4}=20\frac14.$

Thus, the answer is B .

2021 AMC 10B Fall 第 13 题

答案：B

难度评级：1660

解答：

其他年份的第 13 题