Question

在边长为 $1$ 的等边三角形的三条边上，分别以这些边为底作三个全等的等腰三角形。这三个等腰三角形的面积和等于原等边三角形的面积。每个等腰三角形的一条腰长是多少？

Three congruent isosceles triangles are constructed with their bases on the sides of an equilateral triangle of side length $1.$ The sum of the areas of the three isosceles triangles is the same as the area of the equilateral triangle. What is the length of one of the two congruent sides of one of the isosceles triangles?

$\dfrac{\sqrt3}{4}$

$\dfrac{\sqrt3}{3}$

$\dfrac{2}{3}$

$\dfrac{\sqrt2}{2}$

$\dfrac{\sqrt3}{2}$

答案：B

知识点：等边三角形三角形面积勾股定理

难度评级：1560

解答：

等边三角形的面积为 $\dfrac{\sqrt3}{4}$ 。每个等腰三角形的底边为 $1$ ，设其高为 $h$ ，则 $3\cdot\dfrac12 h=\dfrac{\sqrt3}{4}$ ，所以 $h=\dfrac{\sqrt3}{6}$ 。

一条腰是从顶点到一个底边端点的斜边，因此其长度为 $\begin{gathered} \sqrt{\left(\dfrac12\right)^2+\left(\dfrac{\sqrt3}{6}\right)^2}\\ =\sqrt{\dfrac14+\dfrac{1}{12}}\\ =\sqrt{\dfrac13}=\dfrac{\sqrt3}{3}. \end{gathered}$

所以正确答案是 B。

The equilateral triangle has area $\dfrac{\sqrt3}{4}.$ Each isosceles triangle has base $1$ and height $h,$ so $3\cdot\dfrac12 h=\dfrac{\sqrt3}{4},$ giving $h=\dfrac{\sqrt3}{6}.$

A congruent side is the hypotenuse from the apex to a base endpoint: $\begin{gathered} \sqrt{\left(\dfrac12\right)^2+\left(\dfrac{\sqrt3}{6}\right)^2}\\ =\sqrt{\dfrac14+\dfrac{1}{12}}\\ =\sqrt{\dfrac13}=\dfrac{\sqrt3}{3}. \end{gathered}$

Thus, the correct answer is B.

2014 AMC 12A 第 10 题

答案：B

难度评级：1560

解答：

其他年份的第 10 题