Question

等腰三角形 $ABC$ 中， $AB$ 和 $BC$ 长度相等。在下面两幅图中，分别画出与 $\overline{AC}$ 平行的线段，使 $\triangle ABC$ 的阴影部分面积相同。两个未阴影部分的高分别为 $11$ 和 $5$ 个单位。 $\triangle ABC$ 的高 $h$ 是多少？

Isosceles triangle $ABC$ has equal side lengths $AB$ and $BC.$ In the figures below, segments are drawn parallel to $\overline{AC}$ so that the shaded portions of $\triangle ABC$ have the same area. The heights of the two unshaded portions are $11$ and $5$ units, respectively. What is the height $h$ of $\triangle ABC?$

$14.6$

$14.8$

$15$

$15.2$

$15.4$

答案：A

知识点：相似面积比

难度评级：1930

小提示：

阴影面积相等给出两个未阴影三角形面积之间的方程

Equal shaded areas give an equation between the two unshaded triangle areas

大提示：

相似三角形面积按高度比例的平方缩放

Similar triangle areas scale as the square of their heights

视频讲解：

Click to load, then click again to play

文字解答：

设 $\triangle ABC$ 的面积为 $a$ 。左图中的未阴影三角形高为 $11$ ，并且与整个三角形相似，所以它的面积是 $a\left(\frac{11}{h}\right)^2$ 。因此，左图的阴影面积是 $a\left(1-\left(\frac{11}{h}\right)^2\right)$ 。

右图中的阴影三角形高为 $h-5$ ，并且与整个三角形相似，所以它的面积是 $a\left(\frac{h-5}{h}\right)^2$ 。令两图的阴影面积相等，再约去 $a$ ，得到 $1-\left(\dfrac{11}{h}\right)^2 =\left(\dfrac{h-5}{h}\right)^2\text{。}$

化简得 $h^2 - 121 = h^2 - 10h + 25\text{，}$ 也就是 $10h = 146\text{。}$

所以 $h=14.6$ 。

所以正确答案是 A。

Let $a$ be the area of $\triangle ABC.$ The unshaded triangle in the left figure has height $11$ and is similar to the full triangle, so its area is $a\left(\frac{11}{h}\right)^2.$ Therefore, the shaded area there is $a\left(1-\left(\frac{11}{h}\right)^2\right).$

In the right figure, the shaded triangle has height $h-5$ and is similar to the full triangle, so its area is $a\left(\frac{h-5}{h}\right)^2.$ Equating the shaded areas and canceling $a$ gives $1-\left(\dfrac{11}{h}\right)^2 =\left(\dfrac{h-5}{h}\right)^2.$

Simplifying yields $h^2 - 121 = h^2 - 10h + 25.$ This simplifies to $10h = 146,$

so $h=14.6.$

Thus, A is the correct answer.

2023 AMC 8 第 24 题