AMC8 2023

AMC8 2023 · Q24

AMC8 2023 · Q24. It mainly tests Similarity, Area & perimeter.

Isosceles $\triangle ABC$ has equal side lengths $AB$ and $BC$. In the figure 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 of $h$ of $\triangle ABC$? (Diagram not drawn to scale.)

等腰 $\triangle ABC$ 有相等的边长 $AB$ 和 $BC$。在下面的图中，画了与 $\overline{AC}$ 平行的线段，使得 $\triangle ABC$ 的阴影部分面积相同。两个非阴影部分的的高度分别为 $11$ 和 $5$ 个单位。求 $\triangle ABC$ 的高度 $h$？（图未按比例绘制。）

(A) 14.6 14.6

(B) 14.8 14.8

(D) 15.2 15.2

(E) 15.4 15.4

Answer

Correct choice: (A)

正确答案：（A）

Solution

First, we notice that the smaller isosceles triangles are similar to the larger isosceles triangles. We can find that the area of the gray area in the first triangle is $[ABC]\cdot\left(1-\left(\tfrac{11}{h}\right)^2\right)$. Similarly, we can find that the area of the gray part in the second triangle is $[ABC]\cdot\left(\tfrac{h-5}{h}\right)^2$. These areas are equal, so $1-\left(\frac{11}{h}\right)^2=\left(\frac{h-5}{h}\right)^2$. Simplifying yields $10h=146$ so $h=\boxed{\textbf{(A) }14.6}$.

首先注意到较小的等腰三角形与较大的等腰三角形相似。我们可以发现第一个三角形中灰色区域的面积为 $[ABC]\cdot\left(1-\left(\tfrac{11}{h}\right)^2\right)$。类似地，第二个三角形中灰色部分的面积为 $[ABC]\cdot\left(\tfrac{h-5}{h}\right)^2$。这两个面积相等，因此 $1-\left(\frac{11}{h}\right)^2=\left(\frac{h-5}{h}\right)^2$。化简得 $10h=146$，所以 $h=\boxed{\textbf{(A) }14.6}$。

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