AMC8 2010

AMC8 2010 · Q17

AMC8 2010 · Q17. It mainly tests Fractions, Area & perimeter.

The diagram shows an octagon consisting of 10 unit squares. The portion below $\overline{PQ}$ is a unit square and a triangle with base 5. If $\overline{PQ}$ bisects the area of the octagon, what is the ratio $\frac{XQ}{QY}$?

图中所示八边形由 10 个单位正方形组成。$\overline{PQ}$ 下方的部分是一个单位正方形和一个底边长为 5 的三角形。如果 $\overline{PQ}$ 将八边形的面积二等分，则比值 $\frac{XQ}{QY}$ 为多少？

(A) $\frac{2}{5}$ $\frac{2}{5}$

(B) $\frac{1}{2}$ $\frac{1}{2}$

(D) $\frac{2}{3}$ $\frac{2}{3}$

(E) $\frac{3}{4}$ $\frac{3}{4}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

We see that half the area of the octagon is $5$. We see that the triangle area is $5-1 = 4$. That means that $\frac{5h}{2} = 4 \rightarrow h=\frac{8}{5}$. \[\text{QY}=\frac{8}{5} - 1 = \frac{3}{5}\] Meaning, $\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}$

八边形面积的一半为 $5$。三角形面积为 $5 - 1 = 4$。因此 $\frac{5h}{2} = 4 \rightarrow h = \frac{8}{5}$。 \[ \text{QY} = \frac{8}{5} - 1 = \frac{3}{5} \] 因此，$\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}$

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