AMC10 2024 B

AMC10 2024 B · Q10

AMC10 2024 B · Q10. It mainly tests Triangles (properties), Similarity.

Quadrilateral $ABCD$ is a parallelogram, and $E$ is the midpoint of the side $\overline{AD}$. Let $F$ be the intersection of lines $EB$ and $AC$. What is the ratio of the area of quadrilateral $CDEF$ to the area of $\triangle CFB$?

四边形 $ABCD$ 是平行四边形，$E$ 是边 $\overline{AD}$ 的中点。$F$ 是直线 $EB$ 与 $AC$ 的交点。求四边形 $CDEF$ 的面积与 $\triangle CFB$ 的面积之比？

(A) 5:4 5:4

(B) 4:3 4:3

(D) 5:3 5:3

(E) 2:1 2:1

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let $AB = CD$ have length $b$ and let the altitude of the parallelogram perpendicular to $\overline{AD}$ have length $h$. The area of the parallelogram is $bh$ and the area of $\triangle ABE$ equals $\frac{(b/2)(h)}{2} = \frac{bh}{4}$. Thus, the area of quadrilateral $BCDE$ is $bh - \frac{bh}{4} = \frac{3bh}{4}$. We have from $AA$ that $\triangle CBF \sim \triangle AEF$. Also, $CB/AE = 2$, so the length of the altitude of $\triangle CBF$ from $F$ is twice that of $\triangle AEF$. This means that the altitude of $\triangle CBF$ is $2h/3$, so the area of $\triangle CBF$ is $\frac{(b)(2h/3)}{2} = \frac{bh}{3}$. Then, the area of quadrilateral $CDEF$ equals the area of $BCDE$ minus that of $\triangle CBF$, which is $\frac{3bh}{4} - \frac{bh}{3} = \frac{5bh}{12}$. Finally, the ratio of the area of $CDEF$ to the area of triangle $CFB$ is $\frac{\frac{5bh}{12}}{\frac{bh}{3}} = \frac{\frac{5}{12}}{\frac{1}{3}} = \frac{5}{4}$, so the answer is $\boxed{\textbf{(A) } 5:4}$.

设 $AB = CD$ 的长度为 $b$，平行四边形垂直于 $\overline{AD}$ 的高度为 $h$。平行四边形面积为 $bh$，$\triangle ABE$ 的面积为 $\frac{(b/2)(h)}{2} = \frac{bh}{4}$。因此，四边形 $BCDE$ 的面积为 $bh - \frac{bh}{4} = \frac{3bh}{4}$。由 AA 相似性，$\triangle CBF \sim \triangle AEF$。且 $CB/AE = 2$，因此 $\triangle CBF$ 从 $F$ 到底边的垂足长度是 $\triangle AEF$ 的两倍。即 $\triangle CBF$ 的高度为 $2h/3$，面积为 $\frac{(b)(2h/3)}{2} = \frac{bh}{3}$。四边形 $CDEF$ 的面积等于 $BCDE$ 减去 $\triangle CBF$ 的面积，即 $\frac{3bh}{4} - \frac{bh}{3} = \frac{5bh}{12}$。最终，$CDEF$ 与 $\triangle CFB$ 面积比为 $\frac{\frac{5bh}{12}}{\frac{bh}{3}} = \frac{\frac{5}{12}}{\frac{1}{3}} = \frac{5}{4}$，答案为 $\boxed{\textbf{(A) } 5:4}$。

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