AMC10 2025 A

AMC10 2025 A · Q15

AMC10 2025 A · Q15. It mainly tests Triangles (properties), Similarity.

In the figure below, $ABEF$ is a rectangle, $\overline{AD}\perp\overline{DE}$, $AF=7$, $AB=1$, and $AD=5$. What is the area of $\triangle ABC$?

下图中，$ABEF$ 是矩形， $\overline{AD}\perp\overline{DE}$， $AF=7$， $AB=1$， $AD=5$。 $ riangle ABC$ 的面积是多少？

(A) \frac{3}{8} \frac{3}{8}

(B) \frac{4}{9} \frac{4}{9}

(D) \frac{7}{15} \frac{7}{15}

(E) \frac{1}{8}\sqrt{15} \frac{1}{8}\sqrt{15}

Answer

Correct choice: (A)

正确答案：（A）

Solution

Because $ABEF$ is a rectangle, $\angle ABC=90°$. We are given that $\angle ADE=90°$, and since $\angle ECD=\angle ACB$ by vertical angles, $\triangle ECD \sim \triangle ACB$. Let $AC=x$. By the Pythagorean Theorem, $CB=\sqrt{x^2-1}$. Since $AF=BE=7$, $EC=7-\sqrt{x^2-1}$. Because $AC=x$ and $AD=5$, $CD=5-x$. By similar triangles, \[\frac{7-\sqrt{x^2-1}}{x}=\frac{5-x}{\sqrt{x^2-1}}\]. Cross-multiplying, we get that \[7\sqrt{x^2-1}-x^2+1=5x-x^2\], so \[7\sqrt{x^2-1}=5x-1\]. We square both sides, and this is simply a quadratic in $x$: \[24x^2+10x-50=0\], which has a positive root $x=\frac{5}{4}$. Since $AB=1$, we can plug this into the Pythagorean Theorem, with $\frac{5}{4}$ being the hypotenuse, to get $BC=\frac{3}{4}$, and ${1}\cdot \frac{\frac{3}{4}}{2}$ to equal $[ABC]= \boxed{\textbf{(A)} \frac{3}{8}}$

因为 $ABEF$ 是矩形， $\angle ABC=90°$。已知 $\angle ADE=90°$，且由于 $\angle ECD=\angle ACB$ 是垂心角， $\triangle ECD \sim \triangle ACB$。设 $AC=x$。由勾股定理， $CB=\sqrt{x^2-1}$。由于 $AF=BE=7$， $EC=7-\sqrt{x^2-1}$。因为 $AC=x$ 和 $AD=5$， $CD=5-x$。由相似三角形， \[\frac{7-\sqrt{x^2-1}}{x}=\frac{5-x}{\sqrt{x^2-1}}\]。交叉相乘，得 \[7\sqrt{x^2-1}-x^2+1=5x-x^2\]，所以 \[7\sqrt{x^2-1}=5x-1\]。两边平方，这是一个关于 $x$ 的二次方程： \[24x^2+10x-50=0\]，正根 $x=\frac{5}{4}$。由于 $AB=1$，代入勾股定理，斜边 $\frac{5}{4}$，得 $BC=\frac{3}{4}$，面积 $\frac{1\cdot \frac{3}{4}}{2}= \boxed{\textbf{(A)} \frac{3}{8}}$

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