AMC12 2020 A

AMC12 2020 A · Q18

AMC12 2020 A · Q18. It mainly tests Angle chasing, Similarity.

Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^\circ$, $AC = 20$, and $CD = 30$. Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E$, and $AE = 5$. What is the area of quadrilateral $ABCD$?

四边形$ABCD$满足$\angle ABC = \angle ACD = 90^\circ$，$AC = 20$，$CD = 30$。对角线$\overline{AC}$和$\overline{BD}$相交于点$E$，且$AE = 5$。四边形$ABCD$的面积是多少？

(A) 330 330

(B) 340 340

(D) 360 360

(E) 370 370

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Because the diagonals intersect, quadrilateral $ABCD$ is convex. Let $F$ be the foot of the perpendicular from $B$ to $AC$. Because $\triangle ECD$ is a right triangle, $\angle AEB=\angle DEC$ is acute, so $F$ lies between $A$ and $E$. Let $h=BF$ and $x=EF$. See the figure below. Because $\triangle BFE\sim\triangle DCE$, $$\frac{h}{x}=\frac{30}{20-5}=2.$$ Because $\triangle ABF\sim\triangle BCF$, $$\frac{h}{5-x}=\frac{15+x}{h}.$$ Together these imply that $x^2+2x-15=0$. Solving for $x$ yields $x=3$ and $h=6$. The area of $ABCD$ is the area of $\triangle ABC$ plus the area of $\triangle ACD$, and this equals $\frac12\cdot20\cdot6+\frac12\cdot20\cdot30=360$.

答案（D）：由于两条对角线相交，四边形 $ABCD$ 是凸四边形。设 $F$ 为从 $B$ 向 $AC$ 作垂线的垂足。由于 $\triangle ECD$ 是直角三角形，$\angle AEB=\angle DEC$ 为锐角，所以 $F$ 位于 $A$ 与 $E$ 之间。令 $h=BF$，$x=EF$。见下图。因为 $\triangle BFE\sim\triangle DCE$， $$\frac{h}{x}=\frac{30}{20-5}=2.$$ 因为 $\triangle ABF\sim\triangle BCF$， $$\frac{h}{5-x}=\frac{15+x}{h}.$$ 由此可得 $x^2+2x-15=0$。解得 $x=3$ 且 $h=6$。$ABCD$ 的面积等于 $\triangle ABC$ 的面积加上 $\triangle ACD$ 的面积，即 $$\frac12\cdot20\cdot6+\frac12\cdot20\cdot30=360.$$

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