AMC10 2022 B

AMC10 2022 B · Q2

AMC10 2022 B · Q2. It mainly tests Triangles (properties), Pythagorean theorem.

In rhombus $ABCD$, point $P$ lies on segment $\overline{AD}$ so that $\overline{BP} \perp \overline{AD}$, $AP = 3$, and $PD = 2$. What is the area of rhombus $ABCD$?

在菱形$ABCD$中，点$P$在线段$\overline{AD}$上，使得$\overline{BP} \perp \overline{AD}$，$AP = 3$，$PD = 2$。菱形$ABCD$的面积是多少？

(A) 3\sqrt{5} 3\sqrt{5}

(B) 10 10

(D) 20 20

(E) 25 25

Answer

Correct choice: (D)

正确答案：（D）

Solution

Since $ABCD$ is a rhombus, all sides are equal, so \[ AB = AD = AP + PD = 3 + 2 = 5. \] Point $P$ lies on $\overline{AD}$ and $BP \perp AD$. Apply the Pythagorean theorem in right triangle $ABP$: \[ AP^2 + BP^2 = AB^2 \] \[ 3^2 + BP^2 = 5^2 \] \[ 9 + BP^2 = 25 \] \[ BP^2 = 16 \implies BP = 4. \] The area of a rhombus is base times height. Taking $AD$ as the base and $BP$ as the corresponding height, the area is \[ 5 \times 4 = 20. \] Thus, the area of rhombus $ABCD$ is $20$.

由于$ABCD$是菱形，所有边相等，因此$AB = AD = AP + PD = 3 + 2 = 5$。点$P$在$\overline{AD}$上且$BP \perp AD$。在直角三角形$ABP$中应用勾股定理： $AP^2 + BP^2 = AB^2$ $3^2 + BP^2 = 5^2$ $9 + BP^2 = 25$ $BP^2 = 16 \implies BP = 4$。菱形的面积是底乘高。以$AD$为底，$BP$为对应高，面积为 $5 \times 4 = 20$。因此菱形$ABCD$的面积是$20$。

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