AMC12 2022 B

AMC12 2022 B · Q2

AMC12 2022 B · Q2. It mainly tests Pythagorean theorem, Area & perimeter.

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 $20$. \boxed{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 $ABCD$ 的面积为 $20$。 \[\boxed{20}]

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