AMC12 2019 B

AMC12 2019 B · Q18

AMC12 2019 B · Q18. It mainly tests Coordinate geometry, 3D geometry (volume).

Square pyramid $ABCDE$ has base $ABCD$, which measures 3 cm on a side, and altitude $\overline{AE}$ perpendicular to the base, which measures 6 cm. Point $P$ lies on $\overline{BE}$, one third of the way from $B$ to $E$; point $Q$ lies on $\overline{DE}$, one third of the way from $D$ to $E$; and point $R$ lies on $\overline{CE}$, two thirds of the way from $C$ to $E$. What is the area, in square centimeters, of $\triangle PQR$?

正方形金字塔$ABCDE$的底面$ABCD$边长3厘米，高$\overline{AE}$垂直于底面，长6厘米。点$P$在$\overline{BE}$上，从$B$到$E$的三分之一处；点$Q$在$\overline{DE}$上，从$D$到$E$的三分之一处；点$R$在$\overline{CE}$上，从$C$到$E$的五分之二处。$\triangle PQR$的面积有多少平方厘米？

(A) $\frac{3\sqrt{2}}{2}$ $\frac{3\sqrt{2}}{2}$

(B) $\frac{3\sqrt{3}}{2}$ $\frac{3\sqrt{3}}{2}$

(D) $2\sqrt{3}$ $2\sqrt{3}$

(E) $3\sqrt{2}$ $3\sqrt{2}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Place the pyramid in coordinate space with point $A$ at $(0,0,0)$, point $B$ at $(3,0,0)$, point $C$ at $(3,3,0)$, point $D$ at $(0,3,0)$, and point $E$ at $(0,0,6)$. Then $P$ has coordinates $(2,0,2)$, $Q$ has coordinates $(0,2,2)$, and $R$ has coordinates $(1,1,4)$. It follows from the distance formula that $PR=QR=\sqrt{6}$, and $PQ=2\sqrt{2}$. Because $\triangle PQR$ is isosceles, the length of its altitude to $PQ$ can be found from the Pythagorean Theorem to be $2$, and thus the area of $\triangle PQR$ is $2\sqrt{2}$.

答案（C）：将该棱锥放在坐标空间中，令点 $A$ 在 $(0,0,0)$，点 $B$ 在 $(3,0,0)$，点 $C$ 在 $(3,3,0)$，点 $D$ 在 $(0,3,0)$，点 $E$ 在 $(0,0,6)$。则 $P$ 的坐标为 $(2,0,2)$，$Q$ 的坐标为 $(0,2,2)$，$R$ 的坐标为 $(1,1,4)$。由距离公式可得 $PR=QR=\sqrt{6}$，且 $PQ=2\sqrt{2}$。因为 $\triangle PQR$ 是等腰三角形，到 $PQ$ 的高可由勾股定理求得为 $2$，因此 $\triangle PQR$ 的面积为 $2\sqrt{2}$。

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