AMC12 2008 B

AMC12 2008 B · Q25

AMC12 2008 B · Q25. It mainly tests Angle chasing, Polygons.

Let $ABCD$ be a trapezoid with $AB||CD, AB=11, BC=5, CD=19,$ and $DA=7$. Bisectors of $\angle A$ and $\angle D$ meet at $P$, and bisectors of $\angle B$ and $\angle C$ meet at $Q$. What is the area of hexagon $ABQCDP$?

设 $ABCD$ 为梯形，满足 $AB||CD, AB=11, BC=5, CD=19,$ 且 $DA=7$。$\angle A$ 与 $\angle D$ 的角平分线交于 $P$，$\angle B$ 与 $\angle C$ 的角平分线交于 $Q$。求六边形 $ABQCDP$ 的面积。

(A) $28\sqrt{3}$ $28\sqrt{3}$

(B) $30\sqrt{3}$ $30\sqrt{3}$

(D) $35\sqrt{3}$ $35\sqrt{3}$

(E) $36\sqrt{3}$ $36\sqrt{3}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Error creating thumbnail: Unable to save thumbnail to destination Note: In the image AB and CD have been swapped from their given lengths in the problem. However, this doesn't affect any of the solving. Drop perpendiculars to $CD$ from $A$ and $B$, and call the intersections $X,Y$ respectively. Now, $DA^2-BC^2=(7-5)(7+5)=DX^2-CY^2$ and $DX+CY=19-11=8$. Thus, $DX-CY=3$. We conclude $DX=\frac{11}{2}$ and $CY=\frac{5}{2}$. To simplify things even more, notice that $90^{\circ}=\frac{\angle D+\angle A}{2}=180^{\circ}-\angle APD$, so $\angle P=\angle Q=90^{\circ}$. Also, \[\sin(\angle PAD)=\sin(\frac12\angle XDA)=\sqrt{\frac{1-\cos(\angle XDA)}{2}}=\sqrt{\frac{3}{28}}\] So the area of $\triangle APD$ is: \[R\cdot c\sin a\sin b =\frac{7\cdot7}{2}\sqrt{\frac{3}{28}}\sqrt{1-\frac{3}{28}}=\frac{35}{8}\sqrt{3}\] Over to the other side: $\triangle BCY$ is $30-60-90$, and is therefore congruent to $\triangle BCQ$. So $[BCQ]=\frac{5\cdot5\sqrt{3}}{8}$. The area of the hexagon is clearly \begin{align*} [ABCD]-([BCQ]+[APD]) &=\frac{15\cdot5\sqrt{3}}{2}-\frac{60\sqrt{3}}{8}\\ &=30\sqrt{3}\implies\boxed{\mathrm{B}} \end{align*} Note: Once $DY$ is found, there is no need to do the trig. Notice that the hexagon consists of two trapezoids, $ABPQ$ and $CDPQ$. $PQ = \frac{19-7-5 +11}{2} = 9$. The height is one half of $BY$ which is $\frac{5\sqrt{3}}{4}$. So the area is \[\frac{1}{2} \cdot \frac{5\sqrt{3}}{4}(19+9+11+9)=30\sqrt{3}\]

Error creating thumbnail: Unable to save thumbnail to destination 注：图中 $AB$ 与 $CD$ 的长度与题目给定的互换了，但这不影响解题过程。从 $A$ 和 $B$ 向 $CD$ 作垂线，垂足分别为 $X,Y$。则 $DA^2-BC^2=(7-5)(7+5)=DX^2-CY^2$，且 $DX+CY=19-11=8$。因此 $DX-CY=3$。于是 $DX=\frac{11}{2}$，$CY=\frac{5}{2}$。进一步注意到 $90^{\circ}=\frac{\angle D+\angle A}{2}=180^{\circ}-\angle APD$，所以 $\angle P=\angle Q=90^{\circ}$。并且 \[\sin(\angle PAD)=\sin(\frac12\angle XDA)=\sqrt{\frac{1-\cos(\angle XDA)}{2}}=\sqrt{\frac{3}{28}}\] 所以 $\triangle APD$ 的面积为： \[R\cdot c\sin a\sin b =\frac{7\cdot7}{2}\sqrt{\frac{3}{28}}\sqrt{1-\frac{3}{28}}=\frac{35}{8}\sqrt{3}\] 再看另一侧：$\triangle BCY$ 是 $30-60-90$ 三角形，因此与 $\triangle BCQ$ 全等。所以 $[BCQ]=\frac{5\cdot5\sqrt{3}}{8}$。六边形面积显然为 \begin{align*} [ABCD]-([BCQ]+[APD]) &=\frac{15\cdot5\sqrt{3}}{2}-\frac{60\sqrt{3}}{8}\\ &=30\sqrt{3}\implies\boxed{\mathrm{B}} \end{align*} 注：一旦求出 $DY$，就不需要做三角函数。注意该六边形由两个梯形 $ABPQ$ 与 $CDPQ$ 组成。$PQ = \frac{19-7-5 +11}{2} = 9$。高为 $BY$ 的一半，即 $\frac{5\sqrt{3}}{4}$。因此面积为 \[\frac{1}{2} \cdot \frac{5\sqrt{3}}{4}(19+9+11+9)=30\sqrt{3}\]

Topics