AMC12 2021 B

AMC12 2021 B · Q15

AMC12 2021 B · Q15. It mainly tests Triangles (properties), Pythagorean theorem.

The figure is constructed from $11$ line segments, each of which has length $2$. The area of pentagon $ABCDE$ can be written as $\sqrt{m} + \sqrt{n}$, where $m$ and $n$ are positive integers. What is $m + n ?$

该图形由 $11$ 条长度均为 $2$ 的线段构成。五边形 $ABCDE$ 的面积可写为 $\sqrt{m} + \sqrt{n}$，其中 $m$ 和 $n$ 是正整数。$m + n$ 是多少？

(A) 20 20

(B) 21 21

(D) 23 23

(E) 24 24

Answer

Correct choice: (D)

正确答案：（D）

Solution

Draw diagonals $AC$ and $AD$ to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles $ABC$ and $ADE$, they each have area $2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}$. For triangle $ACD$, we can see that $AC=AD=2\sqrt{3}$ and $CD=2$. Using Pythagorean Theorem, the altitude for this triangle is $\sqrt{11}$, so the area is $\sqrt{11}$. Adding each part up, we get $2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{\textbf{(D)} ~23}$.

画对角线 $AC$ 和 $AD$ 将五边形分为三部分。我们可以计算每个三角形的面积并最后求和。对于三角形 $ABC$ 和 $ADE$，每个面积为 $2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}$。对于三角形 $ACD$，可见 $AC=AD=2\sqrt{3}$ 和 $CD=2$。用勾股定理，该三角形的高为 $\sqrt{11}$，因此面积为 $\sqrt{11}$。各部分相加，得 $2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{\textbf{(D)} ~23}$。

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