AMC10 2002 B

AMC10 2002 B · Q17

AMC10 2002 B · Q17. It mainly tests Area & perimeter, Polygons.

A regular octagon ABCDEFGH has sides of length two. Find the area of $\triangle ADG$.

一个边长为二的正八边形 ABCDEFGH，求 $\triangle ADG$ 的面积。

(A) $4 + 2\sqrt{2}$ $4 + 2\sqrt{2}$

(B) $6 + \sqrt{2}$ $6 + \sqrt{2}$

(D) $3 + 4\sqrt{2}$ $3 + 4\sqrt{2}$

(E) $8 + \sqrt{2}$ $8 + \sqrt{2}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

(C) Construct the right triangle $\triangle AOB$ as shown in the figure. Since $AB=2$, we have $AO=\sqrt{2}$ and $AD=2+2\sqrt{2}$. Similarly, we have $OG=2+\sqrt{2}$, so $$ \text{Area}(\triangle ADG)=\frac{1}{2}(2+2\sqrt{2})(2+\sqrt{2})=(1+\sqrt{2})(2+\sqrt{2})=4+3\sqrt{2}. $$

（C）按图所示作直角三角形 $\triangle AOB$。由于 $AB=2$，可得 $AO=\sqrt{2}$，且 $AD=2+2\sqrt{2}$。同理，$OG=2+\sqrt{2}$，因此 $$ \text{面积}(\triangle ADG)=\frac{1}{2}(2+2\sqrt{2})(2+\sqrt{2})=(1+\sqrt{2})(2+\sqrt{2})=4+3\sqrt{2}. $$

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