AMC10 2012 B

AMC10 2012 B · Q12

AMC10 2012 B · Q12. It mainly tests Triangles (properties), Pythagorean theorem.

Point B is due east of point A. Point C is due north of point B. The distance between points A and C is $10\sqrt{2}$ meters, and $\angle BAC = 45^\circ$. Point D is 20 meters due north of point C. The distance AD is between which two integers?

点B在点A的正东方向。点C在点B的正北方向。点A和点C之间的距离是 $10\sqrt{2}$ 米，且 $\angle BAC = 45^\circ$。点D在点C的正北20米。距离AD在哪两个整数之间？

(A) 30 and 31 30 和 31

(B) 31 and 32 31 和 32

(D) 33 and 34 33 和 34

(E) 34 and 35 34 和 35

Answer

Correct choice: (B)

正确答案：（B）

Solution

. Answer (B): Note that $\angle ABC = 90^\circ$, so $\triangle ABC$ is a $45$-$45$-$90^\circ$ triangle. Because hypotenuse $AC = 10\sqrt{2}$, the legs of $\triangle ABC$ have length $10$. Therefore $AB = 10$ and $BD = BC + CD = 10 + 20 = 30$. By the Pythagorean Theorem, $AD = \sqrt{10^2 + 30^2} = \sqrt{1000}.$ Because $31^2 = 961$ and $32^2 = 1024$, it follows that $31 < AD < 32$.

. 答案（B）：注意 $\angle ABC = 90^\circ$，所以 $\triangle ABC$ 是一个 $45$-$45$-$90^\circ$ 三角形。因为斜边 $AC = 10\sqrt{2}$，所以 $\triangle ABC$ 的两条直角边长度都是 $10$。因此 $AB = 10$，并且 $BD = BC + CD = 10 + 20 = 30$。由勾股定理， $AD = \sqrt{10^2 + 30^2} = \sqrt{1000}.$ 因为 $31^2 = 961$ 且 $32^2 = 1024$，所以 $31 < AD < 32$。

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