Question

点 $B$ 在点 $A$ 的正东方向。点 $C$ 在点 $B$ 的正北方向。点 $A$ 与 $C$ 之间的距离为 $10\sqrt 2$ ，且 $\angle BAC = 45^\circ$ 。点 $D$ 在点 $C$ 正北 $20$ 米处。距离 $AD$ 介于哪两个整数之间？

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

$30$ and $31$

$31$ and $32$

$32$ and $33$

$33$ and $34$

$34$ and $35$

答案：B

知识点：勾股定理特殊直角三角形估算

难度评级：1140

解答：

因为 $AB$ 与 $BC$ 垂直，所以又因为 $\angle BAC = 45^\circ$ ， $\triangle ABC$ 是等腰直角三角形，所以 $AB = BC$ ，从而 $2AB^2 = 200$ 。因此 $AB = BC = 10$ 。 $AB^2 + BC^2 = (10\sqrt 2)^2 = 200.$

于是由勾股定理，因为所以 $BD= BC+CD = 30.$ $\begin{align*}AD^2 &= AB^2+BD^2 \\&= 10^2+30^2\\&=1000\end{align*}$ $31^2 < AD^2 < 32^2,$ $31 \lt AD\lt 32$

所以正确答案是 B。

We know $AB$ and $BC$ are perpendicular, so $AB^2 + BC^2 = (10\sqrt 2)^2 = 200.$ Also, as $\angle BAC = 45^\circ,$ we know that $\triangle ABC$ is an isosceles right triangle, so $AB = BC ,$ making $2AB^2 = 200.$ Thus, $AB = BC = 10.$

As such, we know that $BD= BC+CD = 30.$ Thus, by the Pythagorean Theorem, we have that $\begin{align*}AD^2 &= AB^2+BD^2 \\&= 10^2+30^2\\&=1000\end{align*}$ Thus, since $31^2 < AD^2 < 32^2,$ we have $31 \lt AD\lt 32$

Thus, the correct answer is B .

2012 AMC 10B 第 12 题

答案：B

难度评级：1140

解答：

其他年份的第 12 题