首页 > AMC > AMC10实战

AMC10每日一题（2002年真题#06）

　　今天课窝小编为大家整理了AMC10真题练习，希望考生们认真阅读，能够对你的考试有所帮助。

AMC课程,AMC报名信息,AMC数学竞赛,AMC竞赛培训

Problem

A $45^\circ$ arc of circle A is equal in length to a $30^\circ$ arc of circle B. What is the ratio of circle A's area and circle B's area?

$\text{(A)}\ 4/9 \qquad \text{(B)}\ 2/3 \qquad \text{(C)}\ 5/6 \qquad \text{(D)}\ 3/2 \qquad \text{(E)}\ 9/4$

Solutions

Solution 1

Let $r_1$ and $r_2$ be the radii of circles $A$ and $B$ , respectively.

It is well known that in a circle with radius $r$ , a subtended arc opposite an angle of $\theta$ degrees has length $\frac{\theta}{360} \cdot 2\pi r$ .

Using that here, the arc of circle A has length $\frac{45}{360}\cdot2\pi{r_1}=\frac{r_1\pi}{4}$ . The arc of circle B has length $\frac{30}{360} \cdot 2\pi{r_2}=\frac{r_2\pi}{6}$ . We know that they are equal, so $\frac{r_1\pi}{4}=\frac{r_2\pi}{6}$ , so we multiply through and simplify to get $\frac{r_1}{r_2}=\frac{2}{3}$ . As all circles are similar to one another, the ratio of the areas is just the square of the ratios of the radii, so our answer is $\boxed{\text{(A)}\ 4/9}$ .

Solution 2

Let $c_1$ and $c_2$ be the circumference of circles $A$ and $B$ , respectively.

The length of a $45^{\circ}$ arc of circle $A$ is $\frac{c_1}{\frac{360}{45}}=\frac{c_1}{8}$ , and the length of a $30^{\circ}$ arc of circle $B$ is $\frac{c_2}{\frac{360}{30}}=\frac{c_2}{12}$ . We know that the length of a $45^{\circ}$ arc on circle $A$ is equal to the length of a $30^{\circ}$ arc of circle $B$ , so $\frac{c_1}{8}=\frac{c_2}{12}$ . Manipulating the equation, we get $\frac{c_1}{c_2}=\frac{8}{12}=\frac{2}{3}$ . Because the ratio of the areas is equal to the ratio of the circumferences squared, our answer is $\frac{2^2}{3^2}=\boxed{\text{(A)}\ 4/9}$

Solution 3

The arc of circle $A$ is $\frac{45}{30}=\frac{3}{2}$ that of circle $B$ .

The circumference of circle $A$ is $\frac{2}{3}$ that of circle $B$ ( $B$ is the larger circle).

The radius of circle $A$ is $\frac{2}{3}$ that of circle $B$ .

The area of circle $A$ is ${\left(\frac{2}{3}\right)}^2=\boxed{\text{(A)}\ 4/9}$ that of circle $B$ .

问题

甲 $$ 45 ^ \ $保监会$ 圆A的弧的长度等于一 $$ 30 ^ \ $保监会$ 圆B的弧什么是圆A的面积和圆B的面积之比？

$\ text {（A）} \ 4/9 \ qquad \ text {（B）} \ 2/3 \ qquad \ text {（C）} \ 5/6 \ qquad \ text {（D）} \ 3 / 2 \ qquad \ text {（E）} \ 9/4$

解决方案

解决方案1

设 $$ $ R_1$ 和 $R_2$ 是圆的半径 $A$ 和 $B$ 分别。

众所周知，在具有半径的圆中，与 $R$ 角度相反的对向弧 $$ \ $ THETA$ 具有长度 $\ frac {\ theta} {360} \ cdot 2 \ pi r$ 。

在此使用它，圆圈A的弧度具有长度 $\压裂{45} {360} \ cdot2 \ PI {R_1} = \压裂{R_1 \ PI} {4}$ 。圆弧B具有长度 $\ frac {30} {360} \ cdot 2 \ pi {r_2} = \ frac {r_2 \ pi} {6}$ 。我们知道它们是平等的，所以 $\压裂{R_1 \ PI} {4} = \压裂{R_2 \ PI} {6}$ ，我们通过并简化得到 $\压裂{R_1} {R_2} = \压裂{2} {3}$ 。由于所有圆圈彼此相似，因此面积的比率只是半径比率的平方，所以我们的答案是 $\ boxed {\ text {（A）} \ 4/9}$ 。

解决方案2

设 $C_1$ 和 $C_2$ 是圆的周长 $A$ 和 $B$ 分别。

一个的长度 $45 ^ {\保监会}$ 的圆的弧 $A$ 是 $\压裂{C_1} {\压裂{360} {45}} = \压裂{C_1} {8}$ ，和的长度 $30 ^ {\保监会}$ 的圆的弧 $B$ 是 $\压裂{C_2} {\压裂{360} {30}} = \压裂{C_2} {12}$ 。我们知道 $45 ^ {\保监会}$ 圆上圆弧 $A$ 的长度等于 $30 ^ {\保监会}$ 圆弧的长度 $B$ ，所以 $\压裂{C_1} {8} = \压裂{C_2} {12}$ 。操纵方程，我们得到 $\压裂{C_1} {C_2} = \压裂{8} {12} = \压裂{2} {3}$ 。因为面积的比率等于圆周平方的比率，我们的答案是 $\ frac {2 ^ 2} {3 ^ 2} = \ boxed {\ text {（A）} \ 4/9}$