CalcGospel 國際數學圖譜
返回

P1 2021 June Q7

A Level / Edexcel / P1

IAL 2021 June Paper · Question 7

题目

Problem

The line l1l_1 has equation 4y+3x=484y+3x=48.

The line l1l_1 cuts the yy-axis at the point CC, as shown in Figure 3.

Figure 3

(a) State the yy coordinate of CC.

(1)

The point D(8,6)D(8,6) lies on l1l_1.

The line l2l_2 passes through DD and is perpendicular to l1l_1.

The line l2l_2 cuts the yy-axis at the point EE as shown in Figure 3.

(b) Show that the yy coordinate of EE is −143-\dfrac{14}{3}.

(3)

A sector BCEBCE of a circle with centre CC is also shown in Figure 3.

Given that angle BCEBCE is 1.81.8 radians,

(c) find the length of arc BEBE.

(3)

The region CBEDCBED, shown shaded in Figure 3, consists of the sector BCEBCE joined to the triangle CDECDE.

(d) Calculate the exact area of the region CBEDCBED.

(3)

解答

(a)

解法一

思路

展开

CC 在 yy 轴上，所以令 x=0x=0。

答题过程

展开

When x=0x=0,

4y+3(0)= 48y= 12.\begin{align*} 4y+3(0)=&\,48\\ y=&\,12. \end{align*}

(b)

解法一

思路

展开

先求 l1l_1 的斜率，再取负倒数得到 l2l_2 的斜率。然后用 D(8,6)D(8,6) 写 l2l_2，再令 x=0x=0 找 EE 的 yy 坐标。

答题过程

展开

Rearrange l1l_1:

4y+3x= 48y= −34x+12.\begin{align*} 4y+3x=&\,48\\ y=&\,-\frac34x+12. \end{align*}

So the gradient of l1l_1 is −34-\frac34. Therefore the gradient of l2l_2 is 43\frac43.

Using D(8,6)D(8,6),

y−6= 43(x−8).\begin{align*} y-6=&\,\frac43(x-8). \end{align*}

At EE, x=0x=0, so

y−6= 43(0−8)y−6= −323y= −143.\begin{align*} y-6=&\,\frac43(0-8)\\ y-6=&\,-\frac{32}{3}\\ y=&\,-\frac{14}{3}. \end{align*}

This is the required result.

(c)

解法一

思路

展开

圆心是 CC，所以半径是 CECE。由 (a)(b) 得 CC 和 EE 都在 yy 轴上，直接相减可得半径。

答题过程

展开

The radius is

CE= 12−(−143)= 503.\begin{align*} CE =&\,12-\left(-\frac{14}{3}\right)\\ =&\,\frac{50}{3}. \end{align*}

Arc length is rθr\theta, so

BE= 503(1.8)= 30.\begin{align*} BE =&\,\frac{50}{3}(1.8)\\ =&\,30. \end{align*}

(d)

解法一

思路

展开

区域面积等于扇形 BCEBCE 面积加三角形 CDECDE 面积。扇形用 12r2θ\frac12r^2\theta。三角形 CDECDE 可用底 CE=503CE=\frac{50}{3} 和水平高 88。

答题过程

展开

Area of the sector:

12r2θ= 12(503)2(1.8)= 250.\begin{align*} \frac12r^2\theta =&\,\frac12\left(\frac{50}{3}\right)^2(1.8)\\ =&\,250. \end{align*}

Area of triangle CDECDE:

12(503)(8)=2003.\begin{align*} \frac12\left(\frac{50}{3}\right)(8) =\frac{200}{3}. \end{align*}

Therefore the total area is

250+2003= 9503.\begin{align*} 250+\frac{200}{3} =&\,\frac{950}{3}. \end{align*}