Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2024 June Paper 13 Q3

A Level / CIE / P1

CIE 9709 2024 June Paper 13 Paper · Question 3

题目

Problem

The diagram shows a sector of a circle with centre CC. The radii CACA and CBCB each have length rr cm and the size of the reflex angle ACBACB is θ\theta radians. The sector, shaded in the diagram, has a perimeter of 6565 cm and an area of 225 cm2225 \text{ cm}^2.

(a) Find the values of rr and θ\theta.

[4]

(b) Find the area of triangle ACBACB.

[2]
题目中文翻译

图中显示一个圆心为 CC 的扇形。半径 CACA 和 CBCB 的长度均为 rr cm,反角 ACBACB 的大小为 θ\theta 弧度。图中阴影扇形的周长为 6565 cm,面积为 225 cm2225 \text{ cm}^2。

(a) 求 rr 和 θ\theta 的值。

[4]

(b) 求三角形 ACBACB 的面积。

[2]

解答

(a)

解法一

思路

展开

分别用扇形的周长和面积列式。消去角度后求两个半径,再检查 θ\theta 必须是优角。

答题过程

展开

The perimeter and area give

2r+rθ=65,12r2θ=225.2r+r\theta=65,\qquad \frac12r^2\theta=225.

Thus θ=450/r2\theta=450/r^2, and substitution yields

2r2−65r+450=(r−10)(2r−45)=0.2r^2-65r+450=(r-10)(2r-45)=0.

For r=10r=10, θ=4.5\theta=4.5, which is between π\pi and 2π2\pi. For r=22.5r=22.5, θ=8/9\theta=8/9, which is not reflex. Hence r=10r=10 cm and θ=4.5\theta=4.5 radians.

(b)

解法一

思路

展开

三角形的夹角是小于 π\pi 的角 2π−θ2\pi-\theta,并非扇形的优角。

答题过程

展开

Using the two radii and the included angle, the area is

12(10)(10)sin⁡(2π−4.5)≈48.9 cm2.\frac12(10)(10)\sin(2\pi-4.5)\approx48.9\text{ cm}^2.