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CIE 9709 2022 November Paper 12 Q10

A Level / CIE / P1

CIE 9709 2022 Nov Paper 12 Paper · Question 10

题目

Problem

The diagram shows a cross-section RASBRASB of the body of an aircraft. The cross-section consists of a sector OARBOARB of a circle of radius 2.5 m2.5\text{ m}, with centre OO, a sector PASBPASB of another circle of radius 2.24 m2.24\text{ m} with centre PP and a quadrilateral OAPBOAPB. Angle AOB=23πAOB=\dfrac{2}{3}\pi and angle APB=56πAPB=\dfrac{5}{6}\pi.

(a) Find the perimeter of the cross-section RASBRASB, giving your answer correct to 2 decimal places.

(3)

(b) Find the difference in area of the two triangles AOBAOB and APBAPB, giving your answer correct to 2 decimal places.

(2)

(c) Find the area of the cross-section RASBRASB, giving your answer correct to 1 decimal place.

(3)
题目中文翻译

图中显示一架飞机机身的横截面 RASBRASB。该横截面由圆心为 OO、半径为 2.5 m2.5\text{ m} 的圆的扇形 OARBOARB,圆心为 PP、半径为 2.24 m2.24\text{ m} 的另一圆的扇形 PASBPASB,以及四边形 OAPBOAPB 组成。角 AOB=23πAOB=\dfrac{2}{3}\pi,角 APB=56πAPB=\dfrac{5}{6}\pi。

(a) 求横截面 RASBRASB 的周长,答案精确到小数点后 2 位。

(3)

(b) 求三角形 AOBAOB 与 APBAPB 的面积之差,答案精确到小数点后 2 位。

(2)

(c) 求横截面 RASBRASB 的面积,答案精确到小数点后 1 位。

(3)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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周长只包括上方的优弧和下方的劣弧。上方圆心角需用一周减去题给的角。

答题过程

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The arc ARBARB has angle 2π−2π/3=4π/32\pi-2\pi/3=4\pi/3 at OO. The arc ASBASB has angle 5π/65\pi/6 at PP. The perimeter is

 2.54π3+2.245π6= 16.34 m\begin{align*} &\,2.5\frac{4\pi}{3}+2.24\frac{5\pi}{6}\\ =&\,16.34\,\mathrm m \end{align*}

to two decimal places.

(b)

解法一

思路

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两三角形都已知两边与夹角。分别用正弦面积公式,再相减。

答题过程

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In square metres the difference is

 12(2.5)2sin⁡2π3− 12(2.24)2sin⁡5π6= 1.451929…,\begin{align*} &\,\frac12(2.5)^2\sin\frac{2\pi}{3}\\ -&\,\frac12(2.24)^2\sin\frac{5\pi}{6}\\ =&\,1.451929\ldots, \end{align*}

or 1.45 m21.45\,\mathrm m^2 to two decimal places.

(c)

解法一

思路

展开

截面面积等于两个扇形加中间四边形。图中 PP 与 OO 在弦 ABAB 同侧,所以四边形是两三角形之差。

答题过程

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The two sectors have angles 4π/34\pi/3 and 5π/65\pi/6. Using the unrounded triangle-area difference from (b), the area is

 12(2.5)24π3+ 12(2.24)25π6+ 1.451929…= 21.109921… m2.\begin{align*} &\,\frac12(2.5)^2\frac{4\pi}{3}\\ +&\,\frac12(2.24)^2\frac{5\pi}{6}\\ +&\,1.451929\ldots\\ =&\,21.109921\ldots\,\mathrm m^2. \end{align*}

The cross-sectional area is 21.1 m221.1\,\mathrm m^2 to one decimal place.