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CIE 9709 2022 March Paper 12 Q10

A Level / CIE / P1

CIE 9709 2022 March Paper 12 Paper · Question 10

题目

Problem

The diagram shows a circle with centre AA of radius 5 cm5\text{ cm} and a circle with centre BB of radius 8 cm8\text{ cm}. The circles touch at the point CC so that ACBACB is a straight line. The tangent at the point DD on the smaller circle intersects the larger circle at EE and passes through BB.

(a) Find the perimeter of the shaded region.

(5)

(b) Find the area of the shaded region.

(3)
题目中文翻译

图中所示为一个圆心为 AA、半径为 5 cm5\text{ cm} 的圆,以及一个圆心为 BB、半径为 8 cm8\text{ cm} 的圆。两圆在点 CC 处相切,AA、CC、BB 在同一直线上。较小的圆在点 DD 处的切线与较大的圆交于点 EE,并经过点 BB。

(a) 求阴影区域的周长。

(5)

(b) 求阴影区域的面积。

(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

展开

半径垂直于切线,故 ADBADB 是直角三角形。用它求直边和两个扇形的圆心角。

答题过程

展开

The centres are AB=5+8=13AB=5+8=13 cm apart. Since AD=5AD=5 is perpendicular to the tangent, DB=132−52=12DB=\sqrt{13^2-5^2}=12 cm. The point EE lies between DD and BB, so DE=12−8=4DE=12-8=4 cm. The sector angles in radians are

α=arctan⁡(12/5),β=arctan⁡(5/12)\alpha=\arctan(12/5),\qquad \beta=\arctan(5/12)

at AA and BB respectively. The perimeter is 5α+8β+4=13.05\alpha+8\beta+4=13.0 cm to three significant figures.

(b)

解法一

思路

展开

阴影位于直角三角形内。用三角形面积减去两个扇形面积。

答题过程

展开

The area of triangle ADBADB is 12(5)(12)=30\tfrac12(5)(12)=30 square centimetres. Using the unrounded angles from (a), the shaded area is

30−252α−32β=2.666619… cm2,30-\frac{25}{2}\alpha-32\beta=2.666619\ldots\,\mathrm{cm}^2,

which is 2.67 cm22.67\,\mathrm{cm}^2 to three significant figures.