Download presentation

Presentation is loading. Please wait.

1
**ADDITIONAL MATHEMATICS**

Click here for more downloads ADDITIONAL MATHEMATICS CIRCULAR MEASURES 6 Questions

2
Question 1 The diagram shows a sector POQ with center O and a radius 24cm. Point R and OQ is such that OR:RQ = 3.1. Calculate (a) The value of Q in radians (b) The area in cm² of the shaded region answer

3
Question 2 The diagram shows a circle with centre O and a radius of 6cm. PQR is a tangent to the circle at Q. Given PQ=QR=8cm and arc PSR is drawn with O as it centre. Calculate: (a) The angle θ in radians (b) The length in cm of arc PSR (c) The area in cm of the shaded region answer

4
Question 3 The diagram shows a sector POQ of a circle O. Point S lies on OP, point R lies on OQ and SR is perpendicular to OQ. The length of OS is 8cm and POQ=3/8 π radian. Given that OS:OP = 3:5. Calculate (a) Calculate the length of SP (b) Calculate the perimeter of shaded region (c) Calculate area of shaded region answer

5
Question 4 The diagram shows two sectors OPQR and OST with centre O and of equal area. Given that OQS and ORT are straight lines and POQ = QOR rad, calculate: (a) The radius of sector OST (b) The area of the whole shape O answer

6
Question 5 The diagram shows a sector OPQ with centre O and a radius of 10cm. Given then the area of sector is 50cm, calculate (a) The value of θ in radians (b) The length of chord PQ (c) The area of segment bounded by arc PQ and chord PQ answer

7
Question 6 The diagram shows a sector OPQR with centre O and a radius of 9cm. The length of arc PQR is 11.25cm. Calculate : (a) ∠ POR in radians (b) The area of shaded segment answer

8
**Solution to Question 1 (a) (b) Find θ in radians**

OR = ¾ OQ = ¾(24) = 18 cm cos θ = 18/24 = x π/180 = rad (answer) (b) Area of POQ – Area of POR = ½ r²θ – ½ (OR)(PR) = ½ (24)²(0.7227) – ½ (18)(15.87) = cm² (answer) Back to Question 1

9
**Solution to Question 2 (a) (b) tan θ = 8/6 S = rθ θ = 53°13’**

angle POR = 53°13’ x 2 = ° 106.26° x π/180 = rad (answer) (b) S = rθ = 10(1.855) = 18.55cm (answer) (c) ½ r²(θ – sinθ) = ½ (10)²(1.855 – sin 1.855) = cm² (answer) Back to Question 2

10
**Solution to Question 3 (a) OS = 3/5 OP OP = 5/3 OS = 5/3(9) = 15cm**

Given OS = 8 cm SP = OP – OS = 15 – 8 = 7 cm (answer) (b) S = rθ PQ = rθ = 15(3/8 π) = cm Angle SOR = 3/8 π x 180/π = 67.5° sin 67.5° = SR/8cm SR = 7.39 cm To find RQ tan 67.5° = 7.39/OR OR = 3.06 cm RQ = OQ – OR = 15 – 3.06 = 11.94cm Perimeter = arc PQ + SP + SR + RQ = = 44cm(answer) (c) Area of sector POQ – Area of triangle SOR = ½ r²θ – ½ (OR)(SR) = ½ (15)²(3/8 π) – ½ (3.06)(7.39) = cm² (answer) Back to Question 3

11
**Solution to Question 4 (a) (b) Area sector POR = Area sector SOT**

½ r²θ = ½ r²θ ½ (5)²(1.6) = ½ (r²)(0.8) 20 = 0.4r² r² = 50 r = cm (answer) (b) Area sector POR + Area sector SOT = ½ (5)²(0.8) + ½ (7.071)²(0.8) = = 30 cm² (answer) Back to Question 4

12
**Solution to Question 5 Back to Question 5 (a) Area of sector = 50 cm²]**

½ (10)²θ = 50 θ = 1 rad (answer) (b) Let the midpoint of PQ be r 0.5 x 180/π = 28.65° sin 28.65° = Pr / 10 Pr = 4.79 cm Length of chord PQ = 4.79 x 2 = 9.58 cm (answer) (c) Use ½ r²(θ – sinθ) = ½ (10)² (1 – sin 1) = cm² (answer) Back to Question 5

13
**Solution to Question 6 (a) S = rθ 11.25 = 9θ θ = 1.25 rad (answer) (b)**

Use ½ r²(θ – sinθ) = ½ (9)²(1.25 – sin 1.25) = 12.19cm² (answer) Back to Question 6

Similar presentations

OK

Areas of Segments of Circles SWBAT: To find the areas of segments of circles.

Areas of Segments of Circles SWBAT: To find the areas of segments of circles.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on professional development Ppt on idea cellular advertisement Ppt on teaching english as a second language Ppt on agriculture in punjab Ppt on eid Production in the long run ppt on tv Ppt on ms access 2007 Ppt on leverages Hrm ppt on recruitment companies Ppt on conservation of environmental resources