ADDITIONAL MATHEMATICS

CIRCULAR MEASURES

6 Questions

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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

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

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

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

answer

O

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

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

SOLUTION TO QUESTION 1

(a)Find θ in radiansOR = ¾ OQ = ¾(24) = 18 cmcos θ = 18/24 = 41.41 x π/180 = 0.7227 rad (answer)

(b)Area of POQ – Area of POR= ½ r²θ – ½ (OR)(PR)= ½ (24)²(0.7227) – ½ (18)(15.87)= 65.31 cm² (answer)

Back to Question 1

SOLUTION TO QUESTION 2

(a)tan θ = 8/6θ = 53°13’angle POR = 53°13’ x 2 = 106.26°106.26° x π/180= 1.855 rad (answer)

(b)S = rθ = 10(1.855) = 18.55cm (answer)

(c)½ r²(θ – sinθ)= ½ (10)²(1.855 – sin 1.855)= 44.76 cm² (answer)

Back to Question 2

SOLUTION TO QUESTION 3

(a)OS = 3/5 OPOP = 5/3 OS = 5/3(9) = 15cmGiven OS = 8 cmSP = OP – OS = 15 – 8 = 7 cm (answer)

(b)S = rθPQ = rθ = 15(3/8 π) = 17.67 cmAngle SOR = 3/8 π x 180/π = 67.5°sin 67.5° = SR/8cmSR = 7.39 cmTo find RQtan 67.5° = 7.39/OROR = 3.06 cmRQ = OQ – OR = 15 – 3.06 = 11.94cmPerimeter = arc PQ + SP + SR + RQ = 17.67 + 7 + 7.39 + 11.94 = 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)= 121.23cm² (answer)

Back to Question 3

SOLUTION TO QUESTION 4

(a)Area sector POR = Area sector SOT½ r²θ = ½ r²θ½ (5)²(1.6) = ½ (r²)(0.8)20 = 0.4r²r² = 50r = 7.071 cm (answer)

(b)Area sector POR + Area sector SOT= ½ (5)²(0.8) + ½ (7.071)²(0.8)= 10 + 20= 30 cm² (answer)

Back to Question 4

SOLUTION TO QUESTION 5

(a)Area of sector = 50 cm²]½ r²θ = 50½ (10)²θ = 50θ = 1 rad (answer)

(b)Let the midpoint of PQ be r0.5 x 180/π = 28.65°sin 28.65° = Pr / 10Pr = 4.79 cmLength of chord PQ = 4.79 x 2 = 9.58 cm (answer)

(c)Use ½ r²(θ – sinθ)= ½ (10)² (1 – sin 1)= 7.926 cm² (answer)

Back to Question 5

SOLUTION TO QUESTION 6

(a)S = rθ11.25 = 9θθ = 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