Transcript ADDITIONAL MATHEMATICS

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ADDITIONAL
MATHEMATICS
CIRCULAR MEASURES
6 Questions
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
O
answer
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 radians
OR = ¾ OQ
= ¾(24)
= 18 cm
cos 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 = 8/6
θ = 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 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)
= 17.67 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
= 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
SOT
½ r²θ = ½ r²θ
½ (5)²(1.6) = ½ (r²)(0.8)
20 = 0.4r²
r² = 50
r = 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 = 50
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)
= 7.926 cm² (answer)
Back to Question 5
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