Sectors, Segments, Arcs – Extra Practice
1
ˆ B = ,   0. The area of ΔAOB is three times
The diagram shows a circle centre O and radius 1, with AO
the shaded area.
Find the value of .
2
The diagrams show a circular sector of radius 10 cm and angle θ radians which is formed into a cone of slant
height 10 cm. The vertical height h of the cone is equal to the radius r of its base. Find the angle θ radians.
10cm
10cm
h
r
3
The following diagram shows a circle with radius r and centre O. The points A, B and C are on the circle and
AÔC =.
The area of sector OABC is
Find the value of r and of .
4
2
 and the length of arc ABC is .
3
3
4.
The diagram below shows a triangle and two arcs of circles.
The triangle ABC is a right-angled isosceles triangle, with AB = AC = 2. The point P is the midpoint of
[BC].
The arc BDC is part of a circle with centre A.
The arc BEC is part of a circle with centre P.
E
B
D
P
2
A
5
C
2
(a)
Calculate the area of the segment BDCP.
(b)
Calculate the area of the shaded region BECD.
The diagram below shows a circle, centre O, with a radius 12 cm. The chord AB subtends at an angle of 75°
at the centre. The tangents to the circle at A and at B meet at P.
A
12 cm
P diagram not to
scale
O 75º
B
(a)
Using the cosine rule, show that the length of AB is 12 21 – cos 75 .
(2)
(b)
Find the length of BP.
(3)
(c)
Hence find
(i)
the area of triangle OBP;
(ii)
the area of triangle ABP.
(4)
(d)
Find the area of sector OAB.
(2)
(e)
Find the area of the shaded region.
(2)
(Total 13 marks)
6
The following diagram shows two semi-circles. The larger one has centre O and radius 4 cm. The smaller one
has centre P, radius 3 cm, and passes through O. The line (OP) meets the larger semi-circle at S. The semicircles intersect at Q.
(a)
(i)
Explain why OPQ is an isosceles triangle.
(ii)
Use the cosine rule to show that cos OP̂Q =
(iii)
Hence show that sin OP̂Q =
(iv)
Find the area of the triangle OPQ.
1
.
9
80
.
9
(7)
(b)
Consider the smaller semi-circle, with centre P.
(i)
Write down the size of OP̂Q.
(ii)
Calculate the area of the sector OPQ.
(3)
(c)
Consider the larger semi-circle, with centre O. Calculate the area of the sector QOS.
(3)
(d)
Hence calculate the area of the shaded region.
(4)
(Total 17 marks)
7
The following diagram shows a sector of a circle of radius r cm, and angle  at the centre. The perimeter of
the sector is 20 cm.
20  2r
.
r
(a)
Show that  =
(b)
The area of the sector is 25 cm2. Find the value of r.
8
The following diagram shows a semicircle centre O, diameter [AB], with radius 2.
Let P be a point on the circumference, with PÔB =  radians.
(a)
Find the area of the triangle OPB, in terms of .
(2)
(b)
Explain why the area of triangle OPA is the same as the area triangle OPB.
(3)
Let S be the total area of the two segments shaded in the diagram below.
(c)
Show that S = 2( − 2 sin  ).
(3)
(d)
Find the value of  when S is a local minimum, justifying that it is a minimum.
(8)
(e)
Find a value of  for which S has its greatest value.
(2)
(Total 18 marks)
ANSWERS
Q1.
Area of sector =
θ
2
Area of triangle =
(A1)
sin θ
2
(A1)
sin θ  θ sin θ 
 3 

2
2 
2
M1A1
3 = 4 sin 
(A1)
 = 1.28 radians
A1
Note:
N2
Accept 73.1.
[6]
Q2.
h = r so 2r2 = 100  r2 = 50
l = 10 = 2r
2 π 50
=
10
2 π5 2
=
10
 =  2 = 4.44 (3sf)
(M1)
(M1)
(A1)
(A1)
(C4)
Note: Accept either answer.
[4]
Q3.
METHOD 1
Evidence of correctly substituting into A =
1 2
r θ
2
A1
Evidence of correctly substituting into l = r
A1
For attempting to eliminate one variable …
leading to a correct equation in one variable
(M1)
A1
r=4
=

6
(= 0.524, 30)
A1A1
METHOD 2
Setting up and equating ratios
4
2


3  3
r 2 2r
Solving gives r = 4
r =
2  1 2 4 
  or r    
3  2
3 
(M1)
A1A1
A1
A1
N3
=
r=4

 0.524 , 30
6
=
A1

 0.524 , 30
6
N3
[6]
Q4.
(a)
(b)
area of sector ΑΒDC =
1
π(2)2 = π
4
(A1)
area of segment BDCP = π – area of ABC
=π–2
(M1)
(A1) (C3)
BP =
(A1)
2
area of semicircle of radius BP =
1
π( 2 )2 = π
2
area of shaded region = π – (π – 2) = 2
(A1)
(A1) (C3)
[6]
Q5.
Note: Do not penalize missing units in this question.
(a)
AB2
= 122(2 – 2 cos 75°)
= 122 × 2(1 cos 75°)
AB = 12 2(1  cos 75)
= 122 + 122 – 2 × 12 × 12 × cos 75° (A1)
(A1)
(AG)
2
Note: The second (A1) is for transforming the initial expression to any
simplified expression from which the given result can be clearly seen.
(b)
PÔB = 37.5°
BP = 12 tan 37.5°
= 9.21 cm
(A1)
(M1)
(A1)
OR
BP̂A = 105°
BÂP = 37.5°
AB
BP

sin 105 sin 37.5
AB sin 37.5
BP =
= 9.21(cm)
sin 105 
(c)
(i)
(ii)
(d)
(e)
1
 1

 12  9.21  or  12  12 tan 37.5 
2
2


= 55.3 (cm2) (accept 55.2 cm2)
Area ∆OBP =
1
(9.21)2 sin105°
2
= 41.0 (cm2) (accept 40.9 cm2)
Area ∆ABP =
1
π  75

 12 2  75 
 π  12 2 
 or
2
180  360

2
2
= 94.2 (cm ) (accept 30π or 94.3 (cm ))
Area of sector =
Shaded area = 2 × area ∆OPB – area sector
= 16.4 (cm2) (accept 16.2 cm2, 16.3 cm2)
(A1)
(M1)
(A1)
3
(M1)
(A1)
(M1)
(A1)
4
(M1)
(A1)
2
(M1)
(A1)
2
[13]
Q6. (a)
(i)
(ii)
(iii)
OP = PQ (= 3cm)
So  OPQ is isosceles
9  9 16  2 
 
18
 18 
A1
cos OP̂Q =
1
9
AG
Evidence of using sin2 A + cos2 A = 1
1
81


  80 

81 

A1
80
9
AG
1
 OP  PQ sin P
2
80
2

20
  4.47 
Evidence of using formula for area of a sector
eg Area sector OPQ =
QÔP =
 1.4594 
 0.841
2
Area sector QOS =
1 2
 4  0.841
2
= 6.73
(d)
A1
N1
A1
N1
(M1)
1 2
 3 1.4594 
2
= 6.57
(c)
M1
OP̂Q = 1.4594...
OP̂Q = 1.46
(ii)
N0
1
80 9
3 3
,  0.9938 
2
9 2
Area triangle OPQ =
(i)
N0
M1
Evidence of using area triangle OPQ =
eg
N0
(M1)
cos OP̂Q =
sin OP̂Q =
(b)
32  32  4 2
2  3 3
Using cos rule correctly eg cos OP̂Q =
sin OP̂Q = 1 
(iv)
R1
AG
A1
N2
(A1)
A1
A1
Area of small semi-circle is 4.5 (= 14.137...)
A1
Evidence of correct approach
eg Area = area of semi-circle  area sector OPQ  area sector QOS +
area triangle POQ
M1
Correct expression
eg 4.5  6.5675...  6.7285... + 4.472..., 4.5  (6.7285... + 2.095...),
A1
N2
4.5 (6.5675... + 2.256...)
Area of the shaded region = 5.31
A1
N1
[17]
Q7.
(a)
For using perimeter = r + r + arc length
20 = 2r + r

(b)
20  2r
r
Finding A =
(M1)
A1
AG

1 2  20  2r 
2
r 
 10r  r
2  r 

For setting up equation in r
Correct simplified equation, or sketch
eg 10r – r2 = 25, r2 – 10r + 25 = 0
r = 5 cm
N0
(A1)
M1
(A1)
A1
N2
[6]
Q8. (a)
evidence of using area of a triangle
(M1)
1
eg A   2  2  sin θ
2
A = 2 sin 
(b)
A1
N2
METHOD 1
PÔA =   
1
2  2  sin   θ  (= 2 sin (  ))
2
since sin (  ) = sin 
then both triangles have the same area
area OPA =
(A1)
A1
R1
AG
N0
R3
AG
N0
METHOD 2
triangle OPA has the same height and the same base as triangle OPB
then both triangles have the same area
(c)
(d)
1
2
 2  2
2
area  APB = 2 sin  + 2 sin  (= 4 sin )
S = area of semicircle  area APB (= 2  4 sin )
S = 2( − 2 sin )
area semi-circle =
A1
A1
M1
AG
N0
METHOD 1
attempt to differentiate
(M1)
ds
  4 cos θ
dθ
setting derivative equal to 0
(M1)
eg
correct equation
A1
eg 4 cos  = 0, cos  = 0, 4 cos  = 0
=

2
A1
EITHER
evidence of using second derivative
(M1)
N3
S() = 4 sin 
A1
 
S    4
2
A1

it is a minimum because S    0
2
R1
N0
OR
evidence of using first derivative
(M1)
for  <

, S () < 0
2
(may use diagram)
A1
for  >

, S () > 0
2
(may use diagram)
A1
it is a minimum since the derivative goes from negative
to positive
R1
N0
METHOD 2
2  4 sin  is minimum when 4 sin  is a maximum
4 sin  is a maximum when sin  = 1
=
(e)

2
S is greatest when 4 sin  is smallest (or equivalent)
 = 0 (or )
R3
(A2)
A3
N3
(R1)
A1
N2
[18]