Mathematical Methods 11
Arc length and area of sector – practice questions
1. Find the exact value of the perimeter of the shaded figure:
a) Figure 1
2. The diagram alongside shows a sector of a circle of radius 9 cm.
The sector has a perimeter of 25 cm. Work out the value of x in
degrees. Give your answer correct to one decimal place.
3. The diagram alongside shows an equilateral triangle ABC with
sides of length 6 cm.
P is the midpoint of AB.
Q is the midpoint of AC.
APQ is a sector of a circle, centre A.
Calculate the exact value of the area of the shaded region.
4. The diagram alongside shows a sector OABC of a circle with
centre O.
OA = OC = 10.4 cm.
Angle AOC = 120°.
a) Calculate the perimeter of the sector OABC. Give your
answer correct to 3 significant figures.
b) Calculate the area of the shaded region. Give your answer
correct to 3 significant figures.
b) Figure 2
Mathematical Methods 11
Solutions
1.
9
a) The arc length 𝑠 = 𝑟𝜃 = (2) 𝜋
s
9
Hence, the perimeter of the shaded figure = 2(15) + 2 (2 𝜋)
= 30 + 9𝜋 units
𝜋
120° = 120 (180) =
b)
2𝜋
3
4
2
2𝜋
3
The arc length 𝑠 = 𝑟𝜃 = ( ) ( ) =
4𝜋
3
4
Hence, the perimeter of the shaded figure = 3(4) + 3 (3 𝜋)
= 12 + 4𝜋 units
s
𝑥
)
180
𝑥
= 2(9) + 20
2. The arc length 𝑠 = 𝑟𝜃 = 9 (
25 = the perimeter
𝑥
= 25 − 18
20
𝑥 = 7 × 20
𝑥 = 14.0°
=
𝑥
20
s
Since Q is the midpoint of AC so 𝐴𝑄 = 3 𝑐𝑚.
𝜋
Since triangle ABC is equilateral, angle 𝐵𝐴̂𝐶 = 60° or 3 and BQ is the
3.
height of triangle ABC.
Use Pythagoras, length BQ = √62 − 32 = √27
1
Area of triangle ABC = (6)(√27) = 3√27
2
1
1
𝜋
3𝜋
2
3𝜋
3√27 − 2
Area of sector APQ = 2 𝑟 2 𝜃 = 2 (32 ) ( 3 ) =
Hence area of the shaded region =
𝜋
4. 120° = 120 (180) =
=
6√27−3𝜋
2
2𝜋
3
2𝜋
a) The arc length 𝐴𝐵𝐶 = 𝑟𝜃 = 10.4 ( 3 ) =
20.8𝜋
3
Hence the perimeter of the sector OABC = 2(10.4) +
20.8𝜋
3
= 42.6 𝑐𝑚
b) Let OD be the height of triangle AOC.
Since triangle AOC is isosceles, 𝐴𝑂̂𝐷 = 60°.
Using trigonometry in right – angled triangle, 𝑂𝐷 = 10.4(cos 60°) = 5.2 𝑐𝑚
and 𝐴𝐶 = 2𝐴𝐷 = 2(10.4)(sin 60°) = 18.01 𝑐𝑚
1
The area of triangle AOC = 2 (18.01)(5.2) = 46.8 𝑐𝑚2
1
1
2𝜋
The area of the sector OABC = 2 𝑟 2 𝜃 = 2 (10.4)2 ( 3 ) = 113 𝑐𝑚2
Hence the area of shaded region = 113 − 46.8 = 66.2 𝑐𝑚2
D
𝑐𝑚2