3.16H Area and Circumference of Circles: Higher Tier Practice

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 75 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Work out the circumference of a circle with radius 7 cm. Give your answer in terms of π.

(Total for Question 1 is 2 marks)

A circular badge has diameter 10 cm. Work out the area of the badge. Give your answer in terms of π.

(Total for Question 2 is 3 marks)

A sector of a circle has radius 9 cm and angle 120 degrees. Work out the length of the arc of the sector. Give your answer in terms of π.

(Total for Question 3 is 4 marks)

A sector has radius 6 cm and central angle 150 degrees. Work out the area of the sector. Give your answer in terms of π.

(Total for Question 4 is 5 marks)

A small playing surface is made from a rectangle of length 20 cm and width 14 cm with a semicircle attached to one width so that the diameter of the semicircle equals the width. Find the perimeter of the whole shape (the outer boundary). Give your answer in terms of π.

(Total for Question 5 is 6 marks)

Show that the length of the chord where the line y = 3 intersects the circle x² + y² = 49 is 4 √10. Give clear working.

(Total for Question 6 is 8 marks)

A circle has centre O and radius 12 cm. A chord of the circle has length 16 cm. Find the area of the isosceles triangle formed by the two radii to the chord endpoints and the chord. Give your answer exactly in simplest surd form.

(Total for Question 7 is 12 marks)

Mark scheme · 3.16H Area and Circumference of Circles: Higher Tier Practice

Question 1

M1 use circumference = 2 π r or π d
A1 14 π cm cao
Answer: 14 π cm

Question 2

M1 use area = π r² with r = 5
A1 25 π cao
B1 include units: cm²
Answer: 25 π cm²

Question 3

M1 method: arc length = (angle/360) x circumference = (angle/360) x 2 π r
M1 substitute angle = 120, r = 9 to get (120/360) x 2 π x 9
A1 6 π cao
B1 units: cm
Answer: 6 π cm

Question 4

M1 use sector area = (angle/360) x π r²
M1 substitute angle = 150, r = 6 so (150/360) x π x 36
A1 15 π cao
B1 simplification shown: (150/360) = 5/12 and 36 x 5/12 = 15
B1 units: cm²
Answer: 15 π cm²

Question 5

M1 identify which sides are on the outer boundary: two lengths 20, one width 14 and semicircular arc
M1 find semicircle arc length = half circumference = (1/2) x 2 π r = π r with diameter 14 so r = 7
M1 compute semicircle arc = 7 π
M1 sum the straight sides: 20 + 20 + 14 = 54
A1 final exact perimeter 54 + 7 π cao
B1 units: cm
Answer: 54 + 7 π cm

Question 6

dM1 substitute y = 3 into x² + y² = 49 to obtain x² + 9 = 49 or x² = 40
M1 take square root to find x = ± √40
A1 simplify √40 = 2 √10
dM1 chord length = difference between x-coordinates = 2 x where x = 2 √10
M1 compute chord length = 2 x = 4 √10
A1 4 √10 cao
B1 clear statement linking substitution to chord length
B1 correct final presentation, no units required
Answer: 4 √10

Question 7

M1 identify relationship: chord length c = 2 R sin(θ/2); substitute c = 16, R = 12 to get sin(θ/2) = 16/(2x12) = 16/24 = 2/3
M1 state θ/2 satisfies sin(θ/2) = 2/3 and so cos(θ/2) = √1 - (2/3)² = √5/3
M1 use double-angle identity sin θ = 2 sin(θ/2) cos(θ/2)
M1 substitute values to get sin θ = 2 x (2/3) x (√5/3) = 4 √5 / 9
M1 use triangle area formula for two sides R and included angle θ: area = (1/2) R² sin θ
M1 substitute R = 12 and sin θ = 4 √5 / 9 to get area = 1/2 x 144 x (4 √5 / 9)
A1 simplify: 72 x (4 √5 / 9) cao
A1 further simplification: (288 √5) / 9 cao
A1 reduce fraction: 32 √5 cao
B1 show exact surd form and state units cm²
B1 clear logical sequence linking chord to area (AO3 communication)
A1 final answer 32 √5 cm² cao
Answer: 32 √5 cm²

Area and Circumference of Circles: Higher Tier Practice - Worksheets, Questions and Revision