© T Madas

We can use these formulae to find areas of sectors and lengths of arcs The area of a circle The circumference of a circle We can use these formulae to find areas of sectors and lengths of arcs Arc Sector r © T Madas

We can use these formulae to find areas of sectors and lengths of arcs The area of a circle The circumference of a circle We can use these formulae to find areas of sectors and lengths of arcs Area of sector = 50° 6 cm © T Madas

We can use these formulae to find areas of sectors and lengths of arcs The area of a circle The circumference of a circle We can use these formulae to find areas of sectors and lengths of arcs Length of arc = 50° 6 cm © T Madas

Example © T Madas

Find the area of the circular segment, from the information given. 13cm 24 cm © T Madas

Find the area of the circular segment, from the information given. 13cm 24 cm © T Madas

x Find the area of the circular segment, from the information given. 13cm 12 cm x 24 cm 5 cm © T Madas

Find the area of the circular segment, from the information given. 13cm 12 cm 5 cm Area of the triangle © T Madas

Find the area of the circular segment, from the information given. 138.74 cm2 13cm 12 cm 5 cm Area of the triangle 134.76° Area of sector = © T Madas

© T Madas

Two circular sectors have radii of 5 cm and 6 cm and represent one sixth and one eighth of a circle respectively. Calculate the area and the perimeter of each sector, correct to 1 decimal place. area of a circle 1 6 A = r 2 π A = π x 5 2 area of the sector A = π x 5 2 1 6 ≈ 13.1 cm2 5 cm x area of a circle A = r 2 π 1 8 A = π x 6 2 area of the sector A = π 1 8 x 6 2 ≈ 14.1 cm2 x 6 cm © T Madas

circumference of a circle circumference of a circle Two circular sectors have radii of 5 cm and 6 cm and represent one sixth and one eighth of a circle respectively. Calculate the area and the perimeter of each sector, correct to 1 decimal place. circumference of a circle 1 6 C = 2 r π C = 2 x π x 5 length of the arc 5 cm L = 2 x π x 5 1 6 ≈ 5.2 cm x circumference of a circle C = 2 r π 1 8 C = 2 x π x 6 length of the arc L = 2 x π 1 8 x 6 ≈ 4.7 cm x 6 cm © T Madas

© T Madas

42π ⇔ ⇔ ⇔ ⇔ π π π π π π A = r 2 A = A = 18 A = r 2 A = A = 24 The diagram below shows a composite shape consisting of: a semicircle, centre at O and a radius of 6 cm a circular sector, centre at A corresponding to an angle of 60°. Calculate the area of the composite shape in terms of π Round your answer to part (1), to 3 significant figures. area of the semicircle A = 1 2 x π r 2 ⇔ 6 cm O A = 1 2 x π x 62 ⇔ 42π A B 60° A = 18 π area of the sector A = 1 6 x π r 2 ⇔ A = 1 6 x π x 122 ⇔ C A = 24 π 60° = 1 6 of a circle © T Madas

The diagram below shows a composite shape consisting of: a semicircle, centre at O and a radius of 6 cm a circular sector, centre at A corresponding to an angle of 60°. Calculate the area of the composite shape in terms of π Round your answer to part (1), to 3 significant figures. A = 42 π = 42 x π ≈ 132 cm2 [3 s.f.] 6 cm O 42π A B 60° C © T Madas

© T Madas

The radii of the two sectors are r and R as shown in the diagram. The diagram below shows a pond consisting of 2 concentric sectors corresponding to the same angle θ, and two semicircles at each end. The radii of the two sectors are r and R as shown in the diagram. Find a formula for the area A of this pond in terms of r, R and θ. Find the area of a pond with this shape if r = 7 m, R = 9 m and θ = 45° r θ R © T Madas

The diagram below shows a pond consisting of 2 concentric sectors corresponding to the same angle θ, and two semicircles at each end. The radii of the two sectors are r and R as shown in the diagram. Find a formula for the area A of this pond in terms of r, R and θ. Find the area of a pond with this shape if r = 7 m, R = 9 m and θ = 45° π θ 360 R 2 x r θ R © T Madas

The diagram below shows a pond consisting of 2 concentric sectors corresponding to the same angle θ, and two semicircles at each end. The radii of the two sectors are r and R as shown in the diagram. Find a formula for the area A of this pond in terms of r, R and θ. Find the area of a pond with this shape if r = 7 m, R = 9 m and θ = 45° π θ 360 θ 360 R 2 x – π r 2 x r θ R © T Madas

The diagram below shows a pond consisting of 2 concentric sectors corresponding to the same angle θ, and two semicircles at each end. The radii of the two sectors are r and R as shown in the diagram. Find a formula for the area A of this pond in terms of r, R and θ. Find the area of a pond with this shape if r = 7 m, R = 9 m and θ = 45° 2 R – r π θ 360 π θ 360 R – r 2 A = R 2 x – r 2 x + π r θ R © T Madas

The diagram below shows a pond consisting of 2 concentric sectors corresponding to the same angle θ, and two semicircles at each end. The radii of the two sectors are r and R as shown in the diagram. Find a formula for the area A of this pond in terms of r, R and θ. Find the area of a pond with this shape if r = 7 m, R = 9 m and θ = 45° π r 2 x θ 360 – R 2 + R – r 2 A = π r 2 x θ 360 – R 2 + R – r 2 A = © T Madas

The diagram below shows a pond consisting of 2 concentric sectors corresponding to the same angle θ, and two semicircles at each end. The radii of the two sectors are r and R as shown in the diagram. Find a formula for the area A of this pond in terms of r, R and θ. Find the area of a pond with this shape if r = 7 m, R = 9 m and θ = 45° π r 2 x θ 360 – R 2 + R – r 2 A = c πθ 360 π 4 R – r 2 A = R 2 – r 2 + 45π 360 π 4 9 – 7 2 c A = 92 – 72 + π 8 π 4 x 4 c A = x 32 + 4π + π c A = A = 5π ≈ 15.7 m2 © T Madas

© T Madas

area of the given sector The prism shown has a cross section of a circular sector of radius 8 cm corresponding to a central angle of 45°. The thickness of the prism is 3 cm. 1. Calculate the volume of the prism, correct to 3 s.f. 2. Calculate the surface area of the prism, correct to 3 s.f. area of a circle A = r 2 π 45° A = π x 8 2 3 cm area of the given sector 8 cm 45 360 A = π x 8 2 x 8 cm 45° volume of the prism 45 360 V = π x 8 2 x 3 x V = 24π V ≈ 75.4 cm3 [3 s.f.] © T Madas

area of the given sector area of the given sector The prism shown has a cross section of a circular sector of radius 8 cm corresponding to a central angle of 45°. The thickness of the prism is 3 cm. 1. Calculate the volume of the prism, correct to 3 s.f. 2. Calculate the surface area of the prism, correct to 3 s.f. area of the given sector A = π x 8 2 45 360 x 45° 3 cm area of the given sector 8 cm 45 360 A = π x 8 2 x 8 cm 45° © T Madas

The prism shown has a cross section of a circular sector of radius 8 cm corresponding to a central angle of 45°. The thickness of the prism is 3 cm. 1. Calculate the volume of the prism, correct to 3 s.f. 2. Calculate the surface area of the prism, correct to 3 s.f. area of the given sector x 2 A = π x 8 2 45 360 x 2 = 16π x 45° 2πr 3 cm area of the side rectangle x 2 A = 3 x 8 x 2 = 48 8 cm area of the curved surface 8 cm 45° 45 360 A = 2 x π x 8 x 3 = 6π x total surface area A = 16π + 48 + 6π ≈ 117 cm2 [3 s.f.] © T Madas

© T Madas

area of the “larger” sector The diagram below models the action of a single windscreen wiper rotating through an angle of 162° on a rectangular windscreen measuring 141 cm by 72 cm. The wiper AB is 45 cm long and OA = 15 cm. Calculate what percentage of the screen gets wiped. 141 area of a circle A = r 2 π A = π x 60 2 72 area of the “larger” sector B 162 360 162° 45 A = π x 60 2 = 1620π A x 15 O All measurements in cm © T Madas

area of the “larger” sector area of the “inner” sector The diagram below models the action of a single windscreen wiper rotating through an angle of 162° on a rectangular windscreen measuring 141 cm by 72 cm. The wiper AB is 45 cm long and OA = 15 cm. Calculate what percentage of the screen gets wiped. 141 area of a circle A = r 2 π A = π x 60 2 72 area of the “larger” sector B 162 360 45 A = π x 60 2 = 1620π 162° A x 15 O area of the “inner” sector All measurements in cm A = π x 15 2 162 360 = 101.25π x © T Madas

The diagram below models the action of a single windscreen wiper rotating through an angle of 162° on a rectangular windscreen measuring 141 cm by 72 cm. The wiper AB is 45 cm long and OA = 15 cm. Calculate what percentage of the screen gets wiped. 141 area of a circle A = r 2 π A = π x 60 2 72 area of the “larger” sector B 162 360 45 A = π x 60 2 = 1620π 162° A x 15 O area of the “inner” sector All measurements in cm A = π x 15 2 162 360 = 101.25π x area swept by the wiper is 1620π – 101.25π = 1518.75π cm2 ≈ 4771.3 cm2 © T Madas

area swept by the wiper is The diagram below models the action of a single windscreen wiper rotating through an angle of 162° on a rectangular windscreen measuring 141 cm by 72 cm. The wiper AB is 45 cm long and OA = 15 cm. Calculate what percentage of the screen gets wiped. 141 %age = area wiped total area x 100 c %age = 1518.75π 141 x 72 x 100 c 72 %age ≈ 47% B 162° 45 A 15 O All measurements in cm area swept by the wiper is 1620π – 101.25π = 1518.75π cm2 ≈ 4771.3 cm2 © T Madas

Exam Question © T Madas

Two circles of respective radii of 3 cm and 4 cm are overlapping each other in such a way so that their centres are 5 cm apart. Calculate the perimeter of the compound shape, correct to 2 significant figures. C 3,4,5 = Alarm Bells ! 3 cm 4 cm ∆ABC is right angled By trig on ∆ABC : θ A B 5 cm © T Madas

Simplify the diagram by showing only the relevant information Two circles of respective radii of 3 cm and 4 cm are overlapping each other in such a way so that their centres are 5 cm apart. Calculate the perimeter of the compound shape, correct to 2 significant figures. Now some angle calculations: If θ ≈ 53.13° RCBA ≈ 36.87° RCAD ≈ 106.26° RCBD ≈ 73.74° C 36.87° 253.74° 3 cm 4 cm θ A B 5 cm 286.26° reflex RCAD ≈ 253.74° reflex RCBD ≈ 286.26° 106.26° 73.74° D Simplify the diagram by showing only the relevant information © T Madas

Two circles of respective radii of 3 cm and 4 cm are overlapping each other in such a way so that their centres are 5 cm apart. Calculate the perimeter of the compound shape, correct to 2 significant figures. This arc corresponds to an angle of 253.74° on a circle with radius of 3 cm This arc corresponds to an angle of 286.26° on a circle with radius of 4 cm C 253.74° 3 cm 4 cm A B 286.26° D © T Madas

Two circles of respective radii of 3 cm and 4 cm are overlapping each other in such a way so that their centres are 5 cm apart. Calculate the perimeter of the compound shape, correct to 2 significant figures. The smaller of the 2 arcs C 253.74° 3 cm 4 cm A B 286.26° The larger of the 2 arcs D The required perimeter 33.3 cm [2 s.f.] © T Madas

© T Madas

In a circle of radius 5 m the tangents at the endpoints of a chord AB of length 8 m meet at point C. Calculate the perimeter and the area of the finite region bounded by the two tangents and the circle B OBD is right angled OD = 3 m Trig to find RBOD 5 m 4 m θ C 8 m 3 m O D 5 m A © T Madas

In a circle of radius 5 m the tangents at the endpoints of a chord AB of length 8 m meet at point C. Calculate the perimeter and the area of the finite region bounded by the two tangents and the circle B The area of the sector 5 m θ C O D The length of the arc 5 m A area of the sector ≈ 11.59 m2 length of the arc ≈ 11.59 m © T Madas

In a circle of radius 5 m the tangents at the endpoints of a chord AB of length 8 m meet at point C. Calculate the perimeter and the area of the finite region bounded by the two tangents and the circle B OBC is right angled OB = 5 m RBOC = 53.13° Trig to find BC 5 m 4 m θ C 3 m O D 5 m A area of the sector ≈ 11.59 m2 length of the arc ≈ 11.59 m BC = 20/3 m © T Madas

In a circle of radius 5 m the tangents at the endpoints of a chord AB of length 8 m meet at point C. Calculate the perimeter and the area of the finite region bounded by the two tangents and the circle B Area of OBC 5 m θ C area of the sector ≈ 11.59 m2 O D half the shaded area ≈ 5.08 m2 5 m shaded area ≈ 10.16 m2 A area of the sector ≈ 11.59 m2 length of the arc ≈ 11.59 m BC = 20/3 m © T Madas

In a circle of radius 5 m the tangents at the endpoints of a chord AB of length 8 m meet at point C. Calculate the perimeter and the area of the finite region bounded by the two tangents and the circle B length of the arc ≈ 11.59 m Length BC = 20/3 m ≈ 6.67 m 5 m θ E C 11.59 6.67 x 2 ≈ 23.18 ≈ 13.34 O D 5 m Total perimeter ≈ 36.5 m A area of the sector ≈ 11.59 m2 length of the arc ≈ 11.59 m BC = 20/3 m © T Madas

Exam Question © T Madas

L A A circular sector has radius r and an area A . The sector corresponds to an arc of length L . Write an expression for A in terms of r and L . L A r © T Madas

L A θ A circular sector has radius r and an area A . The sector corresponds to an arc of length L . Write an expression for A in terms of r and L . L A θ r © T Madas

Now what? A circular sector has radius r and an area A . The sector corresponds to an arc of length L . Write an expression for A in terms of r and L . Now what? © T Madas

A circular sector has radius r and an area A . The sector corresponds to an arc of length L . Write an expression for A in terms of r and L . 2 © T Madas

Harder Examples © T Madas

curvilinear areas 1 © T Madas

The following pattern is produced inside a square whose side length is 8 cm Calculate the shaded area D C P l a n 45° The shaded region can be split into 8 congruent shapes take one of these shapes add the right angled triangle next to it they make up a 45° sector of a circle whose radius is half the square’s diagonal A B 8 cm © T Madas

Calculate the shaded area The following pattern is produced inside a square whose side length is 8 cm Calculate the shaded area D C Pythagoras: 4 45° 4 x 32 Area of sector: Area of triangle: A B 8 cm © T Madas

Calculate the shaded area The following pattern is produced inside a square whose side length is 8 cm Calculate the shaded area D C Area of sector: Area of triangle: Area of sector: Area of triangle: A B 8 cm © T Madas

Calculate the shaded area The following pattern is produced inside a square whose side length is 8 cm Calculate the shaded area D C Area of sector: Area of triangle: Shaded area: A B 8 cm © T Madas

curvilinear areas 2 © T Madas

Find the area of the shaded region In a square with side length of 8 cm, 2 quarter circular arcs are drawn, as shown below. Find the area of the shaded region The bottom part of the shaded area consists of: D C an equilateral triangle with side length of 8 cm plus 2 circular segments Look at this shaded section: It is a circular sector, one sixth of a circle of radius 8 cm. The area of the circular segment equals the area of this sector less the area of the equilateral triangle. 60° A 8 cm B © T Madas

Find the area of the shaded region In a square with side length of 8 cm, 2 quarter circular arcs are drawn, as shown below. Find the area of the shaded region The area of the segment D C sector = 1/6 circle triangle 60° A 8 cm B © T Madas

Find the area of the shaded region In a square with side length of 8 cm, 2 quarter circular arcs are drawn, as shown below. Find the area of the shaded region The area of the segment D C 32π 3 – 16 3 60° A 8 cm B © T Madas

Find the area of the shaded region In a square with side length of 8 cm, 2 quarter circular arcs are drawn, as shown below. Find the area of the shaded region Now look at one of the non - shaded regions The area of the segment D C The area of one of the non shaded regions equals: One twelfth of a circle of radius 8 cm Less the area of the segment 32π 3 – 16 3 30° 60° A 8 cm B © T Madas

Find the area of the shaded region In a square with side length of 8 cm, 2 quarter circular arcs are drawn, as shown below. Find the area of the shaded region D C 32π 3 – 16 3 30° 60° A 8 cm B © T Madas

Find the area of the shaded region In a square with side length of 8 cm, 2 quarter circular arcs are drawn, as shown below. Find the area of the shaded region D C 16 3 16π 3 16π 3 16 3 32π 3 – 16 3 30° 60° A 8 cm B © T Madas

Find the area of the shaded region In a square with side length of 8 cm, 2 quarter circular arcs are drawn, as shown below. Find the area of the shaded region D C The shaded region is 64 cm2 less the two non shaded regions we just found 16 3 16π 3 16π 3 16 3 A 8 cm B © T Madas

The Curved Surface of a Cone STOP © T Madas

θ L h L A r r r © T Madas

© T Madas

© T Madas

© T Madas

© T Madas

h L r © T Madas

© T Madas