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Amy Ferguson Moncure

IB AA · Unit 2 · 18A Radian Measure

Compound Shapes

Break the figure into sectors, rectangles, or a cone. Convert a degree angle before you use a radian formula.

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Key Ideas Easy Medium Hard

l = rθ

A = 12 r² θ

slant² = height² + radius²

What to Remember

Add the pieces

Trace the outside edge. An internal side is not part of the perimeter.

Area is the sum of the rectangle and the sector, or the difference of two sectors.

A sector becomes a cone

The radius of the sector is the slant height of the cone.

The arc of the sector is the circumference of the base.

rθ = 2πR

Easy

Rectangle and a sector

Add the areas. For the perimeter, skip the side that is inside the figure.

Example
A rectangle 6 cm by 4 cm has a sector of radius 6 cm and angle 1.2 radians on the side of length 6 cm. Find the area and the perimeter of the whole figure.Rectangle 6 cm by 4 cm with a sector of angle 1.26 cm4 cm1.2dashed side is inside

Area = rectangle + sector

Area = 6 × 4 + 12 × 6² × 1.2

Area = 24 + 21.6

Area = 45.6 cm²

The top of the rectangle is inside the figure. The outer radius replaces it and has the same length, so the straight edges still total the rectangle’s perimeter. Then add the arc.

P = 2(h + w) + arc

l = rθ

l = 6 × 1.2 = 7.2 cm

P = 2(4 + 6) + 7.2

P = 27.2 cm

Answer: area 45.6 cm², perimeter 27.2 cm

1
A rectangle 8 cm by 5 cm has a sector of radius 8 cm and angle 0.5 radians on the side of length 8 cm. Find the area and the perimeter of the whole figure.Rectangle 8 cm by 5 cm with a sector of angle 0.58 cm5 cm0.5dashed side is inside
Show Answer

Area = rectangle + sector

Area = 8 × 5 + 12 × 8² × 0.5

Area = 40 + 16

Area = 56 cm²

The top of the rectangle is inside the figure. The outer radius replaces it and has the same length, so the straight edges still total the rectangle’s perimeter. Then add the arc.

P = 2(h + w) + arc

l = rθ

l = 8 × 0.5 = 4 cm

P = 2(5 + 8) + 4

P = 30 cm

Answer: area 56 cm², perimeter 30 cm

2
A rectangle 10 cm by 3 cm has a sector of radius 10 cm and angle 1.5 radians on the side of length 10 cm. Find the area and the perimeter of the whole figure.Rectangle 10 cm by 3 cm with a sector of angle 1.510 cm3 cm1.5dashed side is inside
Show Answer

Area = rectangle + sector

Area = 10 × 3 + 12 × 10² × 1.5

Area = 30 + 75

Area = 105 cm²

The top of the rectangle is inside the figure. The outer radius replaces it and has the same length, so the straight edges still total the rectangle’s perimeter. Then add the arc.

P = 2(h + w) + arc

l = rθ

l = 10 × 1.5 = 15 cm

P = 2(3 + 10) + 15

P = 41 cm

Answer: area 105 cm², perimeter 41 cm

3
A rectangle 5 cm by 6 cm has a sector of radius 5 cm and angle 0.8 radians on the side of length 5 cm. Find the area and the perimeter of the whole figure.Rectangle 5 cm by 6 cm with a sector of angle 0.85 cm6 cm0.8dashed side is inside
Show Answer

Area = rectangle + sector

Area = 5 × 6 + 12 × 5² × 0.8

Area = 30 + 10

Area = 40 cm²

The top of the rectangle is inside the figure. The outer radius replaces it and has the same length, so the straight edges still total the rectangle’s perimeter. Then add the arc.

P = 2(h + w) + arc

l = rθ

l = 5 × 0.8 = 4 cm

P = 2(6 + 5) + 4

P = 26 cm

Answer: area 40 cm², perimeter 26 cm

4
A rectangle 9 cm by 4 cm has a sector of radius 9 cm and angle 0.6 radians on the side of length 9 cm. Find the area and the perimeter of the whole figure.Rectangle 9 cm by 4 cm with a sector of angle 0.69 cm4 cm0.6dashed side is inside
Show Answer

Area = rectangle + sector

Area = 9 × 4 + 12 × 9² × 0.6

Area = 36 + 24.3

Area = 60.3 cm²

The top of the rectangle is inside the figure. The outer radius replaces it and has the same length, so the straight edges still total the rectangle’s perimeter. Then add the arc.

P = 2(h + w) + arc

l = rθ

l = 9 × 0.6 = 5.4 cm

P = 2(4 + 9) + 5.4

P = 31.4 cm

Answer: area 60.3 cm², perimeter 31.4 cm

5
A rectangle 7 cm by 5 cm has a sector of radius 7 cm and angle 0.4 radians on the side of length 7 cm. Find the area and the perimeter of the whole figure.Rectangle 7 cm by 5 cm with a sector of angle 0.47 cm5 cm0.4dashed side is inside
Show Answer

Area = rectangle + sector

Area = 7 × 5 + 12 × 7² × 0.4

Area = 35 + 9.8

Area = 44.8 cm²

The top of the rectangle is inside the figure. The outer radius replaces it and has the same length, so the straight edges still total the rectangle’s perimeter. Then add the arc.

P = 2(h + w) + arc

l = rθ

l = 7 × 0.4 = 2.8 cm

P = 2(5 + 7) + 2.8

P = 26.8 cm

Answer: area 44.8 cm², perimeter 26.8 cm

Medium

Cone, or three arcs

The sector radius becomes the slant height. Each arc on an equilateral triangle is 60°.

Example
A paper sector of radius r cm and angle θ is rolled into a cone of height 8 cm and base radius 6 cm. Find r and θ.Sector rolled into a coneθr8 cm6 cm

The sector radius is the slant height.

r² = 8² + 6²

r² = 64 + 36 = 100

r = 10 cm

The arc is the base circumference.

l = 2πR

l = 2π × 6 = 12π

l = rθ

θ = lr

θ = 12π10

θ = 6π5

Answer: r = 10 cm, θ = 6π5

1
A paper sector of radius r cm and angle θ is rolled into a cone of height 12 cm and base radius 5 cm. Find r and θ.Sector rolled into a coneθr12 cm5 cm
Show Answer

The sector radius is the slant height.

r² = 12² + 5²

r² = 144 + 25 = 169

r = 13 cm

The arc is the base circumference.

l = 2πR

l = 2π × 5 = 10π

l = rθ

θ = lr

θ = 10π13

θ = 10π13

Answer: r = 13 cm, θ = 10π13

2
A paper sector of radius r cm and angle θ is rolled into a cone of height 9 cm and base radius 12 cm. Find r and θ.Sector rolled into a coneθr9 cm12 cm
Show Answer

The sector radius is the slant height.

r² = 9² + 12²

r² = 81 + 144 = 225

r = 15 cm

The arc is the base circumference.

l = 2πR

l = 2π × 12 = 24π

l = rθ

θ = lr

θ = 24π15

θ = 8π5

Answer: r = 15 cm, θ = 8π5

3
A paper sector of radius r cm and angle θ is rolled into a cone of height 4 cm and base radius 3 cm. Find r and θ.Sector rolled into a coneθr4 cm3 cm
Show Answer

The sector radius is the slant height.

r² = 4² + 3²

r² = 16 + 9 = 25

r = 5 cm

The arc is the base circumference.

l = 2πR

l = 2π × 3 = 6π

l = rθ

θ = lr

θ = 6π5

θ = 6π5

Answer: r = 5 cm, θ = 6π5

4
An equilateral triangle has sides of 9 cm. Three arcs, each centred at a vertex, replace the sides. Find the exact perimeter of the curved figure.Equilateral triangle with three arcsside 9 cm
Show Answer

Each arc is centred at a vertex of the equilateral triangle, so the angle is 60°.

60°1×π rad180°

θ = π3

l = rθ

One arc = 9 × π3 = 3π

Perimeter = 3 arcs

Perimeter = 3 × 3π = 9π

Answer: 9π cm

5
An equilateral triangle has sides of 14 cm. Three arcs, each centred at a vertex, replace the sides. Find the exact perimeter of the curved figure.Equilateral triangle with three arcsside 14 cm
Show Answer

Each arc is centred at a vertex of the equilateral triangle, so the angle is 60°.

60°1×π rad180°

θ = π3

l = rθ

One arc = 14 × π3 = 14π3

Perimeter = 3 arcs

Perimeter = 3 × 14π3 = 14π

Answer: 14π cm

Hard

Overlap

The overlap of two equal circles is two segments. Add the arcs for the perimeter.

Example
Two circles each have radius 6 cm. The centre of each circle lies on the other circle. Find the perimeter and the area of the shaded overlap.Two overlapping circleseach radius 6 cmshaded

The centre of each circle lies on the other circle, so each triangle of two radii and the line of centres is equilateral.

The shaded angle at each centre is 120°.

120°1×π rad180°

θ = 2π3

One arc = rθ

One arc = 6 × 2π3 = 4π

Perimeter = two arcs

Perimeter = 8π

Perimeter ≈ 25.1 cm

One segment = 12 r² θ − 12 r² sin θ

sin 2π3 = square root of 32

The shaded area is two segments.

Area = r² (θ − sin θ)

Area = 36 (2π3 − square root of 32)

Area = 24π − 18square root of 3

Area ≈ 44.2 cm²

Answer: perimeter 8π cm, about 25.1 cm; area about 44.2 cm²

1
Two circles each have radius 5 cm. The centre of each circle lies on the other circle. Find the perimeter and the area of the shaded overlap.Two overlapping circleseach radius 5 cmshaded
Show Answer

The centre of each circle lies on the other circle, so each triangle of two radii and the line of centres is equilateral.

The shaded angle at each centre is 120°.

120°1×π rad180°

θ = 2π3

One arc = rθ

One arc = 5 × 2π3 = 10π3

Perimeter = two arcs

Perimeter = 20π3

Perimeter ≈ 20.9 cm

One segment = 12 r² θ − 12 r² sin θ

sin 2π3 = square root of 32

The shaded area is two segments.

Area = r² (θ − sin θ)

Area = 25 (2π3 − square root of 32)

Area = 50π3 − 25square root of 32

Area ≈ 30.7 cm²

Answer: perimeter 20π3 cm, about 20.9 cm; area about 30.7 cm²

2
Two circles each have radius 9 cm. The centre of each circle lies on the other circle. Find the perimeter and the area of the shaded overlap.Two overlapping circleseach radius 9 cmshaded
Show Answer

The centre of each circle lies on the other circle, so each triangle of two radii and the line of centres is equilateral.

The shaded angle at each centre is 120°.

120°1×π rad180°

θ = 2π3

One arc = rθ

One arc = 9 × 2π3 = 6π

Perimeter = two arcs

Perimeter = 12π

Perimeter ≈ 37.7 cm

One segment = 12 r² θ − 12 r² sin θ

sin 2π3 = square root of 32

The shaded area is two segments.

Area = r² (θ − sin θ)

Area = 81 (2π3 − square root of 32)

Area = 54π − 81square root of 32

Area ≈ 99.5 cm²

Answer: perimeter 12π cm, about 37.7 cm; area about 99.5 cm²

3
Two circles each have radius 10 cm. The centre of each circle lies on the other circle. Find the perimeter and the area of the shaded overlap.Two overlapping circleseach radius 10 cmshaded
Show Answer

The centre of each circle lies on the other circle, so each triangle of two radii and the line of centres is equilateral.

The shaded angle at each centre is 120°.

120°1×π rad180°

θ = 2π3

One arc = rθ

One arc = 10 × 2π3 = 20π3

Perimeter = two arcs

Perimeter = 40π3

Perimeter ≈ 41.9 cm

One segment = 12 r² θ − 12 r² sin θ

sin 2π3 = square root of 32

The shaded area is two segments.

Area = r² (θ − sin θ)

Area = 100 (2π3 − square root of 32)

Area = 200π3 − 50square root of 3

Area ≈ 123 cm²

Answer: perimeter 40π3 cm, about 41.9 cm; area about 123 cm²

4
An equilateral triangle has sides of 8 cm. Three arcs, each centred at a vertex, replace the sides. Find the exact perimeter of the curved figure.Equilateral triangle with three arcsside 8 cm
Show Answer

Each arc is centred at a vertex of the equilateral triangle, so the angle is 60°.

60°1×π rad180°

θ = π3

l = rθ

One arc = 8 × π3 = 8π3

Perimeter = 3 arcs

Perimeter = 3 × 8π3 = 8π

Answer: 8π cm

5
A paper sector of radius r cm and angle θ is rolled into a cone of height 15 cm and base radius 20 cm. Find r and θ.Sector rolled into a coneθr15 cm20 cm
Show Answer

The sector radius is the slant height.

r² = 15² + 20²

r² = 225 + 400 = 625

r = 25 cm

The arc is the base circumference.

l = 2πR

l = 2π × 20 = 40π

l = rθ

θ = lr

θ = 40π25

θ = 8π5

Answer: r = 25 cm, θ = 8π5

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