Skip to content
Cambridge IGCSE0607

Compound shapes and parts of shapes

International Mathematics 0607 Chapter Notes

What this chapter covers

Compound shapes and parts of shapes
ShareWhatsAppPost
Compound shapes and parts of shapes notes

Read online free. Download a watermarked copy with a free account.

Read the notes

The full Compound shapes and parts of shapes notes as text: skim, search, and jump between subtopics.

~9 min read

In this chapter

  1. 1Perimeter of compound shapes
  2. 2Area of compound shapes
  3. 3Parts of shapes: halves and quarters of circles
  4. 4Volume of compound solids and parts of solids
  5. 5Surface area of compound solids
  6. 6Frustum: volume and surface area

1. Perimeter of compound shapes

A compound shape is made by joining simple shapes. Its perimeter is the length of the outside edge only. Edges where two shapes join are not part of it.

A semicircle has a curved edge of half the circumference. Add the straight diameter only if it is on the outside. Answers may be asked for in terms of π: leave π in the answer.

curved edge of a semicircle = 1/2 × π d

perimeter = straight edges + curved edges

Examiner insight

“The arc length of the semi-circle had to be found and then the base or diameter added.” “A small number added up the perimeters of the two rectangles ignoring the fact that two sides of the rectangles were not part of the perimeter of the combined shape.”

Worked example 14 marks

Do not use a calculator. A shape is made from a rectangle 14 cm long and 8 cm wide and a semicircle. The semicircle is joined to one of the 8 cm sides, outside the rectangle. Work out the perimeter of the shape. Give your answer in terms of π.

  1. 1

    outside straight edges: 14 + 14 + 8 = 36 cm [1]

  2. 2

    the joined 8 cm side is not on the outside

  3. 3

    curved edge = 1/2 × π × 8 [1]

  4. 4

    = 4π cm [1]

  5. 5

    perimeter = 36 + 4π cm [1]

2. Area of compound shapes

Split a compound shape into rectangles, triangles and circles, or take one shape away from another. Find each area, then add or subtract.

Write the working for each step. Area is in square units and perimeter is in units of length.

rectangle A = l × w

triangle A = 1/2 × b × h

circle A = π r²

Examiner insight

“Some candidates only found the area of one square.” “Often, cm2 was given as the units to both area and perimeter.”

Worked example 14 marks

This pattern is drawn using 10 identical squares and 5 identical isosceles triangles. The total shaded area of the pattern is 231 cm². Each square has side length 3 cm. Each isosceles triangle has base length x cm and height 9.4 cm. Work out the value of x.

  1. 1

    area of squares = 10 × 3 × 3 = 90 cm² [1]

  2. 2

    area of triangles = 231 - 90 = 141 cm² [1]

  3. 3

    5 × 1/2 × x × 9.4 = 141, so 23.5x = 141 [1]

  4. 4

    x = 141 ÷ 23.5 = 6 [1]

3. Parts of shapes: halves and quarters of circles

A part of a shape is a fraction of the whole. A semicircle is one half of a circle. A quarter circle is one quarter.

Take the same fraction of the circle's area. For the perimeter, take the same fraction of the circumference, then add the straight edges that are on the outside.

semicircle A = 1/2 π r²

quarter circle A = 1/4 π r²

Examiner insight

“Most candidates recognised that it was necessary to subtract the area of the triangle OAC from the area of the sector, but some only worked out the area of the sector.” “Many candidates decided to subtract their arc lengths due to being confused with similar problems that required the area of the sector.”

Worked example 14 marks

Do not use a calculator. A square has side 8 cm. A quarter circle of radius 8 cm is drawn inside the square with its centre at one corner. Work out the area of the square that is outside the quarter circle. Give your answer in terms of π.

  1. 1

    area of square = 8 × 8 = 64 cm² [1]

  2. 2

    quarter circle = 1/4 × π × 8² [1]

  3. 3

    = 16π cm² [1]

  4. 4

    shaded area = 64 - 16π cm² [1]

4. Volume of compound solids and parts of solids

A compound solid is made by joining simple solids, such as a cylinder and a hemisphere. Find the volume of each part and add them.

A hemisphere is half a sphere, so its volume is half the volume of a sphere. Take care to use half, not the whole.

V_cylinder = π r² h

V_cone = 1/3 π r² h

V_hemisphere = 1/2 × 4/3 π r³

Examiner insight

“A few candidates gave the volume of a sphere instead of the hemisphere.” “Many candidates found only the volume of the cone or of the cylinder.”

Worked example 14 marks

Do not use a calculator. A solid is made from a cylinder of radius 3 cm and height 10 cm and a hemisphere of radius 3 cm. The hemisphere is joined to one end of the cylinder. Work out the volume of the solid. Give your answer in terms of π.

  1. 1

    cylinder = π × 3² × 10 = 90π [1]

  2. 2

    hemisphere = 1/2 × 4/3 × π × 3³ [1]

  3. 3

    = 18π [1]

  4. 4

    total = 90π + 18π = 108π cm³ [1]

5. Surface area of compound solids

Find the surface area by listing every surface you can touch from the outside, then adding. A circle where two solids join is inside the solid: leave it out.

The curved surface of a cone uses the slant height l, not the vertical height. Find it with Pythagoras. A hemisphere has half the curved area of a sphere.

A_cylinder, curved = 2π r h

A_cone, curved = π r l

l² = r² + h²

A_sphere = 4π r²

Examiner insight

“The extra difficulty was to be aware that Pythagoras was required to find the slant height and the perpendicular height was frequently used instead.” “A common error was to subtract the total surface area of the cone from the total surface area of the cylinder.”

Worked example 15 marks

Do not use a calculator. A solid is a cone joined to the top of a cylinder. Both have radius 5 cm. The cylinder has height 8 cm. The cone has vertical height 12 cm. Work out the total surface area of the solid. Give your answer in terms of π.

  1. 1

    slant height l = √(5² + 12²) = 13 cm [1]

  2. 2

    cone, curved: π × 5 × 13 = 65π [1]

  3. 3

    cylinder, curved: 2 × π × 5 × 8 = 80π [1]

  4. 4

    base of cylinder: π × 5² = 25π (the joined circle is not counted) [1]

  5. 5

    total = 65π + 80π + 25π = 170π cm² [1]

6. Frustum: volume and surface area

A frustum is a cone with a smaller cone cut off its top. Its volume is the large cone minus the small cone.

For its surface area, subtract the curved surface of the small cone from the curved surface of the large cone. Then add the flat circle at the top and the flat circle at the base.

V_frustum = V_large cone - V_small cone

A_frustum = π R L - π r l + π R² + π r²

Examiner insight

“Some of the rest were able to show a cubed ratio but were then unable to subtract the required volumes.”

Worked example 17 marks

A frustum is made by removing a cone of radius 3 cm and height 6 cm from the top of a cone of radius 6 cm and height 12 cm. Work out(a) the volume and(b) the total surface area of the frustum. Give each answer correct to 3 significant figures.

  1. 1

    (a) large cone = 1/3 × π × 6² × 12 = 144π [1]

  2. 2

    small cone = 1/3 × π × 3² × 6 = 18π [1]

  3. 3

    frustum = 144π - 18π = 126π = 396 cm³ [1]

  4. 4

    (b) slant heights: √(6² + 12²) = √(180) and √(3² + 6²) = √45 [1]

  5. 5

    curved = π × 6 × √(180) - π × 3 × √45 = 189.67 [1]

  6. 6

    circles: π × 6² + π × 3² = 141.37 [1]

  7. 7

    total = 331 cm² [1]

Recap

  • Perimeter of a compound shape: add the outside edges only.
  • A semicircle's curved edge is half the circumference. Leave π in the answer when asked.
  • Area of a compound shape: split it and add, or subtract one shape from another.
  • Volume of a compound solid: find each part, then add. A hemisphere is half a sphere.
  • Surface area of a compound solid: add every outside surface. Leave out a circle where solids join.
  • A cone's curved surface uses the slant height, from Pythagoras.
  • Extended: frustum volume = large cone − small cone. Frustum surface area = curved difference + top circle + base circle.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Do not use a calculator. A semicircle has diameter 12 cm. Find its perimeter. Give your answer in terms of π.2 marks
  2. Do not use a calculator. A rectangle is 9 m by 6 m. A rectangle 3 m by 2 m is cut out of one corner. Work out (a) the perimeter and (b) the area of the shape that is left.3 marks
  3. Do not use a calculator. A square has side 10 cm. A circle of radius 5 cm is drawn inside it, touching all four sides. Work out the area of the square outside the circle. Give your answer in terms of π.3 marks
  4. A running track is a rectangle with a semicircle at each end. The rectangle is 90 m long and 60 m wide. The semicircles have diameter 60 m. Work out the distance once round the track. Give your answer correct to 3 significant figures.3 marks
  5. Do not use a calculator. A hemisphere has radius 3 cm. Work out its volume. Give your answer in terms of π.2 marks
  6. A solid is a cylinder of radius 6 cm and height 7 cm with a hemisphere of radius 6 cm joined to one end. Calculate the volume of the solid. Give your answer correct to 3 significant figures.4 marks
  7. A solid is a cone joined to the top of a cylinder. Both have radius 4 cm. The cylinder has height 10 cm and the cone has vertical height 7 cm. Calculate the total surface area of the solid. Give your answer correct to 3 significant figures.5 marks
  8. Do not use a calculator. A cone of radius 9 cm and height 12 cm has a small cone of radius 3 cm and height 4 cm removed from its top. This leaves a frustum. Work out the volume of the frustum. Give your answer in terms of π.4 marks
  9. Do not use a calculator. A lampshade is the curved surface of a frustum. It is open at both ends. The top circle has radius 6 cm and the bottom circle has radius 12 cm. The vertical height of the lampshade is 8 cm. Work out the area of material in the lampshade. Give your answer in terms of π.6 marks
  10. A solid metal cone has radius 9 cm and height 24 cm. It is cut parallel to its base, 16 cm above the base. The small cone at the top is removed to leave a frustum. The frustum is melted and all the metal is made into solid hemispheres of radius 1.5 cm. Work out the number of complete hemispheres that can be made.6 marks

Free notes are just the start.

Unlock every Workbook and Chapter at a Glance, and generate your own worksheets and predicted papers.

Explore plans

Related chapters

Types of numberSetsPowers and rootsFractions, decimals and percentagesOrderingThe four operations