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Cambridge IGCSE Mathematics 0580 · Mensuration

Compound shapes and parts of shapes

Perimeters and areas of shapes made from rectangles, triangles, circles and sectors, and of parts cut from them; surface areas and volumes of solids joined together or cut in half, including the hemisphere. Extended: the frustum of a cone or pyramid.

C5.5E5.5Core and ExtendedFree lesson
What the syllabus says

C5.5 (Core)

  • 1 Carry out calculations and solve problems involving perimeters and areas of:
  • compound shapes
  • parts of shapes.
  • 2 Carry out calculations and solve problems involving surface areas and volumes of:
  • compound solids
  • parts of solids.

Notes and examples

  • Answers may be asked for in terms of π.
  • e.g. find the volume of half of a sphere.

E5.5 (Extended)

  • 1 Carry out calculations and solve problems involving perimeters and areas of:
  • compound shapes
  • parts of shapes.
  • 2 Carry out calculations and solve problems involving surface areas and volumes of:
  • compound solids
  • parts of solids.

Notes and examples

  • Answers may be asked for in terms of π.
  • e.g. find the surface area and volume of a frustum.

Quoted from the Cambridge IGCSE Mathematics 0580 syllabus for 2025 to 2027.

Splitting and subtracting

A compound shape is made from simpler shapes whose areas you already know: rectangles, triangles, trapeziums, circles and sectors. There are two ways to find its area.

  • Split it into simple pieces, find the area of each, and add.
  • Draw the larger simple shape around it, find that area, and subtract the pieces that are missing.

Both give the same answer; choose whichever needs fewer, easier calculations. Before calculating, write any missing lengths on the diagram. Missing lengths come from the given ones: in an L-shape, the two horizontal pieces on the top add up to the whole width at the bottom.

Set the working out clearly, naming each piece, because each correct piece earns method marks even if a later step goes wrong.

PQ12 cm9 cm7 cm5 cm

Split along the dashed line into rectangles P and Q, or subtract the missing corner from the 12 by 9 rectangle.

Worked example 1

Find (a) the area, (b) the perimeter of the L-shape in the diagram.

  1. Missing lengths: the right-hand step across is 12−7=5 cm, and the step up is 9−5=4 cm.

  2. (a) Splitting: P is 7×9=63 and Q is 5×5=25, total 88cm2.

  3. Check by subtracting: 12×9−5×4=108−20=88cm2.

  4. (b) Go round the outside: 12+5+5+4+7+9=42 cm.

Answer: (a) 88cm2 (b) 42 cm

Perimeters of compound shapes

The perimeter is the distance round the outside only. Lines you drew to split the shape are inside it and are not part of the perimeter.

A good habit is to start at one corner and go round the shape, writing down every edge until you get back to the start. That way no edge is missed or counted twice.

When part of the edge is curved, use arc lengths: a semicircle on a side of length d adds 12πd, and the straight side it replaces is no longer part of the perimeter. In terms of π, a perimeter often has a π part and a whole-number part, such as (32+4π) cm.

An interesting fact: the L-shape above has the same perimeter as the full 12 by 9 rectangle, 2×(12+9)=42 cm, because cutting out a corner just moves two edges inwards.

8 cm6 cm

The dashed line is inside the window, so it is not part of the perimeter.

Worked example 2

The window in the diagram is a rectangle 8 cm by 6 cm with a semicircle on top. Find, in terms of π, (a) its area, (b) its perimeter.

  1. The semicircle has diameter 8 cm, so radius 4 cm.

  2. (a) Rectangle =8×6=48. Semicircle =12×π×42=8π.

  3. Area =(48+8π)cm2.

  4. (b) Curved edge =12×2×π×4=4π.

  5. Straight edges: 6+8+6=20 (the top of the rectangle is inside the shape).

  6. Perimeter =(20+4π) cm.

Answer: (a) (48+8π)cm2 (b) (20+4π) cm

Check yourself

A rectangle measures 10 cm by 7 cm. A right-angled triangle with shorter sides 4 cm and 3 cm is cut out of it. Find the area that is left.

Answer it in the portal, where your answer is checked as you type →

Parts of shapes

Many questions ask for a part of a shape: a shaded region, a ring, the part of a square outside a quarter circle. Almost always the method is subtraction: whole minus what is not wanted.

  • A ring (annulus) between circles of radius R and r with the same centre: area =πR2−πr2. Do not work out π(R−r)2; that is a different, smaller number.
  • A square with a quarter circle drawn from one corner: shaded region = square − quarter circle.
  • A semicircle, quarter circle or sector: use the fraction of the whole circle, as in the circles topic.

For the perimeter of a part, trace its boundary. It may include arcs, parts of sides and radii; list each one.

Check yourself

A ring is the region between two circles with the same centre, of radii 5 cm and 3 cm. Find the area of the ring in terms of π.

Answer it in the portal, where your answer is checked as you type →

Compound solids

A compound solid is made by joining simple solids: a cone on top of a cylinder (a pencil, a rocket), a hemisphere on a cylinder (a grain silo), a cone on a hemisphere (an ice cream).

Volume: add the volumes of the parts, or subtract a part that has been removed (a hole drilled through a block, a hollow in a mould).

Surface area: this needs more care. Add the areas that are on the outside only. Where two solids are joined, the joining faces are hidden inside, so they are left out. For a cone on a cylinder of the same radius, the surface is the cone's curved surface, the cylinder's curved surface and the one circular base, and not the circle where they meet.

Read the question for what counts: "the outside surface", "excluding the base", "an open container". When a hollow is cut into a solid, its inner surface becomes part of the surface area.

Parts of solids

A part of a solid is usually a half or another simple fraction.

  • Hemisphere (half a sphere) of radius r: volume 12×43πr3=23πr3; curved surface 12×4πr2=2πr2. A solid hemisphere also has a flat circular face of area πr2, so its total surface area is 3πr2. A bowl has only the curved surface.
  • Half a cylinder (a trough) cut along its length: half the volume, half the curved surface, plus two semicircular ends and, if closed, a flat rectangle across the top.
  • A cylinder or prism filled only partly, to a given depth, holds the volume of the solid with that depth as its height.

Worked example 3

A toy is a solid cone of radius 4 cm and height 9 cm joined to a solid hemisphere of radius 4 cm, flat face to flat face. Find the total volume of the toy in terms of π.

  1. Cone: 13×π×42×9=48π.

  2. Hemisphere: 23×π×43=128π3.

  3. Total =48π+128π3=144π+128π3=272π3cm3 (about 285cm3).

Answer: 272π3cm3

Check yourself

A solid hemisphere has radius 3 cm. Find its volume in terms of π.

Answer it in the portal, where your answer is checked as you type →

Frustums Extended

A frustum is what is left when the top of a cone (or pyramid) is cut off by a cut parallel to the base. Buckets, lampshades and plant pots are frustums.

Volume of a frustum = volume of the large cone − volume of the small cone removed.

Curved surface area of a frustum = curved surface of the large cone − curved surface of the small cone, that is πRL−πrl with each cone's own radius and slant height. Add the circular ends if the frustum is closed (a bucket has a base but no lid).

Often you are given only the frustum, not the complete cone. Because the small cone is an exact scaled copy of the large one (they are similar), its measurements are in proportion: if the top radius is half the base radius, the small cone's height is half the large cone's height. Use this to find the missing height first. If the frustum has height d and radii R and r, and the large cone has height H, then

HR=H−dr

which you can solve for H.

RrNOT TO SCALE

A frustum is a large cone with a similar small cone (dashed) removed from the top.

Worked example 4 Extended

A frustum has base radius 6 cm, top radius 3 cm and height 5 cm. Calculate its volume.

  1. The top radius is half the base radius, so the small cone is half the height of the large cone.

  2. So the frustum is the bottom half of the height: H−5=12H, giving H=10 cm and a small cone of height 5 cm.

  3. Large cone =13×π×62×10=120π.

  4. Small cone =13×π×32×5=15π.

  5. Frustum =120π−15π=105π=329.86…=330cm3 (3 s.f.).

Answer: 105π=330cm3

Check yourself Extended

A cone of radius 6 cm and height 12 cm has a cone of radius 2 cm and height 4 cm cut from its top, parallel to the base. Find the volume of the frustum that is left, in terms of π.

Answer it in the portal, where your answer is checked as you type →

Key points

  • Compound area: split and add, or take the larger shape and subtract.

  • Find missing lengths from the given ones before calculating.

  • Perimeter: only the outside edges; arcs for curved edges; never the internal split lines.

  • Ring: πR2−πr2. Hemisphere: V=23πr3, curved surface 2πr2.

  • Compound solids: add or subtract volumes; for surface area include only faces on the outside.

  • Extended: frustum = large cone − small cone, for both volume and curved surface; use similarity to find the missing height.

Where marks are lost

  • Including the internal dividing line in a perimeter, or the hidden joining face in a surface area.

  • Missing an edge of a compound shape because it was not labelled.

  • Working out a ring as π(R−r)2.

  • Using the whole sphere for a hemisphere, or forgetting the flat face of a solid hemisphere.

  • Adding the area of a hole instead of subtracting it.

  • Extended: subtracting the small cone's height or radius instead of its volume; a frustum is not a cone with radius R−r.

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An original lesson written to the Cambridge IGCSE Mathematics 0580 syllabus (2025 to 2027). The worked examples and questions are our own, not Cambridge questions. Back to every IGCSE Mathematics 0580 lesson.