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IGCSE Mathematics 0580 · Chapter 8 notes · ~28 min read

Mensuration — area & volume

Convert units, find area and perimeter, then master circles, arcs, sectors, prisms, cylinders, cones, spheres and frustums. Every formula is explained with a picture so you understand where it comes from.

UnitsPerimeter & areaCircles, arcs & sectorsSurface areaVolume of prisms, cylinders, cones, spheres
RSby Riyaz Sir
mentorriyaz.in
No tag = Core (everyone)EXTENDED Extended onlyhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Area is measured in squared units and volume in cubed units. To convert, use the length factor, squared for area and cubed for volume.
  • Circle: C = 2πr = πd and A = πr². For arcs and sectors take a fraction θ ÷ 360 of the whole circle. Prism: V = area of cross-section × length.
  • Surface area = the area of the net. Cones, pyramids and spheres (Extended) have formulae that are given on the formula sheet: you only need to use them.

By the end you can

  • Convert between units of length, area, volume and capacity.
  • Find the perimeter and area of rectangles, triangles, parallelograms, trapezia and compound shapes.
  • Find the circumference and area of circles, and the arc length and area of sectors.
  • Find the volume and surface area of cuboids, prisms and cylinders.
  • Find the volume and surface area of pyramids, cones, spheres and hemispheres. EXT
  • Work with frustums, composite solids and the area of a segment. EXT

1Units of measure

Length, area and volume

LengthAreaVolume01 m100 cm1 m²100 cm × 100 cm1 m³100 × 100 × 100 cm1 m = 100 cm1 cm = 10 mm1 km = 1000 m1 m² = 10 000 cm²1 cm² = 100 mm²1 km² = 1 000 000 m²1 m³ = 1 000 000 cm³1 litre = 1000 cm³1 m³ = 1000 litresArea: square the length factor. Volume: cube it.
A metre is 100 cm, but a square metre is 100 × 100 = 10 000 cm², and a cubic metre is 100 × 100 × 100 = 1 000 000 cm³.

Capacity — how much a container holds. 1 litre = 1000 cm³ = 1000 ml, so 1 ml = 1 cm³, and 1 m³ = 1000 litres.

To changeLengthAreaVolume
m to cm× 100× 10 000× 1 000 000
cm to mm× 10× 100× 1000
km to m× 1000× 1 000 000× 1 000 000 000
  • Going to a smaller unit (m to cm) multiply. Going to a bigger unit (cm to m) divide.
  • Square the factor for area and cube it for volume. This is the step most students forget.
Worked example

Change 3.5 m² to cm².

  1. 1 m = 100 cm, so 1 m² = 100² = 10 000 cm²
  2. 3.5 × 10 000 = 35 000 cm²
Worked example

Change 2 400 000 cm³ to m³.

  1. 1 m³ = 1 000 000 cm³
  2. 2 400 000 ÷ 1 000 000 = 2.4 m³
Worked example

A tank measures 50 cm by 40 cm by 30 cm. How many litres of water does it hold when full?

  1. volume = 50 × 40 × 30 = 60 000 cm³
  2. 1 litre = 1000 cm³
  3. 60 000 ÷ 1000 = 60 litres
Common mistakeUsing 1 m² = 100 cm². That is the length factor. Area needs 100² = 10 000.
Exam tipAlways convert to one unit first, then calculate. Mixing metres and centimetres in one formula gives wrong answers.

2Perimeter and area

Four shapes to know

RectangleA = l × wP = 2(l + w)lwTriangleA = ½ × b × hh = perpendicularheightbhParallelogramA = b × hh = upright height(not the slant)bhTrapeziumA = ½(a + b) × ha and b are theparallel sidesbah
Area is measured with the perpendicular (upright) height. Perimeter is the distance all the way round.

Perimeter — the distance around the edge of a shape. Area is the space inside it.

  • Use the perpendicular height (at right angles to the base) in the triangle, parallelogram and trapezium formulae. Never use the slanted side.
  • Rearranging: to find a missing length, use the formula backwards. For a triangle, h = 2A ÷ b.
Worked example

Find the area and perimeter of a rectangle 12 cm by 7 cm.

  1. area = 12 × 7 = 84 cm²
  2. perimeter = 2(12 + 7) = 38 cm
Worked example

Find the area and perimeter of a right-angled triangle with legs 6 cm and 8 cm.

  1. area = ½ × 6 × 8 = 24 cm²
  2. hypotenuse = √(6² + 8²) = √100 = 10 cm
  3. perimeter = 6 + 8 + 10 = 24 cm
Worked example

A parallelogram has base 10 cm, slanted side 6 cm and perpendicular height 5 cm. Find its area and perimeter.

  1. area = base × perpendicular height = 10 × 5 = 50 cm² (the 6 cm is not used)
  2. perimeter = 2(10 + 6) = 32 cm
Worked example

Find the area of a trapezium with parallel sides 7 cm and 11 cm and height 5 cm.

  1. area = ½(7 + 11) × 5
  2. = ½ × 18 × 5 = 45 cm²
Worked example

A triangle has area 36 cm² and base 9 cm. Find its height.

  1. ½ × 9 × h = 36
  2. 4.5h = 36
  3. h = 8 cm
Worked example

A trapezium has area 60 cm² and parallel sides 8 cm and 12 cm. Find its height.

  1. ½(8 + 12) × h = 60
  2. 10h = 60
  3. h = 6 cm
Common mistakeUsing the slanted side as the height of a parallelogram or triangle. If the question gives a slanted side, check that you are using the upright one.

Why the formulae work

TriangleParallelogramTrapeziumbhtriangle = half of arectangleA = ½ × b × hbhmove the cut triangle:parallelogram = rectangleA = b × hbaabtwo copies make aparallelogram: base a + bA = ½(a + b) × h
Each formula comes from a rectangle: cut it, slide it or double it.
  • Triangle: a rectangle cut along a diagonal gives two equal triangles, so the triangle is half the rectangle.
  • Parallelogram: move the triangle on one side to the other side and you get a rectangle with the same base and height.
  • Trapezium: two copies make a parallelogram with base (a + b), so one trapezium is half of it.

Compound shapes

Add the piecesSubtract the holeSemicircle on top185063510818 + 50 = 68 cm²split into rectangles12681084380 − 12 = 68 cm²subtract the holer = 48540 + ½π(4)² = 65.1 cm²rectangle + semicircle
Split the shape into pieces you know, add them or subtract the part that is missing.
  • Method 1: split into rectangles, triangles and circle parts, find each area and add.
  • Method 2: find the area of the big shape and subtract the part that is not there (a hole or a corner).
  • Perimeter: add up only the outside edges. Where two shapes join, the joining line is not part of the perimeter.
Worked example

Find the area of an L-shape that fits in a 10 cm by 8 cm rectangle with a 4 cm by 3 cm corner removed.

  1. big rectangle = 10 × 8 = 80 cm²
  2. missing corner = 4 × 3 = 12 cm²
  3. area = 80 − 12 = 68 cm²
  4. check by adding: 6 × 3 + 10 × 5 = 18 + 50 = 68 ✓
Worked example

A shape is a rectangle 8 cm by 5 cm with a semicircle of radius 4 cm on one of its 8 cm sides. Find its area and perimeter.

  1. area = 8 × 5 + ½ × π × 4² = 40 + 25.13 = 65.1 cm²
  2. arc length = ½ × 2π × 4 = 4π = 12.57 cm
  3. perimeter = 8 + 5 + 5 + 12.57 = 30.6 cm (the 8 cm where the semicircle joins is not counted twice)
Worked example

A square of side 10 cm has a circle of radius 4 cm cut out of it. Find the area that is left.

  1. square = 10 × 10 = 100 cm²
  2. circle = π × 4² = 16π = 50.27 cm²
  3. left = 100 − 50.27 = 49.7 cm²
Common mistakeIncluding the joining line in the perimeter of a compound shape, or forgetting a curved edge. Trace the outside with a pencil once.

3Circles, arcs and sectors

Circumference and area

Circumference C = π × dArea A = π × r²drwrap the circle onto a line:ddd0.14 d3 diameters and a bit: C ≈ 3.14 × dalso C = 2πr (d = 2r)rfill it with squares of side r:r²r²r²3 squares and a bit: A ≈ 3.14 × r²use the π button on your calculator
A circle's circumference is a little more than 3 diameters. Its area is a little more than 3 squares of side r.

π (pi) — the number 3.14159… that you get when you divide the circumference by the diameter of any circle. Use the π button on your calculator.

circumference C = πd = 2πr
d = diameter, r = radius. d = 2r
area A = πr²
square the radius first, then multiply by π
  • Leave in terms of π when the question asks (for example 16π cm²). Otherwise give 3 significant figures from the calculator value of π.
  • Half circle (semicircle): perimeter = πr + 2r (the curved edge plus the diameter). Area = ½πr².
Worked example

Find the circumference and area of a circle with radius 7 cm.

  1. C = 2π × 7 = 14π = 44.0 cm
  2. A = π × 7² = 49π = 154 cm²
Worked example

Find the circumference and area of a circle with diameter 12 cm.

  1. radius = 12 ÷ 2 = 6 cm
  2. C = π × 12 = 37.7 cm
  3. A = π × 6² = 36π = 113 cm²
Worked example

The circumference of a circle is 30 cm. Find its radius and area.

  1. 2πr = 30, so r = 30 ÷ (2π) = 4.77 cm
  2. A = π × 4.775² = 71.6 cm² (use the unrounded value of r on your calculator)
Worked example

A circular lawn has area 100 m². Find its radius.

  1. πr² = 100
  2. r² = 100 ÷ π = 31.83
  3. r = √31.83 = 5.64 m
Common mistakePutting the diameter into πr². If a question gives the diameter, halve it first.

Arc length and sector area

θ = 72°sectorarcrArc and sectorfraction = θ ÷ 360arc length = fraction × 2πrsector area = fraction × πr²example: r = 10 cm, θ = 72°fraction = 72 ÷ 360 = 1/5arc = 1/5 × 2π × 10 = 12.6 cmarea = 1/5 × π × 10² = 62.8 cm²
An arc and a sector are fractions of the circle. The fraction is the angle at the centre divided by 360°.

Arc — part of the circumference of a circle.

Sector — the slice of the circle between two radii and an arc.

arc length = θ360 × 2πr
θ = angle at the centre, in degrees
sector area = θ360 × πr²
the same fraction of the area
  • Step 1: write the fraction θ ÷ 360 (simplify it). Step 2: multiply it by the circumference or the area of the whole circle.
  • Perimeter of a sector = arc length + 2 × radius (do not forget the two straight edges).
Worked example

A sector has radius 6 cm and angle 150°. Find the arc length, area and perimeter.

  1. fraction = 150 ÷ 360 = 512
  2. arc = 512 × 2π × 6 = 5π = 15.7 cm
  3. area = 512 × π × 6² = 15π = 47.1 cm²
  4. perimeter = 15.7 + 2 × 6 = 27.7 cm
Worked example

An arc of length 8.4 cm lies on a circle of radius 6 cm. Find the angle at the centre.

  1. circumference = 2π × 6 = 37.70 cm
  2. fraction = 8.4 ÷ 37.70 = 0.2228
  3. θ = 0.2228 × 360 = 80.2°
Worked example

A sector of a circle of radius 8 cm has area 40 cm². Find its angle.

  1. whole circle = π × 8² = 201.06 cm²
  2. fraction = 40 ÷ 201.06
  3. θ = 40 ÷ 201.06 × 360 = 71.6°
Common mistakeUsing the fraction upside down or using 360 ÷ θ. The fraction must be less than 1: θ on top, 360 on the bottom.
Memory trick"Fraction × whole": a sector is a fraction of the whole circle. If the angle is 90° you are taking a quarter of the circle.

SegmentsEXTENDED

SectorTriangleSegment−=¼ × π × 10²= 78.5 cm²½ × 10 × 10= 50 cm²78.5 − 50= 28.5 cm²segment = sector − triangle (radius 10 cm, angle 90°)
A segment is what is left when you cut the triangle away from the sector.

Segment — the region between a chord and an arc. It looks like a slice of a lens.

  • EXT Area of segment = area of sector − area of triangle. The triangle has two sides equal to the radius, so its area is ½r² sin θ.
  • EXT For θ = 90° the triangle is right-angled and its area is just ½ × r × r.
Worked example

EXT Find the area of the segment cut off by a chord that makes an angle of 90° at the centre of a circle of radius 10 cm.

  1. sector = ¼ × π × 10² = 78.54 cm²
  2. triangle = ½ × 10 × 10 = 50 cm²
  3. segment = 78.54 − 50 = 28.5 cm²
Worked example

EXT A chord makes an angle of 60° at the centre of a circle of radius 8 cm. Find the area of the segment.

  1. sector = 60360 × π × 8² = 33.51 cm²
  2. triangle = ½ × 8 × 8 × sin 60° = 27.71 cm²
  3. segment = 33.51 − 27.71 = 5.80 cm²

4Prisms, cylinders and surface area

Volume of a prism

Volume of a prism = area of cross-section × lengthlhwcuboidV = l × w × hblengthtriangular prismV = ½bh × lengthrhcylinderV = πr² × hthe shaded face is the cross-section
A prism has the same cross-section all the way along its length, so volume = area of cross-section × length.

Prism — a solid with the same cross-section throughout, such as a cuboid, a triangular prism or a cylinder.

volume of a prism = area of cross-section × length
works for every prism and the cylinder
cylinder: V = πr²h
the cross-section is a circle
  • Step 1: find the area of the cross-section (the shape you see if you slice the prism). Step 2: multiply by the length.
Worked example

Find the volume of a cuboid 8 cm by 5 cm by 3 cm.

  1. V = 8 × 5 × 3 = 120 cm³
Worked example

A prism has a triangular cross-section with base 6 cm and height 4 cm. The prism is 10 cm long. Find its volume.

  1. cross-section area = ½ × 6 × 4 = 12 cm²
  2. V = 12 × 10 = 120 cm³
Worked example

Find the volume of a cylinder with radius 5 cm and height 12 cm.

  1. V = π × 5² × 12 = 300π
  2. = 942 cm³
Worked example

A cylinder has volume 500 cm³ and height 8 cm. Find its radius.

  1. πr² × 8 = 500
  2. r² = 500 ÷ (8π) = 19.89
  3. r = 4.46 cm
Worked example

A cylindrical tank has radius 0.5 m and height 1.2 m. How many litres does it hold?

  1. V = π × 0.5² × 1.2 = 0.3π = 0.9425 m³
  2. 1 m³ = 1000 litres
  3. 942 litres
Common mistakeMixing cm and m in one calculation. Convert everything to the same unit before you multiply.

Surface area of cuboids and prisms

top 15end 6front 10end 6back 10bottom 15Cuboid 5 cm × 3 cm × 2 cmtop + bottom: 2 × 15 = 30front + back: 2 × 10 = 20two ends: 2 × 6 = 12total = 62 cm²2(lw + lh + wh)= 2(15 + 10 + 6) = 62Surface area = the area of the net. Add every face.
Surface area is the total area of every face. Draw the net and tick off each face.
  • Surface area = the area of the net. Count every face, including the ones you cannot see.
  • Cuboid: surface area = 2(lw + lh + wh).
Worked example

Find the surface area of a cuboid 5 cm by 3 cm by 2 cm.

  1. top and bottom: 2 × (5 × 3) = 30
  2. front and back: 2 × (5 × 2) = 20
  3. two ends: 2 × (3 × 2) = 12
  4. total = 62 cm²
Worked example

A triangular prism has a right-angled triangle cross-section with sides 3 cm, 4 cm and 5 cm and length 10 cm. Find its surface area.

  1. two triangles: 2 × (½ × 3 × 4) = 12
  2. three rectangles: (3 + 4 + 5) × 10 = 120
  3. total = 132 cm²

Surface area of a cylinder

CylinderNet (opened out)rhcurved surface 2πr × h+ two circles 2 × πr²2πrhrectangleπr²πr²total surface area = 2πrh + 2πr²
Open a cylinder out: the curved part is a rectangle as wide as the circumference of the circle.
  • Cylinder: curved surface = 2πrh and the two circular ends = 2 × πr². Total = 2πrh + 2πr².
Worked example

Find the total surface area of a cylinder with radius 5 cm and height 12 cm.

  1. curved surface = 2π × 5 × 12 = 120π
  2. two ends = 2 × π × 5² = 50π
  3. total = 170π = 534 cm²
Worked example

An open cylinder (no top) has radius 4 cm and height 10 cm. Find its outside surface area.

  1. curved surface = 2π × 4 × 10 = 80π
  2. one base = π × 4² = 16π
  3. total = 96π = 302 cm²

5Pyramids, cones and spheres EXT

Pyramids and cones

PyramidConehV = ⅓ × base area × ha third of the prismwith the same base and heighthrlV = ⅓πr²hcurved surface = πrll² = r² + h² (slant height)
A pyramid or cone is exactly one third of the prism or cylinder with the same base and height.
pyramid: V = ⅓ × base area × h
h = perpendicular height
cone: V = ⅓πr²h
curved surface area = πrl
slant height l² = r² + h²
Pythagoras in the right-angled triangle inside the cone
  • EXT Total surface area of a cone = curved surface + circular base = πrl + πr².
  • EXT The slant height l is the length along the slope. The vertical height h is inside the cone. Do not mix them up.
Worked example

EXT A square-based pyramid has base 6 cm by 6 cm and height 7 cm. Find its volume.

  1. base area = 6 × 6 = 36 cm²
  2. V = ⅓ × 36 × 7 = 84 cm³
Worked example

EXT A cone has radius 5 cm and vertical height 12 cm. Find its slant height, volume and total surface area.

  1. l² = 5² + 12² = 169, so l = 13 cm
  2. V = ⅓ × π × 5² × 12 = 100π = 314 cm³
  3. curved surface = π × 5 × 13 = 65π
  4. base = π × 5² = 25π
  5. total = 90π = 283 cm²
Common mistakeEXT Using the slant height in the volume formula. Volume needs the vertical height h. Slant height only appears in the curved surface area.

Spheres and hemispheres

SphereHemisphererV = 4/3 πr³surface area = 4πr²rcurvedflat circleV = 2/3 πr³curved surface = 2πr²with the flat base = 3πr²The formulae for the sphere, cone and pyramid are on the formula sheet.
A sphere is described by one number, its radius. The formulae are on the formula sheet.
sphere: V = 4/3πr³
surface area = 4πr²
hemisphere: V = ⅔πr³
curved surface = 2πr²
  • EXT A hemisphere is half a sphere. Its total surface area adds the flat circle: 2πr² + πr² = 3πr².
Worked example

EXT Find the volume and surface area of a sphere of radius 6 cm.

  1. V = 4/3 × π × 6³ = 288π = 905 cm³
  2. surface area = 4π × 6² = 144π = 452 cm²
Worked example

EXT Find the volume and total surface area of a solid hemisphere of radius 3 cm.

  1. V = ⅔ × π × 3³ = 18π = 56.5 cm³
  2. curved surface = 2π × 3² = 18π
  3. flat base = π × 3² = 9π
  4. total = 27π = 84.8 cm²
Worked example

EXT A sphere has volume 500 cm³. Find its radius.

  1. 4/3 × π × r³ = 500
  2. r³ = 500 × 3 ÷ (4π) = 119.37
  3. r = ∛119.37 = 4.92 cm

Frustums

r = 3R = 644removedFrustum = big cone − small conesimilar cones: r ÷ R = h ÷ H3 ÷ 6 = 4 ÷ 8big cone: ⅓π(6²)(8) = 96πsmall cone: ⅓π(3²)(4) = 12πfrustum = 96π − 12π= 84π = 264 cm³
A frustum is a cone with its top cut off. Find its volume by subtracting the small cone from the big one.

Frustum — the part of a cone left when a smaller cone is cut off the top by a plane parallel to the base.

  • EXT Frustum = big cone − small cone. The small cone is similar to the big one, so r ÷ R = h ÷ H.
  • EXT Use the similar-cones ratio to find the missing height, then subtract the two volumes.
Worked example

EXT A cone of height 8 cm and base radius 6 cm has a small cone of height 4 cm and radius 3 cm removed from the top. Find the volume of the frustum.

  1. big cone = ⅓ × π × 6² × 8 = 96π
  2. small cone = ⅓ × π × 3² × 4 = 12π
  3. frustum = 96π − 12π = 84π = 264 cm³

Composite solids

Cone + hemisphereCapsule: cylinder + 2 hemispheresr = 3h = 9cone: ⅓π(3²)(9) = 27πhemisphere: ⅔π(3³) = 18πtotal = 45π = 141 cm³8r = 3cylinder: π(3²)(8) = 72π2 hemispheres: 4/3π(3³) = 36πtotal = 108π = 339 cm³
Split the solid into simple solids, find each volume, then add them.
  • EXT Hidden faces: when you join two solids, the joined faces are not part of the surface area.
Worked example

EXT An ice cream is a cone of radius 3 cm and height 9 cm with a hemisphere on top of the same radius. Find its volume.

  1. cone = ⅓ × π × 3² × 9 = 27π
  2. hemisphere = ⅔ × π × 3³ = 18π
  3. total = 45π = 141 cm³
Worked example

EXT A capsule is a cylinder of radius 3 cm and length 8 cm with a hemisphere on each end. Find its volume.

  1. cylinder = π × 3² × 8 = 72π
  2. two hemispheres make one sphere: 4/3 × π × 3³ = 36π
  3. total = 108π = 339 cm³

Problems with a context

  • Costs and amounts: find the area or volume first, then divide by what one unit covers (and round up for whole boxes or tiles).
Worked example

A rectangular lawn is 12 m by 9 m with a circular pond of radius 2 m. One box of seed covers 30 m². How many boxes are needed?

  1. lawn area = 12 × 9 = 108 m²
  2. pond = π × 2² = 12.57 m²
  3. grass = 108 − 12.57 = 95.43 m²
  4. 95.43 ÷ 30 = 3.18
  5. you need 4 boxes (round up)
Worked example

A floor is 4.2 m by 3.5 m. Square tiles of side 30 cm are used. How many tiles are needed?

  1. floor = 4.2 × 3.5 = 14.7 m²
  2. one tile = 0.3 × 0.3 = 0.09 m²
  3. 14.7 ÷ 0.09 = 163.3
  4. 164 tiles

Formula quick card

  • Learn the Core formulae. The Extended solids are given on the formula sheet. Work from this card when you check.
SolidVolumeSurface area
Cuboidl × w × h2(lw + lh + wh)
Any prismcross-section × lengthsum of the faces
Cylinderπr²h2πrh + 2πr²
Cone (Ext)⅓πr²hπrl + πr²
Pyramid (Ext)⅓ × base area × hsum of the faces
Sphere (Ext)4/3 πr³4πr²
Hemisphere (Ext)⅔πr³3πr² with the flat base

6Test yourself

Test yourself — cover the answers first

Quick check

  1. Change 4.5 m² to cm².
  2. Change 3 500 000 cm³ to m³.
  3. How many litres are in 2.5 m³?
  4. Find the area of a triangle with base 14 cm and height 9 cm.
  5. Find the area of a trapezium with parallel sides 5 cm and 9 cm and height 4 cm.
  6. Find the perimeter of a rectangle 9.5 cm by 4 cm.
  7. Find the circumference of a circle with radius 5 cm, in terms of π.
  8. Find the area of a circle with diameter 10 cm, in terms of π.
  9. Find the area of a circle of radius 3.5 cm to 3 significant figures.
  10. Find the circumference of a circle of diameter 8.2 cm to 3 significant figures.
  11. A circle has circumference 44 cm. Find its radius to 3 significant figures.
  12. Find the length of an arc of a circle with radius 9 cm and angle 40°.
  13. Find the area of a sector with radius 12 cm and angle 90° to 3 significant figures.
  14. Find the volume of a cuboid 6 cm by 4 cm by 2.5 cm.
  15. Find the volume of a cylinder with radius 3 cm and height 10 cm in terms of π.
  16. A prism has a cross-section of area 12 cm² and length 7 cm. Find its volume.
  17. Find the surface area of a cube with side 5 cm.
  18. EXT Find the volume of a cone with radius 3 cm and height 7 cm to 3 significant figures.
  19. EXT Find the volume of a sphere of radius 2 cm in terms of π.
  20. EXT Find the surface area of a sphere of radius 10 cm in terms of π.
  21. EXT A cone has radius 6 cm and height 8 cm. Find its slant height.
  22. EXT Find the volume of a square-based pyramid with base 5 cm by 5 cm and height 9 cm.
  23. EXT Find the total surface area of a solid hemisphere of radius 6 cm in terms of π.

Exam-style

  1. A trapezium has parallel sides 9 cm and 15 cm and an area of 84 cm². Find its height. [3]
  2. A rectangle is 12 cm by 8 cm. A circle of radius 3 cm is cut out of it. Find the area of the shape that is left. [4]
  3. A sector has radius 9 cm and angle 140°. Find (a) the arc length, (b) the area of the sector. [4]
  4. A cylindrical tank has radius 0.8 m and height 2 m. Find its capacity in litres, to the nearest litre. [3]
  5. A box is 40 cm by 25 cm by 30 cm. Find its surface area and its capacity in litres. [4]
  6. A prism has a right-angled triangle cross-section with sides 5 cm, 12 cm and 13 cm. The prism is 20 cm long. Find its volume and its total surface area. [5]
  7. EXT A cone has base radius 7 cm and slant height 25 cm. Find its vertical height, its volume and its curved surface area. Give answers to 3 significant figures. [5]
  8. EXT A sphere has volume 36π cm³. Find its radius and its surface area. [4]
  9. EXT A solid is a cylinder of radius 4 cm and height 10 cm with a hemisphere of radius 4 cm on top. Find its volume to 3 significant figures. [4]
  10. EXT A frustum is made by removing a cone of height 6 cm and radius 3 cm from the top of a cone of height 12 cm and radius 6 cm. Find the volume of the frustum to 3 significant figures. [4]
Show answers
  1. 45 000 cm²
  2. 3.5 m³
  3. 2500
  4. 63 cm²
  5. 28 cm²
  6. 27 cm
  7. 10π cm
  8. 25π cm²
  9. 38.5 cm²
  10. 25.8 cm
  11. 7.00 cm
  12. 6.28 cm (2π)
  13. 113 cm²
  14. 60 cm³
  15. 90π cm³
  16. 84 cm³
  17. 150 cm²
  18. 66.0 cm³
  19. 32π ÷ 3 cm³ (33.5 cm³)
  20. 400π cm²
  21. 10 cm
  22. 75 cm³
  23. 108π cm²
  24. area = ½(9 + 15) × h = 12h
    12h = 84
    h = 7 cm
  25. rectangle = 12 × 8 = 96 cm²
    circle = π × 3² = 28.27 cm²
    left = 96 − 28.27 = 67.7 cm²
  26. fraction = 140 ÷ 360 = 7/18
    (a) arc = 7/18 × 2π × 9 = 7π = 22.0 cm
    (b) area = 7/18 × π × 9² = 31.5π = 99.0 cm²
  27. V = π × 0.8² × 2 = 1.28π = 4.0212 m³
    1 m³ = 1000 litres
    4021 litres
  28. surface area = 2(40 × 25 + 40 × 30 + 25 × 30) = 2(1000 + 1200 + 750) = 5900 cm²
    volume = 40 × 25 × 30 = 30 000 cm³
    capacity = 30 000 ÷ 1000 = 30 litres
  29. cross-section = ½ × 5 × 12 = 30 cm²
    volume = 30 × 20 = 600 cm³
    two triangles = 60 cm²
    rectangles = (5 + 12 + 13) × 20 = 600 cm²
    surface area = 660 cm²
  30. h = √(25² − 7²) = √576 = 24 cm
    V = ⅓ × π × 7² × 24 = 392π = 1230 cm³
    curved surface = π × 7 × 25 = 175π = 550 cm²
  31. 4/3πr³ = 36π, so r³ = 27
    r = 3 cm
    surface area = 4π × 3² = 36π = 113 cm²
  32. cylinder = π × 4² × 10 = 160π
    hemisphere = ⅔ × π × 4³ = 128π ÷ 3
    total = 160π + 128π ÷ 3 = 608π ÷ 3 = 637 cm³
  33. big cone = ⅓ × π × 6² × 12 = 144π
    small cone = ⅓ × π × 3² × 6 = 18π
    frustum = 144π − 18π = 126π = 396 cm³
Practise this chapter
Units of measureArea & perimeterCircles, arcs & sectorsSurface area & volume
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