Volume and Surface Area · applications

Applications: Volume and Surface Area

10 question types · Secondary 2 · each worked step by step with a figure that follows the steps

PSLE · O-Level · SAT · GCSE Higher

01

A Lawn with a Semicircular End and a Pond, and the Rolls of Turf to Cover It

methodAdd the Rectangle and the Semicircle, Take Away the Pond, Then Divide by the Area of One Roll and Round Up

A lawn is a rectangle 12 m long and 8 m wide, with a semicircle on one of its 8 m ends. A rectangular pond 3 m by 2 m lies inside the rectangle and is not turfed. (a) Find the area to be turfed, as an exact expression in π and correct to 1 decimal place. (b) Turf is sold in rolls that each cover 1.5 m², at $6.40 a roll. How many rolls are needed to cover the lawn, and what do they cost?

pond4 m12 m8 mrectangle + semicircle − pond
The lawn is the 12 m by 8 m rectangle and a semicircle of radius 4 m, with the pond taken away.
Split the lawn into shapes whose areas we know: the 12 m by 8 m rectangle and the semicircle on its 8 m end, with the pond taken away. The 8 m end is the diameter of the semicircle, so its radius is 4 m.
step 1 of 5

The area of a shape made of simple pieces is the sum of the areas of the pieces, less the area of any piece cut out of it. Turf is bought in whole rolls, so the number of rolls is rounded up.

  1. Split the lawn into shapes whose areas we know: the 12 m by 8 m rectangle and the semicircle on its 8 m end, with the pond taken away. The 8 m end is the diameter of the semicircle, so its radius is 4 m.
  2. The rectangle has area 12 × 8 = 96 m². The semicircle is half a circle of radius 4 m, so its area is 12 × π × 42 = 8π m². The pond has area 3 × 2 = 6 m².
  3. (a) The area to be turfed is 96 + 8π − 6 = 90 + 8π m². Since 8π ≈ 25.13, this is 115.13 m², which is 115.1 m² correct to 1 decimal place.
  4. Each roll covers 1.5 m², so the lawn needs 115.13 ÷ 1.5 ≈ 76.76 rolls. Rolls are bought whole, and 76 rolls cover only 76 × 1.5 = 114 m², which is less than the lawn. So the number is rounded up to 77.
  5. (b) The lawn needs 77 rolls, which cost 77 × $6.40 = $492.80. Check: 77 rolls cover 77 × 1.5 = 115.5 m², just more than the 115.13 m² of lawn.

answer(a) 90 + 8π m², which is 115.1 m² to 1 decimal place; (b) 77 rolls, costing $492.80

techniqueArea of Composite Shapes

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Taking the radius of the semicircle as 8 m, because the end of the lawn is 8 m long. The 8 m end is the diameter, so the radius is 4 m, and the semicircle's area is 8π m², not 32π m².
  • Rounding 76.76 rolls down to 76. Seventy-six rolls cover 114 m² and leave more than 1 m² of bare soil, and part of a roll cannot be bought, so the number of rolls is rounded up.
02

A Swimming Pool with a Sloping Floor, and How Long a Pump Takes to Fill It

methodTreat the Pool as a Prism with a Trapezium Cross-Section, Then Convert Cubic Meters to Liters and Divide by the Pumping Rate

A swimming pool is 25 m long and 10 m wide, and its sides are vertical. Its floor slopes evenly from a depth of 1 m at the shallow end to a depth of 2 m at the deep end. (a) Find the volume of water in the pool when it is full, in cubic meters. (b) The empty pool is filled by a pump that delivers 25 liters of water every second. How long does it take to fill the pool, in hours and minutes?

1 m2 m10 m25 mside wall: a trapezium, sides 1 m and 2 m
The side wall is a trapezium: its parallel sides are the depths 1 m and 2 m, 25 m apart.
Every cut across the pool parallel to its side wall gives the same shape, so the pool is a prism. Its cross-section is the side wall: a trapezium whose parallel sides are the depths, 1 m and 2 m, with the length of the pool, 25 m, between them.
step 1 of 5

A prism has the same cross-section all along its length, and its volume is the area of the cross-section times the length. A pool whose floor slopes from one end to the other is a prism lying on its side: the cross-section is the side wall, which is a trapezium, and the prism runs across the width of the pool.

  1. Every cut across the pool parallel to its side wall gives the same shape, so the pool is a prism. Its cross-section is the side wall: a trapezium whose parallel sides are the depths, 1 m and 2 m, with the length of the pool, 25 m, between them.
  2. The area of the trapezium is half the sum of the parallel sides times the distance between them: 12 × (1 + 2) × 25 = 37.5 m².
  3. (a) The volume is the area of the cross-section times the width of the pool: 37.5 × 10 = 375 m³.
  4. One cubic meter is 1000 liters, so the full pool holds 375 × 1000 = 375 000 liters. The pump delivers 25 liters each second, so it needs 375 000 ÷ 25 = 15 000 seconds.
  5. (b) 15 000 seconds is 15 000 ÷ 60 = 250 minutes, which is 4 hours 10 minutes. Check: 250 minutes is 15 000 seconds, and 15 000 × 25 = 375 000 liters, which is 375 m³.

answer(a) 375 m³; (b) 4 hours 10 minutes

techniquePrisms and Cylinders

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Using the greatest depth for the whole pool, 25 × 10 × 2 = 500 m³. The floor slopes, so the pool is 2 m deep only at the deep end; the trapezium takes the average of the two depths, 1.5 m.
  • Dividing 375 by 25 as if the pump delivered 25 cubic meters a second. The rate is in liters, so the volume must be turned into liters first: 1 m³ is 1000 liters.
03

A Square Pyramid Tent: the Air Inside and the Canvas for Its Sloping Sides

methodTake One Third of the Base Area Times the Height, Then Use Pythagoras from the Top to the Middle of a Base Edge for the Height of Each Face

A tent is a pyramid on a square base of side 3 m, with its top 2 m directly above the center of the base. It has no floor. (a) Find the volume of air inside the tent. (b) Each sloping face is a triangle. Find the slant height of a face, measured from the top straight down the middle of the face, and the area of canvas needed for the four sloping faces.

3 mheight 2 mbase 3 × 3 = 9 m2
The base is a square of side 3 m, of area 9 m².
The base is a square of side 3 m, so its area is 3 × 3 = 9 m².
step 1 of 5

A pyramid holds one third of the prism on the same base with the same height, so its volume is 13 of the base area times the height. The height of each sloping face is the slant height, which is the hypotenuse of a right-angled triangle inside the pyramid.

  1. The base is a square of side 3 m, so its area is 3 × 3 = 9 m².
  2. (a) The volume of the pyramid is 13 × 9 × 2 = 6 m³. So the tent holds 6 m³ of air.
  3. The slant height runs from the top to the midpoint of a base edge. The height of the tent, the slant height and the line from the center of the base to that midpoint make a right-angled triangle, with the right angle at the center of the base. The line from the center to the midpoint is half a side, 1.5 m.
  4. By Pythagoras' theorem, the slant height l satisfies l2 = 22 + 1.52 = 4 + 2.25 = 6.25, so l = √6.25 = 2.5 m.
  5. (b) Each sloping face is a triangle with base 3 m and height 2.5 m, so its area is 12 × 3 × 2.5 = 3.75 m². The four faces need 4 × 3.75 = 15 m² of canvas. The slant height is 2.5 m.

answer(a) 6 m³; (b) slant height 2.5 m, canvas 15 m²

techniquePyramids and Cones

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Using the height of the tent, 2 m, as the height of each triangular face. The face leans inward, so its height is the slant height, 2.5 m, which is the hypotenuse and longer than the upright height.
  • Leaving out the 13 and giving 9 × 2 = 18 m³. That is the volume of the box around the tent; a pyramid holds one third of it.
04

A Spherical Water Tank: What It Holds and the Paint for Two Coats

methodUse the Sphere's Volume and Surface Area Formulas, Convert Cubic Meters to Liters, and Round Up to Whole Cans

A water tower holds its water in a spherical steel tank of radius 2.5 m. Ignore the thickness of the steel. (a) How many liters of water does the tank hold when full? Give the answer correct to 3 significant figures. (b) The whole outside of the tank is given two coats of paint. One liter of paint covers 12 m², and the paint is sold in 5-liter cans. How many cans are needed?

2.5 mV = 4/3 × pi × 2.53= 65.45 m3
The tank holds 43π × 2.53 ≈ 65.45 m³.
With r = 2.5 m, r3 = 15.625. The volume of the tank is 43 × π × 15.625 ≈ 65.45 m³.
step 1 of 5

A sphere of radius r has volume 43π r3 and surface area 4π r2. A volume in cubic meters is changed to liters by multiplying by 1000, and paint is bought in whole cans.

  1. With r = 2.5 m, r3 = 15.625. The volume of the tank is 43 × π × 15.625 ≈ 65.45 m³.
  2. (a) One cubic meter is 1000 liters, so the tank holds about 65.45 × 1000 = 65 450 liters. Correct to 3 significant figures, that is 65400 liters.
  3. The surface area of the tank is 4π r2 = 4 × π × 2.52 = 25π ≈ 78.54 m².
  4. Two coats cover the surface twice, which is 2 × 78.54 = 157.08 m². One liter covers 12 m², so the paint needed is 157.08 ÷ 12 ≈ 13.09 liters.
  5. (b) Three cans hold 3 × 5 = 15 liters, which is more than 13.09 liters, and two cans hold only 10 liters. So 3 cans are needed.

answer(a) 65400 liters, to 3 significant figures; (b) 3 cans

techniqueSpheres · Cones and Spheres

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Mixing up the two formulas, using 4π r2 for the volume or 43π r3 for the area. A volume is in cubic meters and needs r3; an area is in square meters and needs r2.
  • Painting the surface only once. Two coats need twice the area; one coat alone takes 78.54 ÷ 12 ≈ 6.5 liters, which is half the paint needed.
05

Two Solid Chocolate Bunnies of the Same Shape: the Mass and the Foil for the Larger One

methodFind the Length Scale Factor, Then Cube It for the Mass and Square It for the Area of Foil

A chocolate maker sells two solid chocolate bunnies of exactly the same shape. The small bunny is 8 cm tall and has a mass of 60 g. The large bunny is 20 cm tall. (a) Find the mass of the large bunny. (b) Each bunny is wrapped in foil that covers its surface exactly once, and the small bunny needs 90 cm² of foil. How much foil does the large bunny need?

8 cm20 cm60 g90 cm2foil? g? cm2foillength scale factor k = 20/8 = 2.5
The bunnies are similar, with length scale factor k = 20 ÷ 8 = 2.5.
The bunnies are similar, so every length of the large bunny is the same multiple of the matching length of the small one. The length scale factor is k = 20 ÷ 8 = 2.5.
step 1 of 5

When two solids are similar with length scale factor k, every area is multiplied by k2 and every volume by k3. Solid chocolate is the same all the way through, so the mass follows the volume, and the foil follows the surface area.

  1. The bunnies are similar, so every length of the large bunny is the same multiple of the matching length of the small one. The length scale factor is k = 20 ÷ 8 = 2.5.
  2. Volumes are multiplied by k3 = 2.53 = 15.625. Both bunnies are solid chocolate, so the mass is multiplied by the same number.
  3. (a) The large bunny has a mass of 60 × 15.625 = 937.5 g.
  4. Areas are multiplied by k2 = 2.52 = 6.25, and the foil covers the surface of the bunny.
  5. (b) The large bunny needs 90 × 6.25 = 562.5 cm² of foil. Check: the mass has been multiplied by 937.5 ÷ 60 = 15.625 and the foil by 562.5 ÷ 90 = 6.25, and 6.25 × 2.5 = 15.625, as it should be when the lengths are multiplied by 2.5.

answer(a) 937.5 g; (b) 562.5 cm²

techniqueScaling Area and Volume

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Multiplying the mass by the length scale factor, 60 × 2.5 = 150 g. The large bunny is 2.5 times as tall, but also 2.5 times as wide and 2.5 times as deep, so it holds 2.53 times as much chocolate.
  • Using k3 for the foil as well, which gives 90 × 15.625 = 1406.25 cm². Foil covers a surface, and surfaces are multiplied by k2.
06

A Soccer Ball Stitched from Pentagons and Hexagons: Its Seams and Its Corners

methodCount the Edges of the Panels and Halve the Count, Then Apply Euler's Formula V − E + F = 2

A soccer ball is stitched together from 32 leather panels: 12 regular pentagons and 20 regular hexagons. Each seam joins an edge of one panel to an edge of another, and the seams meet at corners. (a) How many seams are there? (b) Use Euler's formula V − E + F = 2 to find the number of corners. The same number of panels meet at every corner: how many?

pentagonhexagon12 × 5 + 20 × 6 = 180 panel edges
One pentagon and the five hexagons round it, laid flat. All the panels have 12 × 5 + 20 × 6 = 180 edges.
Count the edges of the panels, one kind at a time. The 12 pentagons have 12 × 5 = 60 edges and the 20 hexagons have 20 × 6 = 120 edges, which is 60 + 120 = 180 panel edges in all.
step 1 of 5

Euler's formula says that for a solid with flat faces and no holes, V − E + F = 2, where V is the number of corners (vertices), E the number of edges and F the number of faces. A ball stitched from panels is counted in the same way: the panels are the faces, the seams are the edges, and the points where seams meet are the corners.

  1. Count the edges of the panels, one kind at a time. The 12 pentagons have 12 × 5 = 60 edges and the 20 hexagons have 20 × 6 = 120 edges, which is 60 + 120 = 180 panel edges in all.
  2. (a) Each seam joins two panel edges, so counting panel edges counts every seam twice. There are 180 ÷ 2 = 90 seams.
  3. The panels are the faces, so F = 12 + 20 = 32, and the seams are the edges, so E = 90. Euler's formula gives V − 90 + 32 = 2.
  4. So V = 2 + 90 − 32 = 60. The ball has 60 corners.
  5. (b) The panels have 12 × 5 + 20 × 6 = 180 corners between them, and these gather at the 60 corners of the ball. So 180 ÷ 60 = 3 panels meet at each corner. The ball has 60 corners, with 3 panels at each.

answer(a) 90 seams; (b) 60 corners, with 3 panels meeting at each

techniqueEuler’s Formula for Polyhedra

examsGCSE Higher

Common pitfalls

  • Taking 180 as the number of seams. Every seam runs along the edges of two panels, so the count of panel edges is twice the number of seams.
  • Putting F = 12 or F = 20, the number of one kind of panel, into Euler's formula. F counts every face of the solid, and all 32 panels are faces.
07

A Set of Role-Playing Dice: the Twenty-Sided Die, and Which New Dice Could Be Made

methodCount Face Edges and Face Corners and Divide by How Many Faces Share Each, Then Add the Angles That Meet at a Corner

A set of role-playing dice has dice with 4, 6, 8, 12 and 20 faces. On each die every face is the same regular polygon, and the same number of faces meet at every corner. (a) The 20-sided die has equilateral triangles for faces, with five meeting at each corner. Find its number of edges and its number of corners. (b) A designer wants a new die of the same kind, either with six equilateral triangles at every corner or with three regular hexagons at every corner. Explain why neither can be made, and list every regular polygon, with the number of faces at each corner, that could make a die of this kind.

60 deg gapfive triangles300 deg20 × 3 = 60 face edges, two to an edge: E = 30
Each edge of the die is shared by two triangles: 60 ÷ 2 = 30 edges.
Count the edges face by face: 20 triangles have 20 × 3 = 60 edges between them. Each edge of the die is shared by two faces, so the die has E = 60 ÷ 2 = 30 edges.
step 1 of 5

A solid whose faces are all the same regular polygon, with the same number of faces at every corner, is a Platonic solid. Counting the edges and corners of all the faces, then dividing by how many faces share each one, gives the solid's edges and corners. At a corner of a solid the angles of the faces meeting there must add to less than 360°, and that limits which polygons can be used.

  1. Count the edges face by face: 20 triangles have 20 × 3 = 60 edges between them. Each edge of the die is shared by two faces, so the die has E = 60 ÷ 2 = 30 edges.
  2. (a) The triangles also have 20 × 3 = 60 corners between them, and five of them meet at each corner of the die, so V = 60 ÷ 5 = 12. The die has 30 edges and 12 corners. Check with Euler's formula: V − E + F = 12 − 30 + 20 = 2.
  3. At a corner of a solid, the angles of the faces meeting there must add to less than 360°. If they add to exactly 360°, the faces lie flat and cannot fold up into a corner. Six equilateral triangles give 6 × 60° = 360°, and three regular hexagons give 3 × 120° = 360°, so neither can make a die.
  4. At least three faces meet at every corner. Equilateral triangles, with angles of 60°, can meet three, four or five at a time, making 180°, 240° or 300°. Squares, with 90°, can meet only three at a time, making 270°. Regular pentagons, with 108°, can meet only three at a time, making 324°.
  5. (b) A regular polygon with six or more sides has angles of at least 120°, and three of them make at least 360°. So the only possible dice are triangles with 3, 4 or 5 at a corner, squares with 3, and pentagons with 3: 5 shapes in all, which are the 4-, 8-, 20-, 6- and 12-sided dice of the set.

answer(a) 30 edges and 12 corners; (b) six triangles make 6 × 60° = 360° and three hexagons make 3 × 120° = 360°, so each lies flat; the only possibilities are triangles with 3, 4 or 5 at a corner, squares with 3 and pentagons with 3, which is 5 shapes

techniqueWhy There Are Exactly Five Platonic Solids · Euler’s Formula for Polyhedra

examsGCSE Higher

Common pitfalls

  • Dividing the 60 triangle corners by 3, the number of corners of one triangle, to get 20 corners. What matters is how many faces meet at each corner of the die, which is 5.
  • Thinking that hexagons are a good choice because they fit together with no gaps, as in a honeycomb. Fitting with no gap is exactly what makes them lie flat; a corner of a solid needs the angles to add to less than 360° so that the faces can fold up.
08

Three Tennis Balls in a Can: the Fraction of the Can They Fill, and the Air Around Them

methodWrite Both Volumes in Terms of the Ball's Radius, with the Can as Tall as Six Radii

Three tennis balls, each 6.7 cm across, are packed in a cylindrical can that fits them exactly: each ball touches the curved side, and the stack of three touches the base and the lid. (a) What fraction of the can's volume do the balls fill? (b) Find the volume of air in the can, in cm³, correct to 3 significant figures.

r6rballs 6.7 cm acrossr = 3.35 cm; the can is 6r tall
The can has the radius r = 3.35 cm of a ball and the height of three diameters, 6r.
Let the radius of a ball be r cm, so r = 6.7 ÷ 2 = 3.35. The can has the same radius as a ball, and its height is three diameters, which is 6r.
step 1 of 5

A sphere of radius r has volume 43π r3, and a cylinder of radius r and height h has volume π r2 h. When the can fits the balls exactly, both its radius and its height are fixed by the radius of a ball, so both volumes can be written in terms of r alone and compared.

  1. Let the radius of a ball be r cm, so r = 6.7 ÷ 2 = 3.35. The can has the same radius as a ball, and its height is three diameters, which is 6r.
  2. The three balls have volume 3 × 43π r3 = 4π r3. The can has volume π r2 × 6r = 6π r3.
  3. (a) The balls fill 4π r36π r3 = 23 of the can. The r3 cancels, so the fraction is the same whatever the size of the balls.
  4. The air is the can less the balls: 6π r3 − 4π r3 = 2π r3, which is the other third of the can.
  5. (b) With r = 3.35, r3 = 37.595 to 3 decimal places, so the air is 2 × π × 37.595 ≈ 236.2 cm³. That is 236 cm³ correct to 3 significant figures.

answer(a) 23 of the can; (b) 236 cm³

techniqueArchimedes and the Volume of a Sphere · Spheres · Prisms and Cylinders

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Using the 6.7 cm across the ball as the radius. That is the diameter; using it as the radius makes every volume 23 = 8 times too large.
  • Taking the height of the can as 3r, three radii, instead of three diameters. The three balls stand one on another, and each is 2r tall, so the can is 6r tall.
09

A Party Hat Rolled from a Sector of Card: Its Slant Height, Its Area and the Angle to Cut

methodUse Pythagoras for the Slant Height and πrl for the Curved Area, Then Set the Sector's Arc Equal to the Circumference of the Rim

A party hat is a cone with no base. The circle round its rim has radius 7 cm, and the hat is 24 cm tall. It is made by rolling up a sector cut from a sheet of card, with no overlap. (a) Find the slant height of the hat, and the area of card in the sector as an exact expression in π and correct to 1 decimal place. (b) Find the angle at the center of the sector.

25 cm24 cm7 cml2= 72+ 242= 625, l = 25 cm
The slant height is the hypotenuse: l2 = 72 + 242 = 625, so l = 25 cm.
The height of the hat, the radius of the rim and the slant height l make a right-angled triangle, with the slant height as the hypotenuse. By Pythagoras' theorem, l2 = 72 + 242 = 49 + 576 = 625, so l = 25 cm.
step 1 of 5

A cone's curved surface, cut along a slant line and laid flat, is a sector of a circle. The radius of the sector is the cone's slant height, and the arc of the sector is the circumference of the cone's base. The curved area is π r l, where r is the radius of the base and l the slant height.

  1. The height of the hat, the radius of the rim and the slant height l make a right-angled triangle, with the slant height as the hypotenuse. By Pythagoras' theorem, l2 = 72 + 242 = 49 + 576 = 625, so l = 25 cm.
  2. (a) The curved surface of the cone has area π r l = π × 7 × 25 = 175π cm², which is about 549.8 cm². So the slant height is 25 cm, and the sector is 175π cm² of card.
  3. When the sector is rolled up, its two straight edges meet along a slant line of the cone, so the radius of the sector is the slant height, 25 cm. Its arc goes once round the rim, so the arc length is the circumference of the rim, 2 × π × 7 = 14π cm.
  4. A full circle of radius 25 cm has circumference 2 × π × 25 = 50π cm. The sector's arc is 14π50π = 725 of it, so the sector's angle is 725 of 360°.
  5. (b) The angle at the center of the sector is 725 × 360° = 100.8°. Check with the area: the sector is 725 of a circle of radius 25 cm, and 725 × π × 252 = 175π cm², the area found in (a).

answer(a) slant height 25 cm; card 175π cm², which is 549.8 cm² to 1 decimal place; (b) 100.8°

techniqueThe Slant Height of a Cone · Cones and Spheres

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Using the height, 24 cm, as the radius of the sector. The straight edges of the sector become the sloping side of the hat, so the sector's radius is the slant height, 25 cm.
  • Using the rim's radius, 7 cm, as the radius of the sector, or setting the arc equal to the whole circumference of the circle of radius 25 cm. The sector is cut from a circle of radius 25 cm, and only its arc, 14π cm, wraps round the rim.
10

A Grain Silo on a Cone-Shaped Hopper: Its Volume and the Area to Paint

methodAdd the Volumes of the Cylinder and the Cone; for the Paint, Add Only the Outside Surfaces, with the Cone's Slant Height from Pythagoras

A grain silo is a cylinder of radius 2.5 m and height 12 m with a flat roof. It stands on a cone-shaped hopper of the same radius, whose point is 6 m below the bottom of the cylinder. (a) Find the total volume of the silo, as a multiple of π and correct to 3 significant figures. (b) The whole outside of the silo, meaning the roof, the curved wall and the hopper, is to be painted. Find the area to be painted, as a multiple of π and correct to 3 significant figures.

2.5 m12 m6 mcylinder pi × 2.52× 12 = 75 pihopper 1/3 × pi × 2.52× 6 = 12.5 pi
The cylinder holds 75π m³ and the cone-shaped hopper 13π × 2.52 × 6 = 12.5π m³.
The cylinder has volume π r2 h = π × 2.52 × 12 = 75π m³. The hopper is a cone of radius 2.5 m and height 6 m, so its volume is 13 × π × 2.52 × 6 = 12.5π m³.
step 1 of 5

The volume of a combined solid is the sum of the volumes of its parts. Its surface area is the sum of the outside surfaces only: where two parts are joined, the faces that meet are inside the solid and are not counted.

  1. The cylinder has volume π r2 h = π × 2.52 × 12 = 75π m³. The hopper is a cone of radius 2.5 m and height 6 m, so its volume is 13 × π × 2.52 × 6 = 12.5π m³.
  2. (a) The total volume is 75π + 12.5π = 87.5π m³. Since 87.5π ≈ 274.9, that is 275 m³ correct to 3 significant figures.
  3. The curved surface of the hopper needs its slant height l. The radius and the height of the cone are the two shorter sides of a right-angled triangle, so l2 = 2.52 + 62 = 6.25 + 36 = 42.25, and l = 6.5 m.
  4. The roof is a circle of area π × 2.52 = 6.25π m². The curved wall has area 2π r h = 2 × π × 2.5 × 12 = 60π m². The hopper's curved surface has area π r l = π × 2.5 × 6.5 = 16.25π m². The circle where the cylinder meets the hopper is inside the silo, so it is not painted.
  5. (b) The area to be painted is 6.25π + 60π + 16.25π = 82.5π m². Since 82.5π ≈ 259.2, that is 259 m² correct to 3 significant figures.

answer(a) 87.5π m³, which is 275 m³ to 3 significant figures; (b) 82.5π m², which is 259 m² to 3 significant figures

techniqueCombined Volume and Surface Area · Surface Area of a Cylinder · Pyramids and Cones

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Using the height of the hopper, 6 m, in π r l. The curved surface of a cone is measured along its slope, so it needs the slant height, 6.5 m.
  • Adding the two circles where the cylinder and the hopper meet. They are joined to each other inside the silo, so no paint goes on them.
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