# SOLID SHAPES M.K. HOME TUITION. Mathematics Revision Guides Level: GCSE Higher Tier

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1 Mathematics Revision Guides Solid Shapes Page 1 of 19 M.K. HOME TUITION Mathematics Revision Guides Level: GCSE Higher Tier SOLID SHAPES Version: 2.1 Date:

2 Mathematics Revision Guides Solid Shapes Page 2 of 19 SOLID SHAPES The cube. The cube is the most familiar of the regular solids. It has 6 square faces, 8 corners or vertices (singular vertex) and 12 edges. To the right of the figure of the cube is its net. The net of a solid can be formed by cutting along certain edges and unfolding the solid to give a flat map. To make a solid cube from the net, cut out the net, score the edges and cement the tabbed edges to the plain ones so that three faces meet at a vertex. Because all the edges of a cube are equal in length and all the faces are squares, the formulae for its area and volume are especially simple: Volume of cube = l where l is the length of an edge. Surface area of cube = 6l 2, again where l is the length of an edge. Example (1): Find the surface area and volume of a cube whose sides are 5 cm long. The volume of the cube is 5 cm = 125 cm. The surface area is cm 2 = 150 cm 2. Example (2): An open cubic box (without lid) has an inside surface area of.2 m 2. Find its volume. Because we are told that the box has no lid, there are five faces making up the surface area. One face therefore has a surface area of 0.64m 2, and therefore a side of 0.64 metres or 0.8m. The volume is therefore 0.8 m, or 0.512m, or 512 litres.

3 Mathematics Revision Guides Solid Shapes Page of 19 The cuboid. A cuboid is related to the cube, but is slightly less regular, with at least one pair of rectangles for faces rather than having six square faces like the cube. Cuboids are very familiar in everyday life most boxes are of that shape. A cuboid has the same number of faces, vertices and edges as a cube, but now only opposite pairs of faces are equal. One such pair has a total area of 2ab, (length depth). There are two other pairs, one with combined area 2ac, (length height) and a third pair with combined area 2bc, (depth height). The formulae for the cube can be adapted as follows: Volume of cuboid = abc where a, b and c are the length, depth and height. Surface area of cuboid = 2(ab + ac + bc); again a, b and c are the length, depth and height.

4 Mathematics Revision Guides Solid Shapes Page 4 of 19 Example (): Find the surface area and volume of a cuboid whose dimensions are 8 cm 5 cm 4 cm. The volume of the cuboid is cm = 160 cm. The surface area is 2((8 5) + (8 4) + (5 4)) cm 2 = 2 ( ) cm 2 = 184 cm 2. Example (4): A cuboidal room is 5 metres long, 4 metres wide and 2.5 metres high. The ceiling and walls require two coats of paint, with a covering power of 12.5 m 2 per litre. Work out the amount of paint needed after deducting 7 m 2 for doorways and windows. Area of ceiling = length width = 5 4 m 2 = 20 m 2. Areas of walls = (perimeter height) = 2 ((length height) + (width height)) = 2 ((5 2.5) + (4 2.5)) = 45 m 2. This gives a total area of 65 m 2, but we must deduct 7m 2 for doorways and windows, giving the total area to be painted as 58 m 2. This area requires painting twice, so we need paint to cover 116 m 2. Since 1 litre of paint covers 12.5 m 2, we will need 116 litres, or 9.28 litres Two 5-litre cans will therefore be enough!

5 Mathematics Revision Guides Solid Shapes Page 5 of 19 The prism. A prism is any solid of constant cross-section. The side faces are all rectangles (they can be squares) and the end faces are congruent. (The net above is of a triangular prism, seen in the physics lab and a certain brand of Swiss chocolate!) The hexagonal prism is found in the honeycomb, and in an unused pencil. To find the volume of a prism, we multiply the cross-sectional surface area by the length. Volume of prism = surface area of end face length. Example (5a): A triangular prism has two right-angled triangles for its end faces, and is 20 cm long. The sides of the end faces are 9 cm, 12 cm and 15 cm. Find the volume of the prism. Because the triangular end faces are right-angled, with a base of 12 cm and a height of 9 cm, the surface area of a single face = ½ 12 9 cm 2 = 54 cm 2. Since the length of the prism is 20 cm, its volume is therefore cm = 1080 cm.

6 Mathematics Revision Guides Solid Shapes Page 6 of 19 Example (5b): Find the total surface area of the triangular prism in example (5a). We have already calculated the area of a single triangular end face as 54 cm 2. The two of them thus have a total area of 108 cm 2. We now calculate the areas of the rectangular side faces. The base has an area of cm 2 = 240 cm 2. The vertical face has an area of 9 20 cm 2 = 180 cm 2. The sloping face has an area of cm 2 = 00 cm 2. The three side faces have a combined area of ( ) cm 2 = 720 cm 2. Adding the area of the end faces gives the total surface area of the prism as , or 828 cm 2. Some of the calculations could have been simplified by adding the side lengths of the triangular end faces to give the perimeter, and then multiplying the perimeter by the length of the prism. The three side faces have a combined area of ( ) 20 cm 2 = 720 cm 2. The net of the prism shows how we could treat the three rectangular side faces as if they were a single rectangle, of dimensions (length perimeter of ends). Hence: Surface area of prism = (2 surface area of ends) + (length perimeter of ends). (The perimeter of the end face on the net is equal to the combined widths of the rectangular side faces). In the case of the circular prism, otherwise known as the cylinder, the perimeter is the circumference.

7 Mathematics Revision Guides Solid Shapes Page 7 of 19 Example (6) : SwissChoc bars are sold in cardboard boxes as shown on the right. The box is a prism whose end faces are equilateral triangles. Given that the surface area of each triangular end face is 10.8 cm 2, i) calculate the volume of the box to the nearest cm ; ii) calculate its total surface area to the nearest cm 2. i) Since the volume of a prism = surface area of end face length, the volume of the container is ( ) cm = 270 cm. ii) The two end faces have a total area of cm 2 = 21.6 cm 2. The three side faces have a combined area of ( 5) 25 cm 2 = 75 cm 2. Therefore the total surface area of the prism is ( ) cm 2 = 97 cm 2.

8 Mathematics Revision Guides Solid Shapes Page 8 of 19 The cylinder. A cylinder is a special case of a prism, where the end faces are circles. A cylinder has two end faces, and a single continuous curved face, like the label around a can. If we were to cut such a label, its length would be equal to the perimeter, or circumference, of the circle and its height would be the same as that of the cylinder. See the net of a cylinder of radius r and height h shown right. Remember that the circumference is given by the formula C = 2r. Thus, the total surface area of cylinder = 2rh +2r 2 or 2r(r+ h). The 2rh refers to the curved side and the 2r 2 refers to the circular ends. Example (7): A cylinder has a height of 12 cm, and each end face has a radius of 5 cm. Calculate the volume and total surface area, leaving the results in terms of. The volume is given by the formula V = r 2 h, where r = 5 and h = 12. Hence the cylinder has a volume of ( ) cm or 00 cm. The two circular end faces have a total area of 2r 2 or (2 5 2 ) cm 2 or 50 cm 2. The curved side has an area of 2rh or (2 5 12) cm 2 or 120 cm 2. The total surface area of the cylinder is cm = 170 cm 2. Example (8): A cylindrical drum is 1.2m tall and has a diameter of 0.8m. Find its capacity in litres. (1000 litres = 1 cubic metre.) The radius is half the diameter, i.e. 0.4m, and so the volume of the cylinder is given by m = 0.60m or 60 litres. Example (9): A gas storage cylinder is 12m tall with a diameter 9.5m. If 1 litre of paint covers 12m 2, how much paint will be required to paint its top and side? The area of the curved side is 2rh m 2, or m 2, or 58m 2. The area of the top is m 2 or 71m 2. The total area to be painted is 429m 2, requiring about 6 litres of paint.

9 Mathematics Revision Guides Solid Shapes Page 9 of 19 The pyramid. A pyramid is a solid with a polygonal base and sloping triangular faces, which all meet at an apex. Although the example above (with its net) is regular, with a square base and all its triangular faces equilateral, this does not need generally need to be the case see the hexagonal example below. The formula for the volume of a pyramid is V 1 Ah where A is the area of the base and h is the perpendicular height (see diagram right). This holds true whether the pyramid is upright or slanted. If we were to make a cut parallel to the base of the pyramid, we will have two resulting solids, one of which will be a smaller, similar pyramid. The other section, minus the smaller pyramid, is called a frustum (see left). As the section removed is itself a pyramid similar to the main one, the formula for the volume of a frustum can be given as V 1 ( AH ah) where A and H are the base area and (perpendicular) height of the large pyramid, and a and h are the base area and height of the smaller missing section. If the height of neither pyramid is given, then we must apply ratios to the height of the frustum.

10 Mathematics Revision Guides Solid Shapes Page 10 of 19 The tetrahedron. A tetrahedron is a special case of a pyramid with a triangular base. A regular tetrahedron (shown above with its net) has 4 equilateral triangles for its faces, with 4 vertices and 6 edges. Because a tetrahedron is a pyramid, the formula for its volume is base and h is the perpendicular height. V 1 Ah where A is the area of the Example (10): Candles are made in the shape of a square-based pyramid. Their bases are 10cm square and their heights are 18cm. How many such candles can be made out of a cuboidal block of wax measuring 1m 0.6m 0.4m, assuming no wastage of wax? The volume of the cuboidal block is cm = 240,000 cm (Use consistent units) The base area of one candle is = 100 cm 2, and its height is 18 cm, and so the volume of wax 1 used per candle is V cm The number of candles that can be made from the block is

11 Mathematics Revision Guides Solid Shapes Page 11 of 19 Example (11): A square pyramid has a height of 60 cm and a base side length of 0 cm. A cut is made parallel to the base of the pyramid such that a smaller pyramid of base side length of 18 cm is removed. Find the volume of the remaining frustum in litres to significant figures. The base area A of the original pyramid is 0 0 = 900 cm 2, and its height H is 60 cm. The smaller pyramid has a base of side 18 cm, which is 5 Its height is thus 60 6 cm, its base area is 18 = 24 cm of the base of the original pyramid The volume of the frustum is thus V ( AH ah) where A = 900, H = 60, a = 24 and h = ( AH ah) V cm or 14.1 litres to significant figures.

12 Mathematics Revision Guides Solid Shapes Page 12 of 19 The cone. The cone is another special case of a pyramid, this time with a circular base. The volume of a cone is given by 1 V 2 r h where r is the radius of the base and h is the perpendicular height. Again, this holds whether the cone is upright or slanted. The surface area of an upright cone is given by A = rl where r is the radius and l is the slant height. (This formula does not hold for slanted cones.) Again, we can have a frustum of a cone analogous to that of a pyramid. The section removed is itself a smaller similar cone, so the volume formula for a frustum is V ( R H r h) where R and H are the base radius and height of the large cone, and r and h are the base radius and height of the smaller missing section. Working is similar to that of the harder example (11).

13 Mathematics Revision Guides Solid Shapes Page 1 of 19 Example (12): The candlemakers from Example (9) also manufacture a range of upright conical candles, with bases 14cm in diameter and height of 24cm. How many such candles can be made out of a cylinder of wax of diameter 28 cm and height 40 cm assuming no wastage of wax? The volume of the wax cylinder is V = r 2 h, where r = 14 (halve the diameter!) and h = 40. Hence the cylinder has a volume of ( 2 40) cm or 7840 cm. The base radius of one candle is 7 cm, hence the base area is 49cm 2 and its height is 24 cm. The volume of wax used per candle is thus V r h, or The number of candles that can be made from the block is 20. V cm Notice also how all the working was left in terms of rather than using numerical approximations, as in the end we were able to cancel out anyway! Example (1): The finished candles from Example (12) also require their curved surfaces coating with gold paint, though not their bases. Calculate the total area that needs painting to significant figures. One candle has a surface area of A = rl where r is the radius and l is the slant height. We use Pythagoras to find l : l The curved surface area of one cone is therefore 7 cm 2 = 175 cm 2, so the total curved surface area of all 20 of them is cm 2 = 500 cm 2 = cm 2. The total area to be painted = cm 2 to significant figures

14 Mathematics Revision Guides Solid Shapes Page 14 of 19 Example (14): The circular sector (left) forms the net of curved surface of a cone. Its radius is 15 cm and its angle is 288. i) Find the area of the sector. ii) Show that the radius of the base of the cone is 12 cm after the radii are joined together The cone is then completed by having a circular base added to it. iii) Find the total surface area and the volume of the completed cone. i) Since , the sector forms of a circle, and so its area is r The radius of the sector is 15 cm, so the area is is cm 2. (This radius of 15 cm becomes the slant height of the cone.) ii) The arc length is the base of the cone.. cm 2 or 565 cm 2 to the nearest 4 8 2r cm, which, in turn, becomes the circumference of 5 5 The radius of the base of the cone therefore satisfies 2r = 24, and hence r = 12 cm. iii) The area of the base of the cone is 12 2 cm 2 = 144cm 2, so the total surface area is ( = 24 cm 2 = 1018 cm 2. In order to find the volume of the cone, we need to find the vertical height h first, but we know that the slant height l is 15 cm, and the radius r is 12 cm. 2 2 Using Pythagoras, h cm = 9 cm. The volume of the cone is therefore V 1 2 r h cm = 157 cm.

15 Mathematics Revision Guides Solid Shapes Page 15 of 19 The sphere. All points on the outside of a sphere are at a fixed radius from the centre. The volume of a sphere is given by The surface area is given by A = 4r 2. V 4 r where r is its radius. Example (15): A world globe has a diameter of 40cm. Find its surface area and volume. We substitute r = 20 (remember to use the radius! ) into the formulae above to obtain V 4 20 cm =,500 cm. the volume of the sphere is.5 litres to s.f. The area of the sphere is A = cm 2 = 5,00 cm 2. Example (16): i) Show that the total surface area of a hemisphere can be given by the formula A = r 2. ii) Hence calculate the total surface area of a hemispherical paperweight of diameter 8 cm. i) The curved surface of a hemisphere is half the total surface of a sphere, i.e. 2r 2. We then add the area of the circular base, namely r 2, to obtain the total of 2r 2 + r 2 = r 2. ii) Substituting r = 4 (halve the diameter! ) into the formula A = r 2, the total surface area of the paperweight is 48 cm 2 = 151 cm 2. Example (17): A kitchen bowl is hemispherical in shape and is designed to hold litres of water. Find its diameter in centimetres. Since the volume of a sphere is given by half that, i.e. 2 v r V where r is its radius. 4 r v Rearranging to make r the subject, we have r. 2 r Given v = 000 (1 litre = 1000 cm ),, the volume of a hemisphere will be given as = cm. That value is the radius; doubling it gives the diameter, namely 22.5 cm to s.f. V (The corresponding formula to find the radius of a full sphere, given the volume, is r ). 4

16 Mathematics Revision Guides Solid Shapes Page 16 of 19 Example (18): The hemisphere and cone shown below have the same volume, but the radius of the hemisphere is twice that of the cone. Express the vertical height of the cone in terms of its radius. The volume of the hemisphere will be half that of a sphere, i.e. 2 V r where r is its radius. The volume of the cone is 1 2 V r h where r is its radius and h is its vertical height. The volumes of the two solids are equal, so 1 2 r 2 h r r 2 h 2 r (multiplying by ) r 2 h 2r (dividing by ) h 2 r The height of the cone is twice its radius.

17 Mathematics Revision Guides Solid Shapes Page 17 of 19 Example (19): A hollow cylinder of height 4r cm exactly encloses two solid spheres of radius r cm. The volume of space not occupied by the spheres is 500 cm. Find the radius of each sphere. The volume of the cylinder is V = r 2 h, but here h = 4r, so we can rewrite it as V = 4r. The volume of one sphere is volume of both is 8 r. V 4 r, so the combined We are also given that the difference between the volumes is Therefore 4r r 12r 8r 500 4r 500 4r 500 r 125 r 5cm. The radius of the spheres is 5 cm. As in some other earlier examples, all the working was left in terms of.

18 Mathematics Revision Guides Solid Shapes Page 18 of 19 Other regular solids (Not in scope, but interesting nonetheless). The cube and the regular tetrahedron are two solids whose faces are all regular and identical and also arranged in the same way around each vertex. (Incidentally, the technical name for a cube is a regular hexahedron as it has six faces.) There are in fact three other such regular solids; two of them have equilateral triangles as faces, and the final one has regular pentagons. The regular octahedron. The regular octahedron has 8 equilateral triangles as its faces, 6 vertices and 12 edges. (Compare these to the cube s 6 faces, 8 vertices and 12 edges). A regular square-based pyramid forms half of an octahedron. The regular icosahedron. Another regular solid with equilateral triangles for faces is the icosahedron, with 20 of them. It has 12 vertices and 0 edges.

19 Mathematics Revision Guides Solid Shapes Page 19 of 19 The regular dodecahedron. This is the last regular solid with 12 regular pentagons for faces, 20 vertices and 0 edges.

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