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Cross-Section- Surface Area of a Prism- Surface Area of a Cylinder- Volume of a Prism

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Cross-Section- Surface Area of a Prism- Surface Area of a Cylinder- Volume of a Prism S8.6 Volume Things to Learn (Key words, Notation & Formulae) Complete from your notes Radius- Diameter- Surface Area- Volume- Capacity- Prism- Cross-section- Surface area of a prism- Surface area of a cylinder- Volume of a prism- Section 1. Surface area of cuboids: Q1. Work out the surface area of each cuboid shown below: Q2. What is the surface area of a cuboid with the dimensions 4cm, 5cm and 6cm? Section 2. Volume of cuboids: Q1. Calculate the volume of these cuboids: S8.6 Volume Q2. Section 3. Definition of prisms: Label all the shapes and tick the ones that are prisms. Section 4. Surface area of prisms: Q1. Find the surface area of this triangular prism. Q2. Calculate the surface area of this cylinder. S8.6 Volume Q3. Cans are in cylindrical shapes. Each can has a diameter of 5.3 cm and a height of 11.4 cm. How much paper is required to make the label for the 20 cans? Section 5. Volume of a prism: Q1. Find the volume of this L-shaped prism. Q2. Calculate the volume of this prism. Give your answer to 2sf Section 6 . Volume of a cylinder: Q1. Calculate the volume of this cylinder. S8.6 Volume Q2. The diagram shows a piece of wood. The piece of wood is a prism of length 350cm. The cross-section of the prism is a semi-circle with diameter 1.2cm. Calculate the volume of the piece of wood. Give your answer to 3sf. Section 7. Problems involving volume and capacity: Q1. Sam buys a planter shown below. The planter is 2.5m wide, 3m long and 1.5m deep. How much soil does Sam need to fill the planter 75cm deep? Q2. The cuboid shown below has a volume of 60 cm3. The base is 6.2 cm long and 3.7 cm wide. Not to scale 3.7 cm 6.2 cm Calculate the height of the cuboid. S8.6 Volume Q3. The triangular base of a prism is a right triangle with sides ‘a’ and ‘b= 2a’. The height ‘h’ of the prism is 10mm and its volume is equal to 40푚푚3. Find the length of the sides ‘a’ and ‘b’. Section 8. Converting units: Q1. Consider a cuboid that is 1cm by 1cm by 1cm. It has a volume of 1cm3. What is the volume in mm3? Q2. Convert a volume of 382,000 cm3 to m3 . Q3. A swimming pool measures 3m deep and has a base with area 375m2. a) Find the volume of the pool in cm3. b) How many litres of water can the pool hold? S8.6 Volume Answers Section 1. Surface area of cuboids: Q1. Surface area = 2 x Area of the front faces + Perimeter of the front face x length Surface area of 1st cuboid= 2( 10 x 6) + 2(10 x 3) + 2(6 x 3) ( Each individual face is a rectangle) = 120 + 60 + 36 = 216푐푚2 Or Using 2A + PL = 2 (3 x 6) + 10 (3+3+6+6) = 216푐푚2 Surface area of 2nd cuboid= 2( 5 x 11) + 2 ( 2 x 5) + 2 ( 11 x 2) = 110 + 20 + 44 = 174풄풎ퟐ Q2. Opposite faces of a cuboid have same area, so we will double each area of the rectangular face and then add them up to get the total surface area. Therefore, Surface area = 2 ( 4 x 5) + 2( 5 x 6) + 2 ( 6 x 4) = 40 + 60 + 48 = 148푐푚2 Or Surface area = 2A + PL = 2 x (4x 5) + 6 x (4+4+5+5) = 40 + 108 = 148푐푚2 Section 2. Volume of cuboids: Q1. Volume of a cuboid = Length x Width x height Volume of 1st cuboid = 2 x 6 x 4 = 48푐푚3 Volume of 2nd cuboid = 5 x 5 x 2 = 50푐푚3 S8.6 Volume Q2. Let the width of the cuboid be ‘W’. Since the volume = L x W x H Therefore, 8 x W x 3 = 120 24 W = 120 W = 120/24= 5 cm Section 3. Definition of prisms:- A 3-D Polyhedron (A 3-D shape that has straight edges and flat faces) with a constant cross-sectional area. Cuboid (Prism) Cylinder (Not a prism) Triangular Prism Sphere (Not a prism) Hemi-sphere (Not a prism) Cone (Not a prism) Hexagonal Prism Tetrahedron(Not a prism) Cuboid (Prism) Section 4. Surface area of prisms: Q1. Area of the front and the back face = 2 x Area of each triangle = 2 x ½ x base x perpendicular height = base x perpendicular height = 6 x 8 = 48푢푛푡푠2 Perimeter of triangle = 10 +10 +6 = 26 units Surface Area = 2A +PL= 48 + 12 x 26 = 360푢푛푡푠2 S8.6 Volume Q2. Surface area of the cylinder = Area of the top circle + area of the base circle + area of the curved part Area of the top circle = Area of the base = area of the circle with radius (r) of 4cm =휋푟2 = 휋(4)2 = 16 휋 푐푚2 Area of the curved part = Area of the rectangle with length of 6cm and width same as the circumference of the circular parts. Circumference of the circular parts = 2 휋r = 2 휋(4) = 8 휋 Therefore, area of the curved part = 6 x 8 휋 = 48 휋 푐푚2 Hence, Surface area of the cylinder= 16 휋 + 16 휋 + 48 휋 = 80 휋 = 251.3274123 = 251.33푐푚2 (2 d.p.) Q3. Since the labels only cover the curved part of the cylinders, we only need to find the area of the curved parts (rectangles). Radius (r) = ½ of diameter = ½ x 5.3 = 2.65 cm Height of the cylinder= length of the rectangular part= 11.4 cm Width of the rectangle = Circumference of the circle = 2휋푟 = 2휋 × 2.65 = 16.65044106 Hence, Area of one label= 11.4 x 16.65044106 = 189.8150281 푐푚2 Area of the paper required to make labels for 20 cans = 20 x 189.8150281 = 3796.300562 푐푚2 = 3796.3 푐푚2 (1 d.p) Section 5. Volume of a prism: Q1. Volume of prism = Area of cross-section x length of prism Length of this prism = 10mm Area of the cross- section of this prism= area of the front face (L-shaped) S8.6 Volume = Area of the blue vertical rectangle + Area of the white horizontal rectangle = (20 x 60) + ( 20 x 60) = 1200 + 1200 = 1400푚푚2 Volume of the prism = 1400 x 10 = 14000푚푚3 Q2. Area of a semi-circle = ½ of the area of a circle Semi-circle with radius 30cm Length of the prism Area of the cross-section= Area of the semi-circle + area of the rectangle = ½ x 휋 x(30)2 + (60 x 45) = ½ x 휋 x 900 + 2700 = 4113.716694푐푚2 Therefore, Volume of the prism = 4113.716694 x 90 2 = 370234.5025푐푚 3 = 370000푐푚 (3 s.f.) S8.6 Volume Section 6 . Volume of a cylinder: Q1. Volume of a cylinder = 휋푟2ℎ Radius (r) = 4cm, Height (h) = 8cm Therefore, Volume = 휋(4)2(8) = 휋 × 16 × 8 = 128휋 = 402.1238597푐푚3 = 402.124푐푚3(3dp) Q2. The piece of wood is half of a cylinder. Volume = ½ of the volume of the total cylinder Radius (r) = ½ x 1.2= 0.6 cm Height (h) = 350cm Volume = ½ x 휋푟2ℎ = ½ x 휋(0.6)2(350) = 1/2 x 휋 x 0.36 x 350 = 197.9203372푐푚3 3 = 198푐푚 (3 sf) Section 7. Problems involving volume and capacity: Q1. The planter is in the shape of a cuboid. Length = 3m, width = 2.5 m If the depth is 75cm = 0.75m (1cm = 1/100m) Volume of soil required = 3 x 2.5 x 0.75 = 5.625푚3 Q2. Volume = length x width x height Length = 6.2cm, width = 3.7cm, Volume = 60푐푚3 Therefore, 60 = 6.2 x 3.7 x height 60 = 22.94 x height Height = 60 / 22.94 = 2.615518745 = 2.62cm (2dp) Q2. Area of the cross-section= Area of the right angled triangle = ½ x base x height = ½ x a x b = ½ x a x 2a (Substituting b= 2a) S8.6 Volume = 푎2 Length of the prism = 10mm and volume = 40푚푚3 Therefore, 40 = 10 x 푎2 (Volume = area of cross-section x length) 4 = 푎2 (dividing by 10 both sides) a = 2mm (Square rooting both sides) Since b= 2a, b = 2 x2 = 4mm Section 8. Converting units: Q1. 1cm = 10mm 1푐푚3 = 1cm x 1cm x 1cm= 10mm x 10mm x 10mm = 1000푚푚3 Q2. 1cm = 1/ 100 m 1 1 1 382,000푐푚3 = 382000 x x x 푚3 100 100 100 1 = 382 x = 0.382푚3 1000 Q3. a) Volume of the pool = depth x area of the base = 3 x 375 = 1125푚3 b) 1litres = 1000 푐푚3 , 1푐푚3 = 1/1000 litres 1125푚3 = 1125 x 100 x 100 x100 푐푚3 = 1125000000 / 1000 litres = 1125000 litres .
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