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Reas and Volume Areas and Volume. 2 Unit 4:Mathematics Aims Introduce standard formulae to solve surface areas and volumes of regular solids. Objectives.

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Presentation on theme: "Reas and Volume Areas and Volume. 2 Unit 4:Mathematics Aims Introduce standard formulae to solve surface areas and volumes of regular solids. Objectives."— Presentation transcript:

1 reas and Volume Areas and Volume

2 2 Unit 4:Mathematics Aims Introduce standard formulae to solve surface areas and volumes of regular solids. Objectives Be able to use trigonometric methods and standard formula to determine areas and volumes.

3 Areas There is 2 types of areas  Cross Sectional Area  Surface Areas

4 Cross sectional Area PetryalSgwârCylch Triongl Paralelogram SectorAnnulus

5 Area of Regular Hexagon = A = * side 2 A = 2.598 x Side 2 (Approximately) If the side length of regular hexagon is 6 meter, calculate the area of regular hexagon using area formula? Given: Side length = 6 m. Area of Regular Hexagon = A = 2.598 x Side 2 A = 2.598 x 6 2 A = 2.598 x 36 A = 93.53 Therefore, Area of Regular Hexagon is 93.53 Square meter.

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7 Area of parallelogram = 11 x 6 = 66cm² Radius of circle = 4 / 2 =2cm Area of circle = π x 2² =12.56637cm² Area of shape = 66 - 12.56637 =53.4cm²

8 Approximately how much aluminum is needed to cover this soda can that has a radius of 2 inches and a height of 5 inches? SA=2 π rh+2 π r 2 SA= 2(3.14)(2)(5)+2(3.14)(2)(2) SA= 62.8 +25.12 AA SA= 87.92 in. 2 of aluminum

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10 Volume

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12 The volume of every solid, liquid or gas, is how much three-dimensional space it occupies, often quantified numerically. Volumes of a number of simple shapes, such as regular, straight- edged, and circular shapes can be easily calculated using arithmetic formulas. More difficult shapes can be calculated by integral calculus if a formula exists for its boundary.

13 Rectangular prism Find the volume of the rectangular prism of sides 10cm, 15cm, 25cm. Let, a=10cm,b=15cm, c=25cm Formula used: a x b x c Solution: Volume of the rectangular prism = 10 x 15 x 25 = 3750 m 3

14 Sphere Volume of a sphere = 4/3πr 3 Find the volume of a sphere of radius 9.6 m Solution: Volume of a sphere = 4/3πr 3 = 4/3(3.14*9.6 3 ) = 1.33(3.14*884.736) = 1.33*2778.07104 = 3694.83 m 3

15 Cylinder Volume of a cylinder = πr 2 h Locate the volume of a cylindrical canister with radius 7 cm and height 12 cm. Solution: Volume of a cylinder = πr 2 h = 3.14* 7 2 *12 = 3.14*49*12 = 1846.32 cm 3

16 Triangular prism Find the volume of a triangular prism whose length is 3cm, base is 3cm and height is 3cm? Solution: Given: l = 3, b = 3, h = 3 Formula: Volume of a triangular prism = ½ (lb)h = ½ (3×3)3 = ½ (27) = 13.5cm 3

17 Square pyramid Volume of Square Pyramid = (1/3) b²h Locate the surface area and volume of a square pyramid with the given side 3, height 4 and the slant height 5. Solution: Volume of Pyramid = (1/3) b²h = (1/3)* 3² * 4 = 0.33 * 9 * 4 = 12.

18 Cone Volume = 1/3πr 2 h Locate the volume of cone whose base radius is 2.1 cm and height is 6cm using π =22/7 Solution: Volume = 1/3πr 2 h = 1/3*22/7* 2.1 2 *6 = 27.72 cm 3

19 Cube Volume = a³ Locate the volume, surface area and diagonal of a cube with the given side 3. Solution: Volume = a³ = 3³ = 27.

20 Area and volume of hexagon Volume of the hexagonal prism = cross sectional area (csa) x height(h). If the side length of regular hexagon is 8 cm and 3 meters high, calculate the area and volume of a regular hexagon?

21 Area and volume of hexagon Side length = 8 cm. Area of Regular Hexagon = A = (2.598) x Side 2 A = 2.598 x 8 2 A = 2.598 x 64 A = 166.27cm² Volume = csa x height = 166.27 x 300=49881cm³

22 How much paint will this paint can hold if it is 8 inches in diameter and 12 inches in height? V= π r 2 h V= 3.14(4)(4)(12) V= 602.88 in. 3 o baent of paint


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