In this section, we will explore how to calculate the surface area and volume of various solids, including cuboids, prisms, cylinders, spheres, pyramids, and cones. Understanding these calculations is crucial for a wide range of real-world applications. We'll also delve into the significance of π (pi) in these calculations, especially when dealing with circular shapes. The aim is to provide you with a clear understanding of these concepts through practical examples and solutions, preparing you for your CIE IGCSE exams.
Surface Area and Volume of Cuboids A cuboid has six rectangular faces, with opposite faces being congruent.
Surface Area = 2 ( l w + l h + w h ) 2(lw + lh + wh) 2 ( lw + l h + w h ) Image courtesy of Cue Math
Volume = l × w × h l \times w \times h l × w × h Example: Calculate the surface area and volume of a cuboid with dimensions 5 cm by 3 cm by 2 cm.
Solution: Surface Area = 2 ( 5 × 3 + 5 × 2 + 3 × 2 ) = 62 cm 2 . = 2(5 \times 3 + 5 \times 2 + 3 \times 2) = 62 \, \text{cm}^2. = 2 ( 5 × 3 + 5 × 2 + 3 × 2 ) = 62 cm 2 . Volume = 5 × 3 × 2 = 30 cm 3 . = 5 \times 3 \times 2 = 30 \, \text{cm}^3. = 5 × 3 × 2 = 30 cm 3 . Surface Area and Volume of Cylinders A cylinder has two circular bases and a curved surface.
Surface Area = 2 π r h + 2 π r 2 = 2\pi rh + 2\pi r^2 = 2 π r h + 2 π r 2 Volume = π r 2 h = \pi r^2h = π r 2 h Example: Calculate the surface area and volume of a cylinder with radius 4 cm and height 5 cm.
Solution: Surface Area = 2 π × 4 × 5 + 2 π × 4 2 ≈ 226.19 cm 2 . = 2\pi \times 4 \times 5 + 2\pi \times 4^2 \approx 226.19 \, \text{cm}^2. = 2 π × 4 × 5 + 2 π × 4 2 ≈ 226.19 cm 2 . Volume = π × 4 2 × 5 ≈ 251.33 cm 3 . = \pi \times 4^2 \times 5 \approx 251.33 \, \text{cm}^3. = π × 4 2 × 5 ≈ 251.33 cm 3 . Surface Area and Volume of Spheres A sphere is a perfectly round 3D object.
Surface Area = 4 π r 2 = 4\pi r^2 = 4 π r 2 Volume = 4 3 π r 3 = \dfrac{4}{3}\pi r^3 = 3 4 π r 3 Image courtesy of Online Math Learning
Example: Calculate the surface area and volume of a sphere with radius 3 cm.
Solution: Surface Area = 4 π × 3 2 ≈ 113.10 cm 2 . = 4\pi \times 3^2 \approx 113.10 \, \text{cm}^2. = 4 π × 3 2 ≈ 113.10 cm 2 . Volume = 4 3 π × 3 3 ≈ 113.10 cm 3 . = \dfrac{4}{3}\pi \times 3^3 \approx 113.10 \, \text{cm}^3. = 3 4 π × 3 3 ≈ 113.10 cm 3 . Surface Area and Volume of Cones A cone has a circular base and tapers smoothly to a point.
Surface Area = π r ( r + h 2 + r 2 ) = \pi r (r + \sqrt{h^2 + r^2}) = π r ( r + h 2 + r 2 ) Volume = 1 3 π r 2 h = \dfrac{1}{3}\pi r^2h = 3 1 π r 2 h Example: Calculate the surface area and volume of a cone with radius 3 cm and height 4 cm.
Solution: Surface Area = π × 3 × ( 3 + 4 2 + 3 2 ) ≈ 75.40 cm 2 . = \pi \times 3 \times (3 + \sqrt{4^2 + 3^2}) \approx 75.40 \, \text{cm}^2. = π × 3 × ( 3 + 4 2 + 3 2 ) ≈ 75.40 cm 2 . Volume = 1 3 π × 3 2 × 4 ≈ 37.70 cm 3 . = \dfrac{1}{3}\pi \times 3^2 \times 4 \approx 37.70 \, \text{cm}^3. = 3 1 π × 3 2 × 4 ≈ 37.70 cm 3 . Surface Area and Volume of Triangular Prisms Triangular prisms have two triangular bases and three rectangular lateral faces.
Surface Area = Lateral Area + 2 × Base Area = Perimeter × Height + 2 × Base Area = \text{ Lateral Area } + 2 \times \text{ Base Area } = \text{ Perimeter } \times \text{ Height } + 2 \times \text{ Base\,Area } = Lateral Area + 2 × Base Area = Perimeter × Height + 2 × Base Area Image courtesy of Cue Math
Volume = = Base Area × Height = \text{ Base Area } \times \text{ Height } = Base Area × Height Example: Calculate the surface area and volume of a triangular prism where the base triangle has sides of 3 cm, 4 cm, and 5 cm (making it a right-angled triangle), a height (depth) of the prism of 10cm, and the height of the base triangle is 4 cm (given by the right-angled triangle properties).
Solution: Surface Area: 1 2 × base × height = 1 2 × 3 × 4 = 6 cm 2 \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6 \, \text{cm}^2 2 1 × base × height = 2 1 × 3 × 4 = 6 cm 2 Lateral Surface Area : Sum of the areas of the three rectangles formed by the sides of the triangle extended by the prism depth.For sides 3 cm and 4 cm (widths) and depth 10 cm (height), the area is 2 × ( side × depth 2 \times (\text{side} \times \text{depth} 2 × ( side × depth hypotenuse × depth \text{hypotenuse} \times \text{depth} hypotenuse × depth = 2 × 3 × 10 + 4 × 10 + 5 × 10 = 60 + 40 + 50 = 150 cm 2 = 2 \times 3 \times 10 + 4 \times 10 + 5 \times 10 = 60 + 40 + 50 = 150 \, \text{cm}^2 = 2 × 3 × 10 + 4 × 10 + 5 × 10 = 60 + 40 + 50 = 150 cm 2 Total Surface Area :2 × Base Triangle Area + Lateral Surface Area = 2 × 6 + 150 = 12 + 150 = 162 cm 2 2 \times \text{Base Triangle Area} + \text{Lateral Surface Area} = 2 \times 6 + 150 = 12 + 150 = 162 \, \text{cm}^2 2 × Base Triangle Area + Lateral Surface Area = 2 × 6 + 150 = 12 + 150 = 162 cm 2 Volume: Volume = Base Area × Depth = 6 × 10 = 60 cm 3 \text{Base Area} \times \text{Depth} = 6 \times 10 = 60 \, \text{cm}^3 Base Area × Depth = 6 × 10 = 60 cm 3
Surface Area and Volume of Pyramids Pyramids consist of a polygonal base and triangular faces converging at a point.
Surface Area = Base Area + 1 2 × Perimeter × Slant Height = \text{ Base Area }+ \dfrac{1}{2} \times \text{ Perimeter } \times \text{Slant\,Height } = Base Area + 2 1 × Perimeter × Slant Height Volume = $=\dfrac{1}{3} \times \text{ Base Area } \times \text{ Height }
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