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CIE IGCSE Maths Study Notes

4.1.3 Solids

Understanding the attributes of solids is fundamental in geometry, as it forms the basis for recognising and calculating the properties of three-dimensional shapes. This section delves into various solids including cubes, cuboids, prisms, cylinders, pyramids, cones, spheres, hemispheres, and frustums. Additionally, we will explore the concept of faces, surfaces, and edges, which are crucial for identifying and describing these solids.

Solids

Image courtesy of Geeks for Geeks

Cube

A cube is a three-dimensional shape with six square faces, all of the same size, 12 edges, and 8 vertices.

Cube

Image courtesy of CueMath

  • Formulae:
    • Surface area: 6a26a^2 (where aa is the length of a side)
    • Volume: a3a^3

Example: Calculate the surface area and volume of a cube with a side length of 5 cm.

  • Surface area: 6×52=150 cm26 \times 5^2 = 150 \text{ cm}^2
  • Volume: 53=125 cm35^3 = 125 \text{ cm}^3

Cuboid

A cuboid is a three-dimensional shape with 6 rectangular faces, 12 edges, and 8 vertices.

Cuboidal
  • Formulae:
    • Surface area: 2(lw+lh+wh)2(lw + lh + wh) (where ll, ww, and hh are the length, width, and height respectively)
    • Volume: lwhlwh

Example: Calculate the surface area and volume of a cuboid with dimensions 3 cm by 4 cm by 5 cm.

  • Surface area: 2(34+35+45=94 cm22(3\cdot4 + 3\cdot5 + 4\cdot5 = 94 \text{ cm}^2
  • Volume: (345=60) cm3(3\cdot4\cdot5 = 60) \text{ cm}^3

Prism

A prism is a solid with two parallel faces called bases that are congruent polygons and all other faces are parallelograms.

Prism

Image courtesy of Third Space Learning

  • Formulae:
    • Volume: BaseArea×HeightBase Area \times Height

Example: A triangular prism has a base area of 30 cm230 \text { cm}^2 and a height of 5 cm. Calculate its volume.

  • Volume: 30×5=150 cm330 \times 5 = 150 \text{ cm}^3

Cylinder

A cylinder has two parallel circular bases and a curved surface, connecting the bases.

Cylinder
  • Formulae:
    • Surface area: 2πr(h+r)2\pi r(h + r) (where rr is the radius and hh is the height)
    • Volume: πr2h\pi r^2h

Example: Calculate the surface area and volume of a cylinder with a radius of 3 cm and a height of 7 cm.

  • Surface area: 2π×3(7+3)=188.4 cm22\pi \times 3(7 + 3) = 188.4 \text{ cm}^2 (approx.)
  • Volume: π×32×7=198.9 cm3\pi \times 3^2 \times 7 = 198.9 \text{ cm}^3 (approx.)

Pyramid

A pyramid has a polygon base and triangular faces that meet at a point (the apex).

Pyramid
  • Formulae:
    • Volume: 13×BaseArea×Height\frac{1}{3} \times Base Area \times Height

Example: A pyramid with a square base of side 4 cm and a height of 9 cm. Calculate its volume.

  • Volume: 13×42×9=48 cm3\frac{1}{3} \times 4^2 \times 9 = 48 \text{ cm}^3

Cone

A cone has a circular base and a curved surface that tapers to a point.

Cone

Image courtesy of Vedantu

  • Formulae:
    • Surface area: πr(r+h2+r2)\pi r(r + \sqrt{h^2 + r^2}) (where rr is the radius and hh is the height)
    • Volume: 13πr2h\frac{1}{3}\pi r^2h

Example: Calculate the surface area and volume of a cone with a radius of 3 cm and a height of 4 cm.

  • Surface area: π×3(3+42+32)=75.8 cm2\pi \times 3(3 + \sqrt{4^2 + 3^2}) = 75.8 \text{ cm}^2(approx.)
  • Volume: 13π×32×4=37.7 cm3\frac{1}{3}\pi \times 3^2 \times 4 = 37.7 \text{ cm}^3 (approx.)

Sphere

A sphere is a perfectly round geometrical object in three-dimensional space, like a round ball.

Sphere
  • Formulae:
    • Surface area: 4πr24\pi r^2
    • Volume: 43πr3\frac{4}{3}\pi r^3

Example: Calculate the surface area and volume of a sphere with a radius of 5 cm.

  • Surface area: 4π×52=314 cm24\pi \times 5^2 = 314 \text{ cm}^2
  • Volume: 43π×53=523.6 cm3\frac{4}{3}\pi \times 5^3 = 523.6 \text{ cm}^3 (approx.)

Hemisphere

A hemisphere is half of a sphere.

Hemisphere

Image courtesy of Coding Hero

  • Formulae:
    • Surface area: 3πr23\pi r^2
    • Volume: 23πr3\frac{2}{3}\pi r^3

Example: Calculate the surface area and volume of a hemisphere with a radius of 3 cm.

  • Surface area: 3π×32=84.8 cm23\pi \times 3^2 = 84.8 \text{ cm}^2 (approx.)
  • Volume: 23π×33=56.5 cm3\frac{2}{3}\pi \times 3^3 = 56.5 \text{ cm}^3 (approx.)

Frustum

A frustum is a portion of a solid (normally a cone or pyramid) that lies between two parallel planes cutting it.

Frustum

Image courtesy of CueMath

  • Formulae:
    • Volume: 13πh(r12+r22+r1r2)\frac{1}{3}\pi h(r_1^2 + r_2^2 + r_1r_2) (where hh is the height, r1r_1 and r2r_2 are the radii of the two bases)

Example: A frustum of a cone with radii 3 cm and 5 cm and a height of 6 cm. Calculate its volume.

  • Volume: 13π×6(32+52+3×5)=197.9 cm3\frac{1}{3}\pi \times 6(3^2 + 5^2 + 3\times5) = 197.9 \text{ cm}^3(approx.)

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