This document provides information about solid geometry, including definitions of different 3D shapes like prisms, pyramids, cylinders, cones, and spheres. It describes their key properties and provides examples. It also discusses nets, which are 2D shapes that can be folded into 3D solids. The document then covers calculating the surface areas of different solids and provides examples of solving surface area problems for prisms, cylinders, cones, and spheres. It concludes with practice problems for students to calculate surface areas.
3. LEARNING OUTCOMES
State the geometric properties of prisms,
pyramids, cylinders, cones and spheres.
Draw nets for prisms, pyramids,
cylinders and cones.
State and find surface areas of prisms,
pyramids, cylinders, cones and spheres.
4.
5. DEFINITION
Solid geometry is concerned with
three-dimensional shapes.
Some examples of three-dimensional
shapes are:
• Prisms
• Pyramids
• Cylinders
• Cones
• Spheres
6. 12.1 PROPERTIES
SOLIDS DESCRIPTION EXAMPLES
PRISM A solid with two congruent, parallel
bases which are polygons.
PYRAMID A solid with a base which is a polygon
and triangular sides that converge at a
vertex.
CYLINDER A solid with two parallel congruent
circular faces and a curved surface.
CONE A solid with a circular base and a
vertex.
SPHERE A solid having all of its points the same
distance from its centre.
14. 12.3 SURFACE AREA
The surface area of a solid is the
total area of all the faces of the
solid.
• It is measured using squares
• Units include mm²,cm²,m²,km².
15. SOLIDS NETS SURFACE AREA
PYRAMID
Area of four triangular faces +
Area of rectangular base
PRISM
Area of three rectangular faces +
Area of two triangular faces
17. SOLUTION
Area of square base
13 cm
= 10 × 10 = 100cm 2
10 cm
Area of a triangular face
1
= ×10 ×12 = 60cm 2
2
Surface area of the pyramid
= 100 + (4 × 60) = 340cm 2
18.
19. SURFACE AREA OF CYLINDER
r r
l h
l = circumference of the base circle =2πr
Area of curved surface (rectangular) + Area of two circular faces.
= 2πrh + 2πr 2
20. Example
Find the surface area of a cylinder
with a radius of 7 cm and a height of
22
20 cm. (Take π=
7
)
21. SOLUTION
r = 7cm h = 20cm
Surface area of the cylinder
= 2πr + 2πrh
2
22 2 22
= 2( )(7 ) + 2( )(7)(20)
7 7
= 308 + 880 = 1188cm 2
22. SURFACE AREA OF CONE
l
l
r r
Area of sector = π rl
Area of circle = π
2
r
Area of sector + Area of circle = πrl + πr 2
44. Example 1
Find the total surface area of the
following solid. Take π = 3.142 .
45. Example 2
The solid shown below consists of a
cone and a hemisphere with a common
base. What is the total surface area of
=
the solid? Take π3.142 .
“Hemi”
means half.