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Geometry
The environment we live in is 3-dimensional and geometry is the natural language to express
concepts and relationships of space.
We give names to help us identify shapes in space (3D) and in a plane (2D).
We classify objects according to certain attributes.
We discover properties and relationships while dealing with various shapes and relate them to
the world around us.
We abstract plane figures from solid objects by seeing them as faces of solids that can be traced
on paper.
We name, classify, investigate and discover new properties for plane figures.
We learn to construct plane figures and then the nets to construct solids.
In this way, our knowledge of geometry grows in complexity following a 3-step progression:
3D (Manipulative, solids, concrete) 2D (Pictorial, graphic)  1D (Numerical, Abstract)
It is this interplay of the real world of objects and the abstract world of thought that leads to
progress.
SOLID OBJECTS: (3-Dimensional, Real)
1) Basic figures: sphere, cone, cylinder, prism, pyramid.
2) Congruent figures: Spheres, pyramids ….
3) Symmetries: Objects in space
4) Stacking: Objects in space, packing
5) Measurement: volume, surface area
PLANE FIGURES: (2-Dimensional, flat)
1) Basic figures: point, line, ray, segment, angles
2) Polygons: triangles, quadrilaterals…..
3) Circles: centre, radius, diameter, circumference
4) Congruent figures: triangles, squares….
5) Similar figures: map, scale drawing
6) Measurement: perimeter, area
Prisms and Pyramids
Prism:
 A polyhedron with 2 congruent (equal in size and arrangement) faces (bases) that are parallel to
each other.
 The remaining faces are parallelograms.
 A segment between bases is the altitude.
Ex) Triangular Prism
(bases are triangles)
Rectangular Prism.
(bases are rectangles)
Pentagonal Prism
(bases are pentagons)
Pyramid:
 A polyhedron where all faces (but one) have a common vertex.
 The base is a polygon.
 Lateral faces are triangles.
 A line segment from the vertex to the base (perpendicular line) is the altitude.
Square-based pyramid
Altitude
Surface Area
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To find the surface area of an object, it is useful to unfold the object on paper (or for real with a
cereal box) and layout the 3D object as a 2D net.
 A net is useful to find the total surface area as the sum of the areas of all the faces.
Ex1) SA = area of 2 bases (rectangles) and area of 2 sides and area of 2 ends.
Expand the prism into a net (folded out flat) as shown:
10 cm
1
3
4 cm
2
10 cm
2
3
1
5 cm
1
2
3
4 cm
10 cm
5 cm
4 cm
1
2
3
5 cm
4 cm
4 cm
10 cm
Surface Area = 2(4x10) + 2(5x4) + 2(5x10) = 2(40) + 2(20) + 2(50) = 80+40 + 100 = 220 cm2
Ex2)
5 cm
12 cm
15 cm
SA = 2 ( ½ )(5)(12) + 5(15) + 13(15) + 12(15) = 75 + 195 + 180 + 60 = 510 cm2
Ex3) A square based pyramid has Surface Area = Area of base + Area of Lateral Faces
Volume = Lwh÷3
If L=w = 5.0 mm, s = 10.0 mm, h = 6.0 mm, find SA and V
SA = Lw + 4(bs÷2) = (5)(5) + 4[(5)(10) ÷2] = 25 + 4(25) = 125 mm2
s
V = Lwh÷3
h
= (5)(5)(6) ÷3
w
= 50 mm3
L
Ex 4) For a sphere, the SA is found as 4πr2. We know the volume is 4 π r3 ÷ 3.
A sphere has a radius of 25 cm. Find the surface area and the volume to the nearest hundredth.
SA = 4(3.14) (25)2 = 4(3.14) (625) = 7850.00 cm2
V = 4(3.14) (25)3 ÷3 = 4(3.14) (15625) ÷ 3 = 65416.67 cm3
Ex 5) A metal sphere with a diameter of 20 cm is packed in a cubical box. What is the unused volume
of the box (assume the box is a perfect fit for the sphere).
Vsphere = 4 π r3 ÷ 3 = 4(3.14)(20/2)3÷ 3 = 4(3.14)(10) 3 ÷ 3 = 4(3.14)(1000) ÷ 3 = 4186.6 cm3
Vbox = Lwh = (20x20x20) = 203 = 8000 cm3
Difference is = 8000 – 4186.66…. = 3813.3….. = 3813 cm3 unused.
Ex6) A cylinder is made up of a rectangle rolled up and two circles stuck on each end.
The surface area is the sum of the area of the rectangle and the area of the 2 identical circles.
r
h
r
2πr
Area = 2πrh
Area of both circles = 2 x πr2
Total surface area is the sum of the area of these 3 shapes.
Recall that the volume of a cylinder is the area of the base times the height = πr2h
Find the surface area and volume of a right cylinder to the nearest hundredth given it has a height of
21.2 cm and a diameter of 12.6 cm.
21.2 cm
SA = 2πrh + 2 x πr2
= 2(3.14)(6.3)(21.2) + 2(3.14)(6.3)2
= 838.7568 cm2 + 249.2532 cm2
= 1088.01 cm2
12.6 cm
V = πr2h = 3.14(6.3)2(21.2)
= 2642.08 cm3
R = 12.6/2 = 6.3 cm
Ex 7) A cone is a sector of a circle joined at the radii.
Vertex
s
s
s
s2 = h2 + r2
h
r
Base
s- the slant height of the cone is the radius of the circle sector that makes up the cone.
The surface area of a cone can be found by adding the area of the base to the surface area of the circle
sector: Surface Area of a Right Cone = πrs + πr2 = π r(s + r)
A cone with a radius of 2.0 cm and a slant height of 7.0 cm has a surface area:
SA = π (2)(7 + 2) = 3.14(2)(9) = 56.52 cm2
Recall that the volume of a cone is V = πr2h ÷3 or (πr2h)/3
MPM1D Surface area and Volume Practice
Find the surface area and volume for each of the following. Label measures on the provided diagrams.
1) A cylinder with a diameter of 8.0 cm and a height of 10.0 cm.
2) A cone with a base diameter of 12 cm and a vertical height of 15 cm.
3) A sphere with a diameter of 15.00 mm.
Thought Problems:
4) A ball with a diameter of 18 cm is perfectly packaged in a cubical box. What area of cardboard
is needed to construct the box?
5) Extra lively squash balls are packaged in a cylindrical can. The radius of the can is 3.5 cm and
the balls fit perfectly into the can. What surface area of aluminum is needed to make the can?
6) The silo of a barn is made up of a cylinder with one floor and a roof made up of half a sphere,
as shown. Find the volume of the silo.
16 m
7m
7) It’s the year 2020 and conical speakers are all the rage. Your job is to accurately construct the
box to package the speakers. The vertical height of the cone speakers is 20 cm and the cone’s
radius is 5 cm, find the surface area of the material needed to build the box.
35.6 cm
BONUS: A football coach’s head is a sphere and has a hollow spherical core. The outside diameter is
35.6 cm and the thickness of the head is 13.2 cm. Find the volume of material that makes up the head.
13.2 cm
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