Ontario 10 Applied (MFM2P)
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Volume of pyramids and spheres
Lesson

Volume of Pyramids

A pyramid is formed when the vertices of a polygon are projected up to a common point (called a vertex).  A right pyramid is formed when the apex is perpendicular to the midpoint of the base.

We want to be able to calculate the volume of a pyramid. Let's start by thinking about the square based pyramid.

Exploration

Think about a cube, with side length $s$s units.  Now lets divide the cube up into 6 simple pyramids by joining all the vertices to the midpoint of the cube.  

This creates $6$6 square based pyramids with the base equal to the face of one of the sides of the cube, and height, equal to half the length of the side.

$\text{Volume of Cube }=s^3$Volume of Cube =s3

$\text{Volume of one of the Pyramids }=\frac{s^3}{6}$Volume of one of the Pyramids =s36

Now lets think about the rectangular prism, that is half the cube.  This rectangular prism has the same base as the pyramid and the same height as the pyramid.  

Now the volume of this rectangular prism is $l\times b\times h=s\times s\times\frac{s}{2}$l×b×h=s×s×s2= $\frac{s^3}{2}$s32

We know that the volume of the pyramid is $\frac{s^3}{6}$s36 and the volume of the prism with base equal to the base of the pyramid and height equal to the height of the pyramid  is $\frac{s^3}{2}$s32.

$\frac{s^3}{6}$s36 $=$= $\frac{1}{3}\times\frac{s^3}{2}$13×s32

Breaking $\frac{s^3}{2}$s32 into two factors

$\text{Volume of pyramid}$Volume of pyramid $=$= $\frac{1}{3}\times\text{Volume of rectangular prism}$13×Volume of rectangular prism

Using what we found in the diagrams

  $=$= $\frac{1}{3}\times\text{Area of base}\times\text{height }$13×Area of base×height

Previously shown

So what we can see here is that the volume of the pyramid is $\frac{1}{3}$13 of the volume of the prism with base and height of the pyramid.  

Of course this is just a simple example so we can get the idea of what is happening.  

Volume of Pyramid

$\text{Volume of Pyramid }=\frac{1}{3}\times\text{Area of base }\times\text{Height }$Volume of Pyramid =13×Area of base ×Height

Volumes of Spheres

So who cares what the volume of a sphere is. Well...

  • A toy manufacturer needs to figure out how much plastic they need to make super bouncy rubber balls would need to know the volume of the ball. 
  • An astronomer who wants to calculate how much the sun weighs would need to know its volume. 
  • A pharmaceutical company making round pills needs to know the dosage of medicine in each pill, which is found by its volume.
  • A party balloon gas hire company needs to know how much gas fits inside each of their balloons, which is found by its volume.

So apart from having to complete questions at school on volumes of random shapes, volume calculations have a wide variety of applications.  

The Formula

The Volume of a sphere with radius $r$r can be calculated using the following formula:

Volume of Sphere

$\text{Volume of sphere }=\frac{4}{3}\pi r^3$Volume of sphere =43πr3

 

Proof of the formula

Whilst the proof is not typically included as part of the needs of the curriculum, this particular one is a nice introduction to thinking about proofs and abstract proofs. So you don't have to read this if you don't want to but aren't you curious to know where this funny formula came from?

The proof

If four points on the surface of a sphere are joined to the centre of the sphere, then a pyramid of perpendicular height r is formed, as shown in the diagram. Consider the solid sphere to be built with a large number of these solid pyramids that have a very small base which represents a small portion of the surface area of a sphere.

If $A_1$A1, $A_2$A2, $A_3$A3, $A_4$A4, .... , $A_n$An represent the base areas (of all the pyramids) on the surface of a sphere (and these bases completely cover the surface area of the sphere), then,

$\text{Volume of sphere }$Volume of sphere $=$= $\text{Sum of all the volumes of all the pyramids }$Sum of all the volumes of all the pyramids
$V$V  $=$= $\frac{1}{3}A_1r+\frac{1}{3}A_2r+\frac{1}{3}A_3r+\frac{1}{3}A_4r$13A1r+13A2r+13A3r+13A4r   $\text{+ ... +}$+ ... + $\frac{1}{3}A_nr$13Anr
  $=$= $\frac{1}{3}$13  $($(   $A_1+A_2+A_3+A_4$A1+A2+A3+A4  $\text{+ ... +}$+ ... +  $A_n$An  $)$) $r$r 
  $=$= $\frac{1}{3}\left(\text{Surface area of the sphere }\right)r$13(Surface area of the sphere )r
  $=$= $\frac{1}{3}\times4\pi r^2\times r$13×4πr2×r
  $=$= $\frac{4}{3}\pi r^3$43πr3

where $r$r is the radius of the sphere.  

Worked Examples

question 1

Find the volume of the square pyramid shown.

question 2

A small square pyramid of height $4$4 cm was removed from the top of a large square pyramid of height $8$8 cm forming the solid shown. Find the exact volume of the solid.

  1. Give your answer in exact form.

question 3

Find the volume of the sphere shown.

Round your answer to two decimal places.

question 4

A sphere has a radius $r$r cm long and a volume of $\frac{343\pi}{3}$343π3 cm3. Find the radius of the sphere.

Round your answer to two decimal places.

Enter each line of working as an equation.

Outcomes

10P.MT3.04

Solve problems involving the surface areas of prisms, pyramids, and cylinders, and the volumes of prisms, pyramids, cylinders, cones, and spheres, including problems involving combinations of these figures, using the metric system or the imperial system, as appropriate

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