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.
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.
$\text{Volume of Pyramid }=\frac{1}{3}\times\text{Area of base }\times\text{Height }$Volume of Pyramid =13×Area of base ×Height
The volume of a cone has the same relationship to a cylinder as we just saw that a pyramid has with a prism.
That is:
$\text{Volume of Right Cone }=\frac{1}{3}\times\text{Area of Base }\times\text{Height of cylinder}$Volume of Right Cone =13×Area of Base ×Height of cylinder
$V=\frac{1}{3}\pi r^2h$V=13πr2h
The mathematical derivation of the formula for the volume of a cone is beyond this level of mathematics, so for now it is suffice to know the rule and how to use it.
Finding out the amount of snowcone that would fit inside one of those conical (cone-like) cups, you can use the volume equation to find the volume inside the cone, right?
Funnels are in the shapes of cones. If you want to find out how much *volume* of water/liquid flows out of it per unit time (say liters per second for example), you will have to know what the volume of a cone is. you can find this volume flow rate, you can even write down an exact equation to describe the flow rate at any given time, that begins for instance with water filled to the brim, showing a slow flow rate (this is at early times), and for later times your equation will show that you have a faster slow.
Traffic cones are cones. Maybe you need to work out the volume of plastic it takes to make one, ten or a thousand.
A whole lot of things are approximately cones, reactor cooling tower caps, thimbles, and so on. These actually are cones, they just have been truncated (i.e. a small cone at the tip has been subtracted from a larger cone), icicles, stalactites/stalagmites in caves, and paper-made megaphones.
The front of pens have a cone shape to them.
So who cares what the volume of a sphere is. Well...
So apart from having to complete questions at school on volumes of random shapes, volume calculations have a wide variety of applications.
The Volume of a sphere with radius $r$r can be calculated using the following formula:
$\text{Volume of sphere }=\frac{4}{3}\pi r^3$Volume of sphere =43πr3
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?
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.
Find the volume of the square pyramid shown.
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.
Give your answer in exact form.
Find the volume of the cone shown.
Round your answer to two decimal places.
Find the volume of the sphere shown.
Round your answer to two decimal places.
A sphere has a radius $r$r cm long and a volume of $\frac{343\pi}{3}$343π3 cm^{3}. Find the radius of the sphere.
Round your answer to two decimal places.
Enter each line of working as an equation.
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