When calculating the volume of a pyramid we can substitute the values of the length, width and perpendicular height into the formula V = 1/3 lwh. In my experience this is often provided for the students with little explanation as to why a volume of a pyramid is exactly one third the volume of a cuboid.
I wanted my class of 14 and 15-year-olds to derive the formula for the volume of a pyramid for themselves. This way they will be less likely to forget the formula in future as it is something they created through their own knowledge of cuboids and algebraic manipulation.
I start the lesson by asking students to discuss in pairs how they would describe the method for finding the volume of a cuboid to a ten-year-old. To encourage them to think about the different cross-sections students they are not allowed to describe the volume as simply the product of length, width and height. The volume of a cuboid is later described as product of cross-sectional area of a face and its depth. The formula V = lwh works because each of the faces of a cuboid can be its cross-section.
At this point students are asked to calculate the volume and surface area of the three cuboids below. This is done in their books so the method can be referred to later if needed.
To introduce the volume of a pyramid I present the diagram below.
I explain the height of the square based pyramid is exactly one half the height of the cube. Students are to consider how many of these pyramids would fit perfectly inside the cube. I encourage them to discuss this in pairs and use mini-whiteboards to aid their processing.
The almost immediate response from about a quarter of the class is that four pyramids would fit inside. I encourage students to take more time to consider the problem before they commit to a solution.
Some students realised there must be a pyramid on each square face of the cube. However, drawing this in 3D proved difficult to visualise so some students decided to sketch the net.
I now ask the class to write on their whiteboards the volume of the pyramid as a fraction of the volume of the cube. All students wrote 1/6. I write on the main whiteboard our findings so far.
Volume of a Pyramid = 1/6 lwh
We discuss that the pyramid shares the same base length and width of the cube but not the perpendicular height. I ask the students to write the height of the cube in terms of the height of the pyramid and simplify the result.
h = 2x
Volume of Pyramid = 1/6 lw2x
Volume of Pyramid = 1/3 lwx
x is the perpendicular height of the pyramid.
As we progress through the remainder of the lesson, we apply the formula for calculating the volume of a pyramid and other composite solids. The questions become more challenging as students need to apply Pythagoras’ Theorem to calculate the perpendicular height as well as calculating the total volume of composite solids.
Calculating the volume of a frustum is the final problem which we do in the plenary. Click here to view the video.
About a third of the class found it difficult to visualise the frustum as the difference between two similar rectangular-based pyramids. I help by sketching the big pyramid and smaller top pyramid as two separate shapes on the main board. It is now clear to the students the difference between the two forms the volume of the frustum
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