1. Pyramids

2. Do you remember that polyhedra are solids enclosed by polygons?
Now, let’s study the properties of pyramids.
Here are some
examples.
cube
cuboid
prism
pyramid

3. Pyramids
All the surfaces (excluding the base) of a pyramid are triangular with a common vertex.
The figures below are examples of pyramids.
vertex
triangle
square
The base (yellow regions) of a pyramid is a polygon.

4. Except the base, all other surfaces of a pyramid are called lateral faces.
The common edge of two adjacent lateral faces is called a slant edge.
The perpendicular distance between the vertex and the base is the height of the pyramid.

5. The figure shows a right pyramid, it has the following properties:
The length of each slant edge is the same, i.e. VA = VB = VC = VD.
All the lateral faces are isosceles triangles, i.e. VAB, VBC, VCD and VDA are all isosceles triangles.
The foot of perpendicular from the vertex to the base is equidistant to each corner of the base, i.e. OA = OB = OC = OD.
If all the slant edges of a pyramid are equal in length, the pyramid is called a right pyramid.

6. If the base of a right pyramid is a regular polygon, the pyramid is called a regular pyramid.
The regular pyramid has all the properties of a right pyramid.

7. Volumes of Pyramids
The figures show a regular square pyramid and a cube.
The cube can be divided into 6 pyramids as shown. These pyramids are identical to the one shown on the left.

8. For each pyramid, the volume is V, the base area is A and the height is h.
(a) In terms of V,
volume of the cube =
6  volume of a pyramid 6V
(b) In terms of A and h,
volume of the cube =
2Ah
base area A 
height 2h

9. For each pyramid, the volume is V, the base area is A and the height is h.
From the results obtained in (a) and (b), we have:
2Ah
6V = 3 Ah 1
V =

10. cm 9 40 1  pyramid the of Volume = cm 120 = For any pyramids,
A = 40 cm2, h = 9 cm
Refer to the figure on the right.
base area = 40 cm2
9 cm 3 cm 9 40 1 
pyramid
the of
Volume = 3 cm
120 =

11. Frustum of a pyramid
If the top of a pyramid is cut away by a plane which is parallel to the base of the pyramid,
the remaining part is called a frustum of the pyramid.
The removed part is also a pyramid of smaller volume.

12. In this case,
Volume of the frustum
of a pyramid
volume of the
larger pyramid
smaller pyramid = -
volume of
frustum ABCDSPQR
pyramid VABCD
pyramid VPQRS = -

13.  Refer to the figure on the right. pyramid of Volume VPQRS cm 8 30 1B C D P Q S R
8 cm
pyramid of
Volume
VPQRS 3 cm 8 30 1  =
base area = 30 cm2 3 cm 80 =
If the volume of pyramid VABCD is 200 cm3, then
frustum ABCDSPQR of
volume =
volume of pyramid VABCD - volume of pyramid VPQRS 3 cm
80)
(200 - = 3 cm
120 =

14. Follow-up question
The figure shows a frustum ABCDHEFG. Its lower base is a square of side 7 cm. The volume of pyramid VEFGH is 75 cm3 and the height of pyramid VABCD is 12 cm. Find the volumes of (a) pyramid VABCD, (b) frustum ABCDHEFG. A V B C D E F G H
7 cm
12 cm
Solution 3 2 cm 12 7 1
pyramid of
ume
Vol
(a)  =
VABCD 3 cm
196 =

15. Follow-up question (cont’d)
The figure shows a frustum ABCDHEFG. Its lower base is a square of side 7 cm. The volume of pyramid VEFGH is 75 cm3 and the height of pyramid VABCD is 12 cm. Find the volumes of (a) pyramid VABCD, (b) frustum ABCDHEFG. A V B C D E F G H
7 cm
12 cm
Solution 3 cm
75)
(196
frustum of
Volume
(b) - =
ABCDHEFG 3 cm
121 =

16. The figure shows a pyramid VABCDE and its net.
Total Surface Areas of Pyramids
The figure shows a pyramid VABCDE and its net.
The pyramid has a base (orange region) and a number of lateral faces (green regions).
Its total surface area can be found by summing its base area and the areas of all the lateral faces.

17. Refer to the figure on the right.
area of △VBC = 40 cm2
8 cm A V B C D
The base is a square.
Total surface area of right pyramid VABCD
= total area of lateral faces + base area
= 4  area of △VBC + area of ABCD
= (4  ) cm2
= 224 cm2

18. Follow-up question
area = 65 cm2
Find the total surface area of the right pyramid in the figure.
area = 40 cm2
Solution
8 cm
14 cm
Total surface area of the right pyramid
The base is a rectangle.
= total area of lateral faces + base area
= [(2   65) + 14  8] cm2
= ( ) cm2