
A pyramid is a geometric solid that has a polygon as its base and faces that converge at a point called the apex. In other words the faces are not perpendicular to the base.
The triangular pyramid and square pyramid take their names from the shape of their base. We call a pyramid a “right pyramid” if the line between the apex and the centre of the base is perpendicular to the base. Cones are similar to pyramids except that their bases are circles instead of polygons. Spheres are solids that are perfectly round and look the same from any direction.
Examples of a square pyramid, a triangular pyramid, a cone and a sphere:
Square pyramid | ![]() | Surface area=area of base +area of triangular sides=b2+4(12bhs)=b(b+2hs) |
Triangular pyramid | ![]() | Surface area=area of base +area of triangular sides=(12b×hb)+3(12b×hs)=12b(hb+3hs) |
Right cone | ![]() | Surface area=area of base +area of walls=πr2+12×2πrh=πr(r+h) |
Sphere | ![]() | Surface area=4πr2 |
Find the surface area of the following triangular pyramid (correct to one decimal place):
To find the height of the base triangle (hb) we use the theorem of Pythagoras:
The surface area of the triangular pyramid is 105,6 cm2.
Find the surface area of the following cone (correct to 1 decimal place):
To find the slant height, h, we use the theorem of Pythagoras:
The surface area of the cone is 233,2 cm2.
Find the surface area of the following sphere (correct to 1 decimal place):
If a cone has a height of h and a base of radius r, show that the surface area is:
πr2+πr√r2+h2
The cone has two faces: the base and the walls. The base is a circle of radius r and the walls can be opened out to a sector of a circle:
This curved surface can be cut into many thin triangles with height close to a (where a is the slant height). The area of these triangles or sectors can be summed as follows:
Area of sector=12×base×height (of a small triangle)=12×2πr×a=πraa can be calculated using the theorem of Pythagoras:
a=√r2+h2Find the total surface area of the following objects (correct to 1 decimal place if necessary):
We first need to find hb by constructing the vertical (perpendicular) height and using the theorem of Pythagoras:
(hb)2=(b)2−(b2)2=62−(62)2=36−9=27hb=√27 cmNow we can find the surface area:
surface area=area of base+area of triangular sides=12b(hb+3hs)=12(6)(√27+10)≈45,6 cm2The figure here is a cone. The vertical height of the cone is H=9,16 units and the slant height of the cone is h=10 units; the radius of the cone is shown, r=4 units. Calculate the surface area of the figure. Round your answer to two decimal places.
Therefore the surface area for the cone is 175,93 square units.
The figure here is a sphere. The radius of the sphere is r=8 units. Calculate the surface area of the figure. Round your answer to two decimal places.
Therefore the surface area is 804,25 square units.
The figure here shows a pyramid with a square base. The sides of the base are each 7 units long. The vertical height of the pyramid is 9,36 units, and the slant height of the pyramid is 10 units. Determine the surface area of the pyramid.
The surface area for the pyramid is 189 square units.
Square pyramid | ![]() | Volume=13×area of base×height of pyramid=13×b2×H |
Triangular pyramid | ![]() | Volume=13×area of base×height of pyramid=13×12bh×H |
Right cone | ![]() | Volume=13×area of base×height of cone=13×πr2×H |
Sphere | ![]() | Volume=43πr3 |
This video gives an example of calculating the volume of a sphere.
Find the volume of a square pyramid with a height of 3 cm and a side length of 2 cm.
We are given b=2 and H=3, therefore
V=13×22×3=13×12=4 cm3The volume of the square pyramid is 4 cm3.
Find the volume of the following triangular pyramid (correct to 1 decimal place):
The height of the base triangle (hb) is:
82=42+h2b∴hb=√82−42=4√3 cmThe area of the base triangle is:
area of base triangle=12b×hb=12×8×4√3=16√3 cm2The volume of the triangular pyramid is 105,3 cm3.
Find the volume of the following cone (correct to 1 decimal place):
The volume of the cone is 103,7 cm3.
Find the volume of the following sphere (correct to 1 decimal place):
The volume of the sphere is 268,1 cm3.
A triangular pyramid is placed on top of a triangular prism, as shown below. The base of the prism is an equilateral triangle of side length 20 cm and the height of the prism is 42 cm. The pyramid has a height of 12 cm. Calculate the total volume of the object.
First find the height of the base triangle using the theorem of Pythagoras:
Next find the area of the base triangle:
area of base triangle=12×20×10√3=100√3 cm2Now we can find the volume of the prism:
∴volume of prism=area of base triangle×height of prism=100√3×42=4 200√3 cm3The area of the base triangle is equal to the area of the base of the pyramid.
∴volume of pyramid=13(area of base)×H=13×100√3×12=400√3 cm3Therefore the total volume of the object is 7 967,4 cm3.
With the same complex object as in the previous example, you are given the additional information that the slant height hs = 13,3 cm. Now calculate the total surface area of the object.
Because the base triangle is equilateral, each face has the same base, and therefore the same surface area. Therefore the surface area for each face of the pyramid is 133 cm2.
Each side of the prism is a rectangle with base b=20 cm and height hp=42 cm.
area of one prism side=b×hp=20×42=840 cm2Because the base triangle is equilateral, each side of the prism has the same area. Therefore the surface area for each side of the prism is 840 cm2.
Therefore the total surface area (of the exposed faces) of the object is 3 092,2 cm2.
This video shows an example of calculating the volume of a complex object.
The figure below shows a sphere. The radius of the sphere is r=8 units. Determine the volume of the figure. Round your answer to two decimal places.
Vsphere=43πr3=43π(8)3=43π(512)=20483π=2 144,6605...
Therefore the volume for the sphere is 2 144,66 units3.
The figure here is a cone. The vertical height of the cone is H=7 units and the slant height is h=7,28 units; the radius of the cone is shown, r=2 units. Calculate the volume of the figure. Round your answer to two decimal places.
Therefore the volume of the cone is 29,32 units3.
The figure here is a pyramid with a square base. The vertical height of the pyramid is H=8 units and the slant height is h=8,94 units; each side of the base of the pyramid is b=8 units. Round your answer to two decimal places.
Therefore the volume of the square pyramid is: 170,67 units3.
Find the volume of the following objects (round off to 1 decimal place if needed):
We are given the radius of the cone and the slant height. We can use this to find the vertical height (H) of the cone:
H2=132−52=144H=12Now we can calculate the volume of the cone:
V=13×π(r)2×H=13π(5)2(12)=100π≈314,2 cm3We first need to find the vertical (perpendicular) height of the triangle (h) using the theorem of Pythagoras:
h2=b2−(b2)2=36−9=27h=√27 cmNow we can find the volume:
V=13×12bh×H=13×12(√27)(6)×(10)=10√27≈52,0 cm3V=43πr3=43π(10)3≈4 188,8 cm3
Find the surface area and volume of the cone shown here. Round your answers to the nearest integer.
The surface area of the cone is:
Acone=πr(r+h)=π(5)(5+20)=392,69908... cm2≈393 cm2For the volume we first need to find the perpendicular (or vertical) height using the theorem of Pythagoras:
H2=202−52H=√400−25=√375Now we can calculate the volume of the cone:
Vcone=13πr2h=13π(5)2(√375)=506,97233... cm3≈507 cm3Therefore the surface area is 393 cm2 and the volume is 507 cm3.
Calculate the following properties for the pyramid shown below. Round your answers to two decimal places.
Surface area
We first calculate the vertical (perpendicular) height of the base triangle:
h2b=42−22=16−4hb=√12Now we can calculate the surface area of the pyramid:
Apyramid=12b(hb+3hs)=12(6)(√12+3(9))=91,39230... cm2Therefore the surface area of the triangular pyramid is: 91,39 cm2.
Volume
We first need to find the vertical height (H):
H2=92−32=81−9H=√72Now we can find the volume:
Vpyramid=13×12(b)(hb)×H=16(6)(√12)×(√72)=29,39387... cm3Therefore the volume of the pyramid is: 29,39 cm3.
The solid below is made up of a cube and a square pyramid. Find its volume and surface area (correct to 1 decimal place):
The height of the cube is 5 cm. Since the total height of the object is 11 cm, the height of the pyramid must be 6 cm.
We will work out the volume first:
Volume=volume of cube+volume of square pyramid=s3+13bH=(5)3+13(5)2(6)=175 cm3For the surface area we note that one face of the cube is covered by the pyramid. We also note that the base of the pyramid is covered by the cube. So we only need to find the area of 5 sides of the cube and the four triangular faces of the pyramid.
For the triangular faces we need the slant height. We can calculate this using the theorem of Pythagoras:
h2s=H2+(b2)2=(6)2+(2,5)2hs=√42,25The surface area is:
Surface area =5(sides of cube)+4(triangle faces of pyramid)=5(s2)+4(12bhs)=5(52)+4(12(5)(√42,25))=125+10√42,25=190 cm2The surface area is 190 cm2 and the volume is 175 cm3.