13.3 Right pyramids, right cones and spheres (EMA7Q)

Pyramid

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:

f81bb2ab481bcbae06a9e4462071f3ea.png
99975557670d787583c45ede31be2cb3.png
995bbc98c818e78237a54ede442fc463.png
fcae6f01f12894f664a3b13248a4ce0d.png

Surface area of pyramids, cones and spheres (EMA7R)

Square pyramid

8c99684aa96b8e3dcd5f4c9baae3df67.pngSurface area=area of base +area of triangular sides=b2+4(12bhs)=b(b+2hs)

Triangular pyramid

d10ceba5f44c08a0b8cd3e2a2df9c838.pngSurface area=area of base +area of triangular sides=(12b×hb)+3(12b×hs)=12b(hb+3hs)

Right cone

e8392758046f2c700093902a1a809c5e.pngSurface area=area of base +area of walls=πr2+12×2πrh=πr(r+h)

Sphere

c9ea0239cfc34d13446edc18667b6c47.pngSurface area=4πr2

Worked example 8: Finding the surface area of a triangular pyramid

Find the surface area of the following triangular pyramid (correct to one decimal place):

069117fb6b4b7f74d540e67030876e62.png

Find the area of the base

area of base triangle=12bhb

To find the height of the base triangle (hb) we use the theorem of Pythagoras:

40b4accd1ca2599732f2bc308eea5003.png

Find the area of the sides

area of sides=3(12×b×hs)=3(12×6×10)=90 cm2

Find the sum of the areas

93+90=105,6 cm2

Write the final answer

The surface area of the triangular pyramid is 105,6 cm2.

Worked example 9: Finding the surface area of a cone

Find the surface area of the following cone (correct to 1 decimal place):

627fbe07065b02ec1273932906d4fae9.png

Find the area of the base

area of base circle=πr2=π×42=16π

Find the area of the walls

area of sides=πrh

To find the slant height, h, we use the theorem of Pythagoras:

2fc711706a1b59b5bbacd3e3ad326b9d.png

Find the sum of the areas

total surface area=16π+8π53=233,2 cm2

Write the final answer

The surface area of the cone is 233,2 cm2.

Worked example 10: Finding the surface area of a sphere

Find the surface area of the following sphere (correct to 1 decimal place):

a6ca40ffeb4a1404807965af71847a6b.png
surface area of sphere=4πr2=4π(5)2=100π=314,2 cm2

Worked example 11: Examining the surface area of a cone

If a cone has a height of h and a base of radius r, show that the surface area is:

πr2+πrr2+h2

Sketch and label the cone

ae7eece94bcb753bf8338edf1cc7b6eb.png

Identify the faces that make up the cone

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:

e54318b96193ba8d9a72e4b02bfd69fd.png

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=πra

Calculate a

a can be calculated using the theorem of Pythagoras:

a=r2+h2

Calculate the area of the circular base (Ab)

Ab=πr2

Calculate the area of the curved walls (Aw)

Aw=πra=πrr2+h2

Find the sum of the areas A

A=Ab+Aw=πr2+πrr2+h2=πr(r+r2+h2)

Exercise 13.4

Find the total surface area of the following objects (correct to 1 decimal place if necessary):

e4224ec2bcfa24663a08d047c602eba5.png
Surface area=area of base+area of walls=πr(r+hs)=π(5)(5+13)282,7cm2
069117fb6b4b7f74d540e67030876e62.png

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=369=27hb=27 cm

Now we can find the surface area:

surface area=area of base+area of triangular sides=12b(hb+3hs)=12(6)(27+10)45,6 cm2
87fb711ce20463c6b62831b7931b6f90.png
surface area=area of base+area of triangular sides=b(b+2hs)=6(6+2(12))=180 cm2
197698918f7f1b08d74a37e7f2c0c227.png
surface area=4πr2=4π(10)21 256,6 cm2

The 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.

ff09073e07aa1d503ecc66b63fecfd80.png
Acone=πr(r+h)=π(4)(4+10)=56π=175,9291...

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.

c9d6416977cdd913cd5c1592c98de70a.png
Asphere=4πr2=4π(8)2=256π=804,2477...

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.

6e4695aa5235d042bbe48ec8ca3ec075.png
Asquare pyramid=b(b+2hs)=(7)(7+2(10))=189

The surface area for the pyramid is 189 square units.

Volume of pyramids, cones and spheres (EMA7S)

Square pyramid

23da4349df92b529dc49e3a0bee65cf7.pngVolume=13×area of base×height of pyramid=13×b2×H

Triangular pyramid

fc309fbc970e05e7a137139e655909d0.pngVolume=13×area of base×height of pyramid=13×12bh×H

Right cone

48165c0ece92cd4f11f0a725c4db9bdd.pngVolume=13×area of base×height of cone=13×πr2×H

Sphere

ad6f55dab33291d980ecfd5daea07d15.png

Volume=43πr3

This video gives an example of calculating the volume of a sphere.

Video: 2GST

Worked example 12: Finding the volume of a square pyramid

Find the volume of a square pyramid with a height of 3 cm and a side length of 2 cm.

Sketch and label the pyramid

6718f60833199a452680f54daab7cc23.png

Select the correct formula and substitute the given values

V=13×b2×H

We are given b=2 and H=3, therefore

V=13×22×3=13×12=4 cm3

Write the final answer

The volume of the square pyramid is 4 cm3.

Worked example 13: Finding the volume of a triangular pyramid

Find the volume of the following triangular pyramid (correct to 1 decimal place):

2283c5653898afeb38d1375cbe6a0991.png

Sketch the base triangle and calculate its area

05f5fc91bd2b378a6a6f33501d62d3f6.png

The height of the base triangle (hb) is:

82=42+h2bhb=8242=43 cm

The area of the base triangle is:

 area of base triangle=12b×hb=12×8×43=163 cm2

Sketch the side triangle and calculate pyramid height H

0279bc7cf58b08c8a44405d103670aef.png

Calculate the volume of the pyramid

V=13×12bhb×H=13×163×130=105,3 cm3

Write the final answer

The volume of the triangular pyramid is 105,3 cm3.

Worked example 14: Finding the volume of a cone

Find the volume of the following cone (correct to 1 decimal place):

115ea932606ae315d3b9bce84977c9c6.png

Find the area of the base

area of circle=πr2=π×32=9π cm2

Calculate the volume

V=13×πr2×H=13×9π×11=103,7 cm3

Write the final answer

The volume of the cone is 103,7 cm3.

Worked example 15: Finding the volume of a sphere

Find the volume of the following sphere (correct to 1 decimal place):

6a180dbf1bf072d7296367996d5c3038.png

Use the formula to find the volume

volume=43πr3=43π(4)3=268,1 cm3

Write the final answer

The volume of the sphere is 268,1 cm3.

Worked example 16: Finding the volume of a complex object

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.

2b420a55c517408aa1c7c0154b446711.png

Calculate the volume of the prism

First find the height of the base triangle using the theorem of Pythagoras:

8910dd96d1c7aed337132c4f40b87a3a.png

Next find the area of the base triangle:

area of base triangle=12×20×103=1003 cm2

Now we can find the volume of the prism:

volume of prism=area of base triangle×height of prism=1003×42=4 2003 cm3

Calculate the volume of the pyramid

The area of the base triangle is equal to the area of the base of the pyramid.

volume of pyramid=13(area of base)×H=13×1003×12=4003 cm3

Calculate the total volume

total volume=4 2003+4003=4 6003=7 967,4 cm3

Therefore the total volume of the object is 7 967,4 cm3.

Worked example 17: Finding the surface area of a complex object

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.

Calculate the surface area of each exposed face of the pyramid

area of one pyramid face=12b×hs=12×20×13,3=133 cm2

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.

Calculate the surface area of each side of the prism

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 cm2

Because 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.

Calculate the total surface area of the object

total surface area=area of base of prism+area of sides of prism+area of exposed faces of pyramid=(1003)+3(840)+3(133)=3 092,2 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.

Video: 2GSV

Exercise 13.5

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.

3470df1c3cbfdc72cc10d3c2caada13e.png

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.

a6ab1d9b7dc114a5b36721aab8f57dc7.png
Vcone=13×πr2H=13×π(2)2(7)=13π(4)(7)=283π=29,3215...

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.

88f6dcc584dbccab21c1f53981fb2563.png
Vsquare pyramid=13b2H=13(8)2(8)=13(64)(8)=5123170,6666...

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):

e4224ec2bcfa24663a08d047c602eba5.png

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=13252=144H=12

Now we can calculate the volume of the cone:

V=13×π(r)2×H=13π(5)2(12)=100π314,2 cm3
bf6c0af2e3a6694c6b699977c0ac031f.png

We first need to find the vertical (perpendicular) height of the triangle (h) using the theorem of Pythagoras:

h2=b2(b2)2=369=27h=27 cm

Now we can find the volume:

V=13×12bh×H=13×12(27)(6)×(10)=102752,0 cm3
9ca331964ef720423883bc9ac1052296.png
V=13×b2×H=13(6)2(12)=144 cm3
197698918f7f1b08d74a37e7f2c0c227.png

V=43πr3=43π(10)34 188,8 cm3

Find the surface area and volume of the cone shown here. Round your answers to the nearest integer.

ab9ff648c2b5747454fb044714b7fe25.png

The surface area of the cone is:

Acone=πr(r+h)=π(5)(5+20)=392,69908... cm2393 cm2

For the volume we first need to find the perpendicular (or vertical) height using the theorem of Pythagoras:

H2=20252H=40025=375

Now we can calculate the volume of the cone:

Vcone=13πr2h=13π(5)2(375)=506,97233... cm3507 cm3

Therefore 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.

b36b4c8101b63c483b5676c987c17013.png

Surface area

We first calculate the vertical (perpendicular) height of the base triangle:

h2b=4222=164hb=12

Now we can calculate the surface area of the pyramid:

Apyramid=12b(hb+3hs)=12(6)(12+3(9))=91,39230... cm2

Therefore the surface area of the triangular pyramid is: 91,39 cm2.

Volume

We first need to find the vertical height (H):

H2=9232=819H=72

Now we can find the volume:

Vpyramid=13×12(b)(hb)×H=16(6)(12)×(72)=29,39387... cm3

Therefore 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):

ecbf80a1256326109945304bb59f96e8.png

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 cm3

For 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=42,25

The 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+1042,25=190 cm2

The surface area is 190 cm2 and the volume is 175 cm3.