13.2 Right prisms and cylinders (EMA7M)

Right prism

A right prism is a geometric solid that has a polygon as its base and vertical faces perpendicular to the base. The base and top surface are the same shape and size. It is called a “right” prism because the angles between the base and faces are right angles.

A triangular prism has a triangle as its base, a rectangular prism has a rectangle as its base, and a cube is a rectangular prism with all its sides of equal length. A cylinder has a circle as its base. Examples of right prisms and a cylinder are given below: a rectangular prism, a cube and a triangular prism.

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Surface area of prisms and cylinders (EMA7N)

Surface area

Surface area is the total area of the exposed or outer surfaces of a prism.

This is easier to understand if we imagine the prism to be a cardboard box that we can unfold. A solid that is unfolded like this is called a net. When a prism is unfolded into a net, we can clearly see each of its faces. In order to calculate the surface area of the prism, we can then simply calculate the area of each face, and add them all together.

For example, when a triangular prism is unfolded into a net, we can see that it has two faces that are triangles and three faces that are rectangles. To calculate the surface area of the prism, we find the area of each triangle and each rectangle, and add them together.

In the case of a cylinder the top and bottom faces are circles and the curved surface flattens into a rectangle with a length that is equal to the circumference of the circular base. To calculate the surface area we therefore find the area of the two circles and the rectangle and add them together.

Below are examples of right prisms and a cylinder that have been unfolded into nets:

Rectangular prism

5f1f23505e095d6d8ee25fefccc03aeb.png0e9abb941c948ab43f9d07890c2b9428.pngbb375ceffeda2562134a69536f40fa5f.png

Rectangular prism

A rectangular prism unfolded into a net is made up of six rectangles.

Cube

62baec7eae7924a0c233a4cc273e2696.png52ebaf59f995e05925917d6dac6526a0.png1bddf1110a311ba257ab0b9b95f1b370.png

Cube

A cube unfolded into a net is made up of six identical squares.

Triangular prism

5aa931be519312ff3aa4c928654bc31c.png4af611bb8f86289851d50d143ae7ebb3.pngc77b31f1c53e40aa7d44ec56407783d4.png

Triangular prism

A triangular prism unfolded into a net is made up of two triangles and three rectangles. The sum of the lengths of the rectangles is equal to the perimeter of the triangles.

Cylinder

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Cylinder

A cylinder unfolded into a net is made up of two identical circles and a rectangle with a length equal to the circumference of the circles.

This video explains how we can unfold solids into nets.

Video: 2GS2

It may be useful to have some nets of the different polyhedra available for learners to see how they fold up to form the polyhedra.

You can download and print out nets of various polyhedra from senteacher to use in your classroom.

Worked example 2: Finding the surface area of a rectangular prism

Find the surface area of the following rectangular prism:

feae5c060473a0bda20bac750c5379eb.png

Sketch and label the net of the prism

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Find the areas of the different shapes in the net

large rectangle=perimeter of small rectangle×length=(2+5+2+5)×10=14×10=140 cm22×small rectangle=2(5×2)=2(10)=20 cm2
large rectangle2×small rectangle=perimeter of small rectangle×length=(2+5+2+5)×10=14×10=140 cm2=2(5×2)=2(10)=20 cm2

Find the sum of the areas of the faces

large rectangle+2×small rectangle=140+20=160large rectangle+2×small rectangle=140+20=160

Write the final answer

The surface area of the rectangular prism is 160160 cm2cm2.

Worked example 3: Finding the surface area of a triangular prism

Find the surface area of the following triangular prism:

d2d6406d9b8adb92d433f8a926c3788c.png

Sketch and label the net of the prism

a7aeaffca0bbc91ff938f90f312fb8e5.png

Find the area of the different shapes in the net

To find the area of the rectangle, we need to calculate its length, which is equal to the perimeter of the triangles.

To find the perimeter of the triangle, we have to first find the length of its sides using the theorem of Pythagoras:

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Find the sum of the areas of the faces

surface area=area large rectangle+(2×area of triangle)=216+2(12)=240 cm2
surface area=area large rectangle+(2×area of triangle)=216+2(12)=240 cm2

Write the final answer

The surface area of the triangular prism is 240240 cm2cm2.

Worked example 4: Finding the surface area of a cylinder

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

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Sketch and label the net of the cylinder

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Find the area of the different shapes in the net

area of large rectangle=circumference of circle×length=2πr×l=2π(10)×30=1 884,9555... cm2area of circle=πr2=π(10)2=314,1592... cm2
area of large rectanglearea of circle=circumference of circle×length=2πr×l=2π(10)×30=1 884,9555... cm2=πr2=π(10)2=314,1592... cm2
surface area=area large rectangle+(2×area of circle)=1 884,9555...+2(314,1592...)=2 513,3 cm2
surface area=area large rectangle+(2×area of circle)=1 884,9555...+2(314,1592...)=2 513,3 cm2

Write the final answer

The surface area of the cylinder is 2 513,32 513,3 cm2cm2.

Exercise 13.2

Calculate the surface area of the following prisms:

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Area of large rectangle=perimeter of small rectangle×length=(10+7+10+7)×6=34×6=204 cm2Area of 2× small rectangle=2(7×10)=2(70)=140 cm2Area of large rectangle+2×(small rectangle)=204+140=344 cm2
Area of large rectangleArea of 2× small rectangleArea of large rectangle+2×(small rectangle)=perimeter of small rectangle×length=(10+7+10+7)×6=34×6=204 cm2=2(7×10)=2(70)=140 cm2=204+140=344 cm2
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There are three different sized rectangles that make up the sides of this triangular prism. We need to find the area of each one of them. All of the rectangles have a height of 11 but each rectangle has a different base.

area of 2× triangle=2(12b×h)=(32)(7)=39,5979...area of rectangle 1=b×h=(7)(11)=77area of rectangle 2=b×h=(9)(11)=99area of rectangle 3=b×h=(32)(11)=62,2253...Atriangular prism=39,5979...+77+99+62,2253...=277,82 cm2
area of 2× trianglearea of rectangle 1area of rectangle 2area of rectangle 3Atriangular prism=2(12b×h)=(32)(7)=39,5979...=b×h=(7)(11)=77=b×h=(9)(11)=99=b×h=(32)(11)=62,2253...=39,5979...+77+99+62,2253...=277,82 cm2
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Area of large rectangle=circumference of circle×length=2π×r×l=2π×(2)×5=20πArea of circle=πr2=π(2)2=4πSurface area=area large rectangle+2(area of circle)=20π+2(4π)=28π87,96 cm2
Area of large rectangleArea of circleSurface area=circumference of circle×length=2π×r×l=2π×(2)×5=20π=πr2=π(2)2=4π=area large rectangle+2(area of circle)=20π+2(4π)=28π87,96 cm2
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Area of large rectangle=circumference of circle×length=2π×r×l=2π×(5)×10=100πArea of circle=πr2=π(5)2=25πSurface area=area large rectangle+2(area of circle)=100π+2(25π)=150π471,24 cm2
Area of large rectangleArea of circleSurface area=circumference of circle×length=2π×r×l=2π×(5)×10=100π=πr2=π(5)2=25π=area large rectangle+2(area of circle)=100π+2(25π)=150π471,24 cm2
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There are 4 rectangles and 2 squares that make up this rectangular prism. The square has a side length of 5. The rectangles have a base of 5 and a height of 11.

Arectangular prism=4×area rectangle+2×area square=4(b×h)+2(s2)=4(11×5)+2(52)=4(55)+2(25)=270
Arectangular prism=4×area rectangle+2×area square=4(b×h)+2(s2)=4(11×5)+2(52)=4(55)+2(25)=270
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We first need to find the missing side of the triangle. We can do this using the theorem of Pythagoras.

x2=52+(102)2x2=52+52=25+25x=50
x2x2x=52+(102)2=52+52=25+25=50

Now we can find the area of the triangular prism:

perimeter of triangle=10+50+50=24,1421...area of large rectangle=perimeter of triangle×length=24,1421...×20=482,8427...area of triangle=12b×h=12×5×10=25surface area=area large rectangle+2(area of triangle)=482,8427...+2(25)=532,84 cm2
perimeter of trianglearea of large rectanglearea of trianglesurface area=10+50+50=24,1421...=perimeter of triangle×length=24,1421...×20=482,8427...=12b×h=12×5×10=25=area large rectangle+2(area of triangle)=482,8427...+2(25)=532,84 cm2

If a litre of paint covers an area of 22 m2m2, how much paint does a painter need to cover:

a rectangular swimming pool with dimensions 4 m×3 m×2,5 m4 m×3 m×2,5 m (the inside walls and floor only);

We need to find the surface area of the pool. In this case we have a rectangular prism but with one rectangle missing (which would be the top of the pool).

surface area=area of bottom of pool+2(area of long sides)+2(area of short sides)=(4×3)+2(4×2,5)+2(3×2,5)=12+20+15=47 m2
surface area=area of bottom of pool+2(area of long sides)+2(area of short sides)=(4×3)+2(4×2,5)+2(3×2,5)=12+20+15=47 m2

The painter needs one litre of paint for every 22 m2m2 of area. So we must divide the surface area by 2 to find the total amount of paint needed. Therefore, the painter will need 472=24 l472=24 l of paint (rounded up to the nearest litre).

the inside walls and floor of a circular reservoir with diameter 44 mm and height 2,52,5 mm.

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We need to find the surface area of the reservoir. In this case we have a cylinder but with one circle missing (which would be the top of the reservoir).

We are given the diameter of the reservoir. The radius is half the diameter and so r=2 mr=2 m.

surface area=area of bottom of reservoir+area of inside of reservoir=(πr2)+(circumference of base×height of reservoir)=(π(2)2)+(2(π)(2)×2,5)=14π44 m2
surface area=area of bottom of reservoir+area of inside of reservoir=(πr2)+(circumference of base×height of reservoir)=(π(2)2)+(2(π)(2)×2,5)=14π44 m2

The painter needs one litre of paint for every 22 m2m2 of area. So we must divide the surface area by 2 to find the total amount of paint needed. Therefore, the painter will need 442=22 l442=22 l of paint (rounded up to the nearest litre).

Volume of prisms and cylinders (EMA7P)

Volume

Volume is the three dimensional space occupied by an object, or the contents of an object. It is measured in cubic units.

The volume of right prisms and cylinders is simply calculated by multiplying the area of the base of the solid by the height of the solid.

The video below shows several examples of calculating the volume of a right prism.

Video: 2GSB

Rectangular prism

2cd04c92e584166cc32f0dfff11e3ce8.pngVolume=area of base×height=area of rectangle×height=l×b×hVolume=area of base×height=area of rectangle×height=l×b×h

Triangular prism

b1c06d8cb93bc6d94eb3aae3dcdc3987.pngVolume=area of base×height=area of triangle×height=(12b×h)×HVolume=area of base×height=area of triangle×height=(12b×h)×H

Cylinder

e12f4a370695c4995caef7119cf0a7b5.pngVolume=area of base×height=area of circle×height=πr2×hVolume=area of base×height=area of circle×height=πr2×h

Worked example 5: Finding the volume of a cube

Find the volume of the following cube:

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Find the area of the base

area of square=s2=32=9 cm2
area of square=s2=32=9 cm2

Multiply the area of the base by the height of the solid to find the volume

volume=area of base×height=9×3=27 cm3
volume=area of base×height=9×3=27 cm3

Write the final answer

The volume of the cube is 27 cm3.

Worked example 6: Finding the volume of a triangular prism

Find the volume of the triangular prism:

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Find the area of the base

area of triangle=12b×h=(12×8)×10=40 cm2

Multiply the area of the base by the height of the solid to find the volume

volume=area of base×height=12b×h×H=40×20=800 cm3

Write the final answer

The volume of the triangular prism is 800 cm3.

Worked example 7: Finding the volume of a cylinder

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

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Find the area of the base

area of circle=πr2=π(4)2=16π cm2

Multiply the area of the base by the height of the solid to find the volume

volume=area of base×height=πr2×h=16π×15754,0 cm3

Write the final answer

The volume of the cylinder is 754,0 cm3.

Exercise 13.3

Calculate the volumes of the following prisms (correct to 1 decimal place):

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V=l×b×h=6×7×10=420 cm3
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V=12×b×h×H=12×10×5×20=500 cm3

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V=πr2h=π(5)2(10)=250π785,4 cm3

The figure here is a triangular prism. The height of the prism is 7 units; the triangles, which both contain right angles, have sides which are 2, 21 and 5 units long. Calculate the volume of the figure. Round to two decimal places if necessary.

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V=area of base×height=[12bh](H)=[12(2)(21)](7)=(21)(7)32,06

The figure here is a rectangular prism. The height of the prism is 12 units; the other dimensions of the prism are 11 and 8 units. Find the volume of the figure.

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Vrectangular prism=area of base×height=(bh)(H)=(8×11)(12)=1 056

The picture below shows a cylinder. The height of the cylinder is 11 units; the radius of the cylinder is r=4 units. Determine the volume of the figure. Round your answer to two decimal places.

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Vcylinder=(area of circle)(H)=[πr2](H)=[π(4)2](11)=[16π](11)=176π552,92