Surface Area And Volume Of Solid Shapes SS1 Mathematics Lesson Note

MENSURATION OF SOLID SHAPES

Properties Of Solid Shapes

a) A Cube

A cube has the following properties.

i. It has 12 straight edges

ii. It has 8 vertices

iii. It also has 6 square faces

iv. Its net consists of 6 square faces joined together

b)  A Cuboid

A cuboid has the following properties.

i.It has 12 straight  edges

ii  It has 8 vertices

iii. It also has 6 rectangular faces

iv. Its net consist of 6 rectangular faces

c) A Triangular Prism

A triangular prism has the following properties:

i. It has 6 vertices

ii. It has 9 straight edges

iii. It also has 3 rectangular faces and two triangular faces which are the end faces

iv. Its net consist of 3 rectangles and 2 triangles joined together

d) A Cylinder

Properties:

i. A cylinder has 2 circular faces

ii.It has 1 curved surface

iii. It has 2 curved edges

iv. Its net consists of two circular faces and 1 rectangular face i.e. its net consist of 2 circles and 1 rectangle.

e) A Cone

A cone has the following properties:

i. It has one vertex

ii. It has 2 curved edges

iii. It has 1 curved surface

iv. It also has 1 circular face

v. Its net consists of a sector of a circle and a circle

f) Rectangular based pyramids

A rectangular-based pyramid has the following properties:

i. It has 8 straight  edges

ii. It has 5 vertices

iii. It has 4 triangular faces

iv. It has 1 rectangular face

v. Its net consists of 4 triangles and 1 rectangle

SURFACE AREA AND VOLUME OF COMMON SOLID SHAPES

A prism is a solid which has a uniform cross-section. Cubes, cuboids, and cylinders are examples of prisms.  In general,

Volume of prism  = area of uniform cross-section X perpendicular height = area of base x height

NOTE: The general formula for solid shapes is as follows: 

1. a) Cube

Volume  = l3

Surface area = 6l^2

1. b) Cuboid

Volume  =lbh

Surface area  = 2 (lb + lh + bh)

1. c) Cylinder

Volume = πr^2 h

Curved surface area = 2πrh

Total surface area = 2πrh  + 2π r^2

= 2πr  ( h + r)

SURFACE AREA OF A CONE

A sector of a circle can be bent to form the curved surface of an open cone. In the figure below, the sector OA x B is of radius l and arc A X B subtends angle θ at O.  This sector is bent to form a cone of base radius r and slant height

The following points should be noted

The area of the sector is equal to the area of the curved surface of the cone.

The length of arc A x B is the same as the circumference of the circular base of the cone.

Curved surface area of cone  =θ   x   πl2 ÷ 360°

Also,   

    θ   x   2πl    = 2 πr

             360

Divide both sides by 2π

θ  x  2πl     = 2 πr

360   2π          2π

 θ x l  =r

360

divide both sides by l

θ   =   r

360    l

substitute r/l for  θ in equation i) above:

                        360

Curve surface area of cone  =r   x πl2

l

                                 = Πrl

 Hence, 

Total surface area  = curved surface area of a cone + area of circular base 

=   πr l   +π r2

=  πr ( l + r)

Example: 

1. A paper cone has a diameter of 8cm and a height of 3cm

 a). Sketch the cone and hence use Pythagoras theorem to calculate its slant height.

 b). Calculate the curved surface area of the cone in terms of π

1. c) If the cone is cut and opened out into the sector of a circle. What is the angle of

the sector?

1. d) Assuming that the paper cone is closed at its base, what will be the total surface area of the closed paper cone?

Solution:

From the given information about the paper cone, 

 Diameter = 8cm

:. Radius  = diameter ÷  2

   =    8cm      = 4cm

           2

using Pythagoras’ theorem in the right-angled triangle OBC

l2   = /OB/2   + /BC/ 2

l2  = 32   + 42

l2 = 9 + 16

l2 = 25

Take the square root of both sides

√ l2     =√ 25

   l     = 5cm

: The slant height of the paper cone is 5cm

1. b)  Curve surface area of the cone = πrl

=  π   x  4  x 5 cm

= 20 πcm2

1. c) If the paper cone is cut and opened out into the sector of a circle  as shown in the figure above, then 

area of sector of circle  =  curved surface area of the cone

i.e θx π   x (5) 2  = 20 x π

    360

                     5

θx π   x 25   = 20 x π

360

    12  

               5 θ  = 72 x 20

Divide both sides by 5

5 θ  =72 x 20

               5

5 θ  = 72 x 4

θ  = 288°

Volume of Pyramids and Volume of Cone

In general, 

Volume = 1/3 x base area x height 

SQUARE-BASED PYRAMID, RECTANGULAR-BASED PYRAMID AND CONE 

:. The volume of square based pyramid  = 1/3 x b2 x h

volume of the based pyramid  = 1/3 x l x b x h

volume of the cone   = 1/3  x Πr2 x h

Examples

1. A pyramid 8cm high stands on a rectangular base 6cm by 4cm. Calculate the volume of the pyramid.

2. A right pyramid on a base 4cm square has a slanted edge of 6cm. Calculate the volume of the pyramid.

3. Calculate the volume of a cone 14cm in base diameter and 24cm high.

Solutions:

1. Volume of a rectangular based pyramid = 1/3 x l x b x h 

    = 1/3  x 6 x 4 x 8 cm3

    =  8 x8 cm3

       = 64cm3

2) Considering the square base ABCD 

/DB/ 2=  /DC/ 2 + /CB/2

Pythagoras rule: 

/DB/2  = 42  + 42

/B/2 = 16 + 16.

:. √/DB/   =  √ 32

/DB/  = 4 √2 cm

but

/ EB/  = ½  /DB/ 

Since t is the midpoint of / DB/

Then  /EB/  = ½ X 4 X √ 2

 = 2 √2 cm.

Now

Consider right angle    OEB

  OE 2 + EB 2  =  ( OB)2

 OE 2+  ( 2√2) 2  =  ( 6) 2

OE 2  + 4 x 2 = 36

OE 2  + 8  = 36

OE 2  = 36 – 8

OE2 = 28

OE  = √28

OE  = √4 x 7

OE = 2 x √ 7 cm

OE  = 2 √7cm

But OE  =height of the pyramid  = 2√7

:.volume of square of based pyramid  = 1/3  x b2 x h 

1/3  x 42  x 2 x √7  cm3

1/3  x 16 x 2  x  √7    cm3

=  32  x  √7 cm3

    3

 32  x  2.646cmm3

   3

=  32 x.0.882cm3

= 28. 224cm3

= 28.2cm3 to 1 d.p.

3) Since 

Diameter  = 14cm

Radius  = diameter

                     2

=  14  cm.=7cm

       2

:.  Volume of cone  = 1/3 πr2 h

                                = 1/3  x 22/7    x ( 7 ) 2 x 24

                                = 1/3  x 22/7 x 49 x 24 cm3

                                = 22 x 56cm3

                                = 1232 cm3

ASSIGNMENT 

1. Calculate the volume of a cylinder which has a radius of 21cm and a height of 6cm.   A. 8500cm3    B. 8316cm3    C. 7632cm3     D 7500cm3  E. 8000cm3
2. Calculate the total surface of the cylinder in question 1.  

A, 5346cm2        B, 4653cm3      C. 3000cm2    D. 3564 cm2    E 3800cm2

3. Calculate the volume of a cone which has a base diameter of 7cm and a height of 6cm   A. 77cm3    B. 70cm3    C. 88cm3    D. 90cm3    E. 65cm3
4. Calculate the curved surface area of the cone in question 3 above.     

A,  152cm2    B. 150cm2    C. 132cm2    D 142cm2    E. 160cm2

5. Calculate the total surface area of a cuboid which is 8cm by 5cm by 3cm.    

A.198cm2    B. 178cm2   C 188cm2   D 168cm2   E. 158cm2.

6. A water tank is 1.2m square and 1.35m deep. It is half full of water. How many times can a 9-litre bucket be filled from the tank?
7. A measuring cylinder of radius 3cm contains water to a height of 49cm. If this water is poured into a similar cylinder of radius 7cm, what will be the height of the water column?
8. A solid cone has a circular base of radius 7cm. the vertical height of the cone is 15cm. the cone is melted and recast into a metal cube of side xcm. Calculate correct to 3. s.f. the value of x. 
9. A cylindrical container with a diameter of 80cm and height 50of cm is full of liquid. The liquid is then poured into another cylinder with a diameter of 90cm. calculate the depth of the water.