Lesson Notes By Weeks and Term - Senior Secondary School 1

SUBJECT: MATHEMATICS

CLASS:  SS 1

DATE:

TERM: 3rd TERM

REFERENCE BOOKS

• New General Mathematics SSS 1 by M.F. Macrae et al 
• Essential Mathematics SS 1

 
WEEK TWO                        Date……………………

TOPIC: Volumes of frustums of cone, rectangular based pyramid and other pyramids

(a) Volumes of frustums of cone, rectangular based pyramid and other pyramids 

(b) Proofs of angles sum of a triangle = 180o

(c) The exterior angle 

(a) Volumes of frustums of cone, rectangular based pyramid and other pyramids

Many composite solids can be made by joining basic solids together.  In the figure below, the composite solids are made as follows:

(a)     a cube and a square based pyramid.

(b)    A  cylinder and cone

Examples:

1. The figure below shows a composite solid consisting of a cube of edge 28cm and a square-based pyramid of height 16cm.  Calculate the volume of the solid

2. The outer radius of a cylindrical metal tube is R and t is the thickness of the metal.

(a)    Show that the volume V, of metal in a length, I units, of the tube is given by

            V  =  П lt  (2R – t)

(b)    Hence calculate V when R = 7.5, t = 1 and 1 = 20 

Solutions

1. From the diagram of the composite solid given in Question(1)

    Volume of     =     Vol. of         +     Vol.

    Composite solid     square-based                     of 

                Pyramid                    cube

            =  1/3  b2h  +  l3

            =  1/3  x  28  x  28  x  16 + 283 cm3

            =    784  x   16    +   28  x  28  x   28 cm3

            =    12544  +  784  x  28  cm3

                    3

            =    12544   +   21952 cm3

                    3

            =    12544 +  65856 cm3

                            3

            =    26133 1/3 cm3

            =    26133cm3

Vol. of the             Vol. of                        Vol. of

Cylindrical metal    =    outside        -    inside

Tube                cylinder        cylinder

        =    П R2l   -  П r2l   ..................1

    But 

        R  =  t  +  r         ...................2

    Where 

        R      =      radius of outside cylinder 

        t    =    thickness of the cylindrical metal tube 

        r    =    radius of inside cylinder

    From equation (2) 

        r  =  R  -  t

        And substituting П R – t for r in equ (1):

    Vol of the cylindrical        =  ПR2l  -  Пr2l

    Meta.tube            =  ПR2l –П (R – t)2l

                    =  ПR2l – П(R2 – 2Rt + t2) l

                    = ПR2l– ПR2l + 2ПRtl  -  Пt2l

                    =  2 П Rtl – Пt2l

                    =  Пlt (2R – t)

(b)    When R  =  7.5, t  = 1 and l  =  20,  then

    Vol. of the cylindrical

    Metal tube    =    П l t (2R – t)

            =    22/7  x 20  x 1  (2 x 7.5  - 1)

            =    22/7 x 20 x (15 – 1)

            =    22/7  x 20  x 14

            =    44  x 20

            =    880

If a cone or pyramid  standing on a horizontal table is cut through parallel to 

the table, the top part is smaller cone or pyramid.  The other part is called a frustum.

To find the volume or surface area of a frustum, it is necessary to consider the frustum, as a 

complete cone (or pyramid) with the smaller cone (or pyramid) removed.

Examples:

1. Volume of a right circular cone is 5 litres.  Calculate the volumes of the two parts into which the cone is divided by a plane parallel to the base, one-third of the way down from the vertex to the base.  Give your answers to the nearest ml.

Solutions:

From the question,

h    =    1

    H        3

    H  =  3h

Also using similar triangles:

    r    =    h    =     1

    R    H          3

Thus:

    R  =  3r

Vol. of frustum         Vol. of        -    Vol. of

Of cone            =    big cone        small cone

            =     1 П R2 l t    -   1 П r2 h

1. 3

But vol. of big cone   =  5 litres

        =  5  x  1000 ml

Since 1 litre   =  1000 ml

i.e.

Volume of     =    5000 ml

Big cone   

        1/3 П R2H  =  5000 ml

        П R2H  =  3  x  5000 ml

        П R2H  =  15000 ml  ………. (1)

Also,

      From 3h  =  H

        h    =  H/3

and     3r  =  R

            r    = R/3

Thus, vol. of small    =   1/3 Пr2h

    Cone       

    =     x  π  x  x

    =  

    =  

Since from equation (1) above  ПR2H = 15000

Then

Vol. of small   =  = 

Thus:

Vol. of frustum of cone   =     Vol. of  big cone  -    Vol. of  small cone

    =   

             =  

             =   

             =     4814.8m

           4815ml

(b) Proofs of angles sum of a triangle = 180o

The sum of angles on a straight line is 180o. in the diagram below, x and y are adjacent angles on a straight line. 

When two or more angles add up to 180o they are called supplementary angles so x + y = 180o (supplementary angles)

Examples 

Find the unknown angles in the following diagrams: 

50o + 70o + a + 2a = 180o (sum of angles on a straight line)

120o + 3a = 180o3a = 180o – 120o = 60o

a = 603 = 20o

2a = 2 x 20o = 40o

EVALUATION

Essential Mathematics for Senior Secondary School 1 Exercise 15.1 No. 1

(c) The exterior angle

The exterior angle of a triangle is equal to the sum of the two opposite interior angles.

Given: Any triangle ABC

To prove: ACD = x1 + y1

Construction: Draw CE parallel to BA. 

Proof: Let ACE = x and ECD = y 

    x1 = x (alternate angles, BA//CE)

    y1 = y (corresponding angles, BA//CE)

    butACD = x + y

    ∴∠AACD = x1 + y1

Example 

1. In the diagram below, APR is a straight line. Work out the value of x and hence find QPR.

Solution 

3x + 2x + 20 = 200 – 4x (Ext. theorem)

5x + 4x = 200 – 20 

9x = 180 

X = 20o

∴ 200o – 4x = 200o – 4 x 20o

    = 200o – 80o = 120o

∴∠QPR = 180o – 120o (sum of angles on a straight line)

1. The ratio of the angles of a triangle is 3:4:5. Find the smallest and the largest angles.

Solution 

The angles are in the ratio 3:4:5,

i.e. 3 + 4 + 5 = 12 parts 

but the sum of angles of a triangle is 180o.

1st angle = 312 ×180o = 45o

2nd angle = 412 ×180o = 60o

3rd angle = 312 ×180o = 75o

The smallest angle = 45o and the largest angle is 75o

Check: 45o + 60o + 75o = 180o

EVALUATION

1. Three angles of a triangle x, 2x and 3x. find the value of x and hence find the angles. 
2. The ratio of the angles of a triangle is 2, 3 and 4. Find the angles. 

GENERAL EVALUATION

1. A lampshade in the form of a frustum of a cone has a height of 12cm and an upper and lower diameters of 10cm and 20cm.
2. what is the curved surface area of the frustum?
3. What is the volume of the frustum?  
4. Give both answers in terms of π

2.Afrustum of a pyramid is 16cm square at the bottom, 6cm square at the top and 12 cm high.  Find the volume of the frustum.

READING ASSIGNMENT 

NGM SS Bk 1 pages 173-175 Ex 15c Nos 6 and 9 pg 175.

WEEKEND ASSIGNMENT 

1. Calculate the volume in cm3 of the material in a cylindrical pipe 1.8m long, the internal and external diameters being 16cm and 18 respectively.

1. A composite solid consisting of a cone on top of a cylinder.  The height of the cone is 25cm. The height and base diameter of the cylinder are 40cm and 30 respectively. Calculate to 3.s.f. the volume of the solid, taking  π to be 3.14 (see the figure below).

1. A storage container is in the form of a  frustum of a right pyramid 4m square at the top and 2.5m square at the bottom.  If the container is 3m deep. What is its capacity in m3?
2. Three angles of a triangle are (5x – 7)o, (2x + 15)o and (2x + 1)o. find the value of x and hence find the largest and the smallest angles. 
3. The sides PQ and PR of ∆PQR are produced to T and S respectively, such that TQR = 131o and QRS = 98o. findQPR.

THEORY 

1. A right pyramid on a base 10m square is 15m high.

(a) Find the volume of the pyramid.

(b) If the top 6m of the pyramid are removed, what is the volume of the remaining frustum?

2. The cone in the figure below is exactly half full of water by volume.  How deep is the water in the cone?