Area And Volume Of Frustum SS1 Mathematics Lesson Note

VOLUMES OF FRUSTUMS OF CONE, RECTANGULAR-BASED PYRAMID AND OTHER PYRAMIDS 

Many composite solids can be made by joining basic solids together.  

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:

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

    Vol. 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

1. b) 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 the 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

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

the table, the top part is a 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:

The 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

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

PROOFS OF ANGLES: SUM OF A TRIANGLE = 180°

The sum of angles on a straight line is 180°, x and y are adjacent angles on a straight line. 

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

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)

    but = x + y

    ∴∠AACD = x1 + y1

Example:

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 the angles of a triangle is 180°.

1st angle = 3 ÷ 12 ×180°= 45°

2nd angle = 4 ÷ 12 ×180° = 60°

3rd angle = 5 ÷ 12 ×180° = 75°

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

Check: 45° + 60° + 75° = 180°

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.
2. 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 
3. 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?

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. 

The sides PQ and PR of ∆PQR are produced to T and S respectively, such that TQR = 131o and QRS = 98o. findQPR.  

4. 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?

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