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					Volume of a pyramid
Prerequisite knowledge :

1. Pythagoras’ Theorem.
2. Area of some plane figures
  e.g. square, rectangle, triangle, circle, sector
3. Ratio and proportion

Objectives :
 Let the students know and apply the mensuration
 concepts
Egyptian Pyramid
These are pyramids
                          V
                                     Vertex

                                     
                      
Slant edges   
                                     height
                  D
                      
                                                   C

        
    A                             
                                  B
           x
    x


x



        cube


               Three congruent pyramids
      x                               1
              Volume of the pyramid =   x3
x                                     3
                                       1 2
       x                          =       x x
                                       3
                                      1
                                  =       base area  height
                                      3



    For any pyramid,

                             1
     Volume of pyramid =        base area  height
                             3
Example 1
The figure shows a pyramid with a rectangular base ABCD of
area 192 cm2, VE = 15 cm and EF = 9 cm, find the volume of
the pyramid.

                                          V
                    Solution :
         V                                           VF2 = (152 - 92 ) cm2
                                 15 cm                VF =    15 - 9 cm
                                                                 2    2




                                  E           F          = 12 cm
     D              C                 9 cm
             F
A                                Volume of the pyramid
         E 9 cm B
    15 cm                                    1
                                      =         base area  height
                                             3
                                         1
                                      =(    192  12) cm3
                                         3
                                      = 768 cm3
B

                                        Pyramid B


                                         A frustum

     Pyramid A




                     =                     -



Volume of the frustum = Volume of Pyramid A - Volume of Pyramid B
 Example 2
  The base ABCD and upper face EFGH of the frustum are squares of side 16
  cm and 8 cm respectively. Find the volume of the frustum ABCDEFGH.


                    V                  Solution :
                                                              1
                                       Volume of VEFGH = (       ( 8  8 )  6) cm3
                                6 cm                          3
                H           G
                                                           = 128 cm3
            E           F
12 cm                                                        1
                                       Volume of VABCD = (  ( 16  16 )  12) cm3
                                                          3
                D
                                C
                                                          = 1024 cm3

                                       Volume of frustum ABCDEFGH
        A                   B
                                              = (1024 - 128 ) cm3

                                              = 896 cm3
C


                            V
        V

                        D       C
    D           C   V               V
A           B           A       B


                            V
 Total surface area of pyramid VABCD =




        +           +             +              +



            lateral faces                             Base



Total surface area of a pyramid
 = Base area + The sum of of the area of all lateral faces
Example 3
  The figure shows a pyramid with a rectangular base ABCD of area 48
  cm2. Given that area of  VAB = 40 cm2 , area of  VBC = 30 cm2,
  find the total surface area of the pyramid.



         V              Solution :
                        Total surface area of pyramid VABCD

                        = Area of ABCD + (Area VAB + Area VDC +
     D            C                       Area VBC + Area VAD )

 A            B         = Area of ABCD + (Area VAB  2) +
                                         (Area VBC  2)

                         = 48 cm2 + ( (40  2) + (30  2)) cm2

                         = 188 cm2
How to generate a cone?

                    …...




                 …...
How to calculate the curved
      surface area ?


                             l

        l
                       2πr
    r
            Cut here
           Curved surface area
                                                               l
Remark :   Area of sector = 1/2r2 (θ/2)= 1/2 r2 θ or 1/2 r l
                                                                   θ

                                                               r
   After cutting the cone,
   Curved surface area = Area of the sector
   Curved surface area = 1/2 ( l ) ( 2π r )
                                      =πrl
                  Curved surface area = πr l
Volume of a cone

r

                                       h
                                 1
                                   h
                                 3
                         r

    Volume of a cone = 1 πr2 h
                       3
       How to calculate
total surface area of a cone?


         l
                           l   +         r
    r




        Total surface area =πr2 + πr l
                 Examples
1 a)   If h = 12cm, r= 5 cm, what
       is the volume?

       Answer:

                1
       Volume = 3 πr2h

                 = 1 π (52) ( 12)
                   3

                 = 314 cm3
b) what is the total surface area?
  Based Area = π52
             = 25πcm2

  Slant height = 122 + 5 2

                = 13 cm

  Curved surface area = π(5) ( 13)
                       = 65π cm2
  Total surface area = based area + curved surface area

                     = 25π+65π= 90π
                     = 282.6cm2 (corr.to 1 dec.place)
Volume of Frustum
              Volume of Frustum


                          =                   -
                                           R       r

Volume of frustum = volume of big cone - volume of small cone


                       1            1
                   =   3
                        πR3     -   π r3
                                    3
                       1
                   =   3
                           π( R3 - r3 )
Start
Now     Exit
Q1
              The volume of a pyramid of
              square base is 96 cm3. If its
              height is 8 cm, what is the
              length of a side of the base?
               A. 2 cm
               B. 2 3cm             Answer
               C. 6cm
     Answer
      is C                           Help
               D. 12cm
               E. 36cm               To Q2
Q2
          A
                      In the figure, the volumes of
                      the cone AXY and ABC are
                      16 cm3 and 54 cm3
    X         Y       respectively, AX : XB =
                        A. 2 : 1
B                 C
                        B. 2 : 3             Answer
                        C. 8 : 19
        Answer                                Help
                        D. 8 :27
         is A
                        E.   3   16 : 3 38    To Q3
Q3            V              In the figure, VABCD
                             is a right pyramid with
                             a rectangular base. If
                  D
                             AB=18cm, BC=24cm
                             and CV=25cm, find
                         C
              M
A
         B
                      a) the height (VM) of
                         the pyramid,
                                             Answer
                      b) volume of the
    a) 20cm                                   Help
                         pyramid.
    b) 2880cm3
                                              To Q4
Q4
         A
                    The figures shows a right
             50cm
                    circular cone ABC. If AD=
                    48cm and AC= 50cm, find
B              C    (a) the base radius (r) of the cone,
                    (b) the volume of the cone.
                    (Take  = 22 )
                                7
    Answer


     Help
 Let V is the volume of the pyramid and y be the length of
 a side of base
       1
 V = 3  base area  height
                                 what is the length of       a
       1
96 = 3  y2  8                  side of the base?
288 = 8y2

                                                Back to Q1
36 = y2
y=6
Therefore, the length of a                        To Q2
side of base is 6 cm
                              Hints: Using the concept
  AX : XB = ?                 of RATIOS
   AX       16                                A
 ( AB )3 = 54

  ( AX )3 = 278
    AB                                   X            Y
    AX      2
    AB  = 3                             B                C
 AB = AX + XB and AX = 2, AB = 3
                                         Back to Q2
 3 = 2 + XB
 XB = 1
Therefore, AX : XB = 2 : 1                  To Q3
a) the height (VM) of the     b) volume of the
pyramid                       pyramid.
      AC2 =182 + 242             Volume of the pyramid is:
      AC2 = 900
                                  1 ×base area ×height
      AC = 30cm                   3
      MC = 1 AC =15cm         = 1 ×18 ×24 ×20
              2                   3
      252 = VM2 + MC2         = 2880cm3
      625 = VM2 + 152         Therefore, the volume of
      625 - 225 = VM2         the pyramid is 2880cm3
      VM2 = 400
                                              Back to Q3
      VM = 20cm
      Therefore, the height
      (VM) of the pyramid
      is 20 cm                                   To Q4
(a) the base radius (r)          (b) the volume of the cone
                                      The volume (V) of cone is:
   The radius is r, therefore:
                                           1
   502   =   482   +   r2              V = 3  r2 h
                                           1    22
   2500 = 2304 + r2                      = 3  7  142  48
   196 = r2                              = 704 cm3
   r = 14                             The volume is 704 cm3
   The radius is 14cm.                  A
                                                 (Take  = 22/7)
         Back to Q4                           50cm

                                 B               C

				
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