Thursday, May 30, 2013

Volume and Solved Problems

 

Volume 

 Volume is the total space that is enclosed by the 3D figures like cuboid, cube, cone and so on. It is an indication of the space it will occupy or liquid or gases it can store.


cube
Cube

A Cube has 6 equal square faces. A good example is a box with equal height, length and breadth.  

Volume of cube =  l x l x l  ,
       where l is the length of any side. 
  
Total surface area of cube  = 6 X l   

  


Cuboid

A Cuboid has 6 faces just like cube but they are not all equal. For a cuboid at least one of the length, breadth or height is different .

Cuboid with length l , height h, and breadth b









 volume of cuboid  = l x h x b 
      
                  where, l is the length,
                            h is the height,
                      and b is the breadth of the cuboid

Total surface area = 2 ( l x h   + l x b + b x h)

                  where, l is the length,
                            h is the height,
                      and b is the breadth of the cuboid



Cone


Cone


A cone has circular base and then it tapers with increasing height to a point.
A good example is ice cream cone, party hats.
   
  
Volume of cone  = (1/3) x π x  x h 

Where,π  is approximately equal to (22/7) or 3.14
          h is the height of cone,
          r is radius of circular base

Lateral Surface Area = π x r x l 

Total surface area = π x r x l + π x 

                  where,  π is approximately equal to (22/7) or 3.14
                              r is radius of circular base 
                              l  is the slant height of cone







Cylinder:   A cylinder has circular base and then it extends to a certain 
height "h".  Good example is soda can, cookie jars, sanitation pipes.


Volume of cylinder: 
π x  x h

Where,


      π = 3.14 or 22/7
     r  is the radius of the 
           circular base
     h is the height of the cylinder
Lateral Surface area is the surface area of the side( in figure it  is the area in yellow color). It does not include the area of the base and the top 
(area in the blue color).

Lateral Surface area = 2
π x r x h 

Total surface area  = Lateral surface area     +     area of the circular base and top 

Total surface area  = 2 x π x r x h + 2 x π x 
                       = 2 π x r ( h + r)     
  


Sphere

A sphere has round 3D figure. 
Sphere
A good example is cricket ball or tennis ball
  
  
Volume of sphere  = (4/3)π x 

Where,
          π  is approximately equal to                                   
                (22/7) or 3.14
            r is radius of sphere



Total surface area = π 
              where,
                             π is approximately equal to (22/7) or 3.14
                             r is radius of circular base 


Hemisphere

An hemisphere, as the name suggests is half sphere. A sphere cut in half. 


  
  
Hemisphere
Volume of hemisphere  = (2/3)π x 

Where,
      π  is approximately equal to                                   
            (22/7) or 3.14
            r is radius of sphere

Total surface area = 3 x π x          










Solved geometry problems:

Q1 Which is true for a hexagonal pyramid ?

(1) It has six faces and each face is a hexagon
(2) It has a hexagonal base with six triangular faces meeting at a point
(3) It has two hexagonal faces and six rectangular faces
(4) It has six hexagonal faces joined by six rectangular faces

Answer: (2) It has a hexagonal base with six triangular faces meeting at a point

Explanation : The keyword here is "pyramid". A pyramid has a polygon base at one end and an apex at another ( ex cone). An hexagonal pyramid has  hexagonal base and then tapers to a point or an apex.


Q2


Answer: (1) 60

Explanation: Let consider the front face. It has 5 cubes in one rows and there are three such rows. Therefore in front face there are ( 5 x 3) 15 cubes. And there are 4 such rows of (5 x 3) cubes, That's makes a total of 4 x 15 = 60 cubes.


Q3


Answer: (3)  36

Solution: Volume of a cylinder =  (π x  r² x h) 
Since, diameter and hence radius is increases by 25 % but the volume remained the same, the height should have been decreased. Let the new height be h2. And the  volume = (π x (1.25 r)² x h2). Equating both the equation, we have,
(π x  r² x h) = (π x (1.25 r)² x h2)
=> h = (25/16)h2
=> (h2/h)= 16/25
=> (h - h2/h)  x 100 = (25-16/25) x 100 = 9 x 4 = 36
=> % height decrement = 36   
 


Q4. The internal length, breadth and height of a rectangular box A is   20cm, 18 cm and 15 cm respectively and that of the box B are 18cm, 12cm and 5 cm respectively. The volume of Box A is how many times that of B?

(1) 4
(2) 5
(2) 6
(5) 3

Answer:
(1) 5
Solution:

Volume of a rectangular box = height x length x width
Therefor volume of Box A = 20 x 18 x 15 = 5400
And the volume of the box B = 18 x 12 x5 = 1080

Divide the volume of A by the volume of B to get the answer = 5400/1080 = 5

Q5. The Whole surface of a cube is 150 sq. cm. Then the volume of the cube is 
(a) 125  cm³          (b) 216 cm³
(c)  343 cm³          (d) 512 cm³

Answer:(a) 125  cm³    
Solution:
A cube has 6 equal square faces that makes the whole surface area. Let the area of the square be X sq. cm.
Then,                         6 x  X = 150
                          =>    X       = 150/6 = 25 sq. cm.

Since, area of the square = l x l =25
                            =>           l = 5 cm

 And the volume = 5 x 5 x 5 = 125 cm³

Q6. If each edge of a square be doubled, then the increase percentage of its area 
is
(a) 200%         (b) 250%
(c) 280%         (d) 300%

Answer: (d) 300%
Solution:
Let the edge of the square be X cm then the area = X x X =
                   Then if the edge is double then = (2x)(2x) = 4x² 
Therefore. the percentage increase in area is = {(4x²-)/x²} x100 = 300%

Q7.A solid metallic spherical ball of diameter 6 cm is melted and recast into a cone with a diameter of the base as 12 cm.The height of the cone is
(a) 2cm  (b) 3cm
(c) 4cm  (d)  6 cm

Answer:3 cm
Solution:
Volume of a sphere = (4/3)π r³ = (4/3)(22/7)(6/2) ³ = (4/3)(22/7)(3) ³
Again, Volume of the cone =  (1/3)(22/7)(12/2)²(h) = (1/3)(22/7)(6)²(h)
Since same metal is used in recast the cone from the metal ball 
then, the
                   (4/3)(22/7)(3) ³ =  (1/3)(22/7)(6)²(h)
              =>                  h    =  (4)(3)³/(6)²
                                          =  (4)(27)/(36)                          
                                          = 3 cm  
  
                                              

                   

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