PRESENTATION ON “ SURFACE AREAS AND VOLUME” ( CLASS IX). CREATED BY: AMIT.N.YADAV AMIT GARG (TGT MATHS) (KV JRC BAREILLY ).
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PRESENTATION ON
“SURFACE AREAS AND VOLUME”
(CLASS IX)
CREATED BY:
AMIT.N.YADAV
AMIT GARG (TGT MATHS)
(KV JRC BAREILLY)
In our day to day life ,we come across objects
like a Wooden box, a Match box, a Tea packet ,a
Chalk box, a Dice , a Book etc.
All these objects are made of six rectangular
plane regions. These objects are in the shape of
a cuboid.
“cuboid is a solid bounded by six rectangular plane regions”
As we have seen that the surface of a cuboid
consists of six rectangular faces.
“Surface area of a cuboid equals the sum of the areas of its six rectangular faces”
Consider a cuboid whose Length is l cm,
Breadth is b cm and Height h cm.
Total surface Area of the Cuboid
= sum of the areas of all its six faces
= lb+lb+bh+bh+lh+lh
= 2lb+ 2bh + 2lh
= 2 (lb+bh+lh) cm2
“A cuboid whose length, breadth and height are all equal , is called a cube”
Surface area of a cube
= 2(a*a + a*a + a*a)
= 2(a2 + a2 + a2)
= 6 a2
If out of the six faces of a cuboid , we only find the
sum of the areas of four faces leaving the bottom
and top faces. This sum is called the lateral surface
area of the cuboid.
L.S.A of the cuboid
=Area of face AEHD + Area of face BFGC + Area of face ABFE + Area of face DHGC
= b*h + b*h + l*h + l*h
= 2bh + 2lh
= 2 ( l+ b) * h
L.S. Area of a cube
= 2(a*a + a*a)
= 2(a2 + a2)
= 4a2
Examples…
Example 1) Find the Surface Area of a match box
whose length , breadth and height are 16 cm,8
cm , and 6 cm respectively.
Solution) since match box is in the form of a cuboid.
Here l=16 cm , b= 8 cm , h= 6 cm
Surface Area of Match Box=2(lb + bh + lh)
=2(16*8+8*6+16*6) cm2
=2(128+48+96) cm2
=544cm2
Example 2) Find the surface Area of a cube whose edge is 11 cm
Solution)Here l=11cm
surface area of the given cube = 6l2
=6 (11)2 cm2
= 6 * 121 Cm2
=726 cm2
Example 3)Three cubes each of side 5 cm are joined end to end. Find the surface area of the resulting cuboid.
Solution)
The dimension of the cuboid so formed are as under
l=15cm , b=5cm , h=5cm
Surface Area of cuboid= 2(lb + bh + lh)
= 2(75+25+75) cm2
= 350 cm2
Example 4) A swimming pool is 20m in length ,15m in breadth and 4m in depth. Find the cost of cementing its floor and walls at the rate of Rs 20 per square metre.
Solution) we have l=20m , b=15m , h=4m
Area of the four walls = 2(l + b) * h
= 2(20 + 15) * 4 m2
= 280 m2
Area of the floor of the swimming pool= l * b
=(20*15) m2
= 300 m2
Total Area to be cemented = (280 + 300) m2
= 580 m2
Cost of cementing of 1 m2 = Rs 20
Cost of cementing the floor and the walls
= Rs(20 * 580)
= Rs 11600
Questions for practices…
Question 1)The dimensions of the cuboid are in the ratio 1:2:3 and its total surface area is 88m2.Find the dimension.
Question 2) The sum of length breadth and depth of a cuboid is 19cm and the length of its diagonal is 11cm. Find the surface area of the cuboid.
Question 3) Find the lateral surface area and total surface area of a cube of edge 10cm.
Question 4) Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.
Surface Area of Right circular Cylinder
=
Consider a right circular cylinder of radius “r”
and height “h”
Area of the lateral surface of the cylinder
=Area of the rectangle
= l * b
= 2Πr * h
= 2 Π r h square units
b=h
l=2Πr
r
Other method of Finding Surface area of cylinder with the help of paper
h
B=h
L=2πr
Surface area of cylinder = Area of rectangle= 2 π r h
Thus , for a cylinder of radius “r” and height “h” ,
we have
L.S.A = 2 Π r h square units
Each base Area = Π r2
Total Surface Area = (2 Π r h + 2Π r2 )
= 2 Π r (h + r ) Square units
r
r
h
Outer Curved Surface area of cylinder
Activity -: Keep bangles of same radius one over another. It will form a cylinder.
Formation of Cylinder by bangles
Circumference of circle = 2 π r
It is the area covered by the outer surface of a cylinder.
Circumference of circle = 2 π r
Area covered by cylinder = Surface area of cylinder
= (2 π r) x( h)
Curved surface
circular surfaces
Total Surface area of a solid cylinder
=
area of two circular surfaces
Area of curved surface +
= (2 π r) x( h) + 2 π r2
= 2 πr ( h+ r)
Examples…
Example1) The curved surface area of a right
circular cylinder of height 14 cm is 88 cm2.Find
the diameter of the base of the cylinder.
Solution)
Let r be radius and h be the height of the cylinder.
2 π r h = 88 and h = 14
2 * 22/7 * r * 14 = 88
88 r = 88
r = 1
Diameter of the base = 2r = 2 cm
Example 2) A rectangular sheet of paper 44 cm * 18 cm is rolled along its length and a cylinder is formed. Find the radius of the cylinder.
Solution) Let r be the radius of the base and h be the height. Then , h = 18 cm
2 π r = 44
r = 7cm
Hence, radius of the cylinder is 7 cm.
44 cm
18 cm
18 cm
Example 3) The ratio between the curved surface area and the total surface area of a right circular cylinder is 1:2 . Find the ratio between the height and the radius of the cylinder .
Solution)
Let h be the height and r be the radius and of the cylinder.
2 π r h / 2 π r h + 2 π r2 = 1/2
2 π r h / 2 π r (h + r) = ½
h / h + r = ½
2h = h + r
h = r
h:r = 1 :1
Questions for practices…
Question 1) Curved surface area of a right circular cylinder is
4.4m2 . If the radius of the base of the cylinder is 0.7 m , find
its height.
Question 2) In a hot water heating system , there is a cylindrical pipe of length 28m and diameter 5cm.Find the total radiating surface in the system.
Question 3) A cylindrical pillar is 50cm in diameter and 3.5m in height. Find the cost of painting the curved surface of the pillar at the rate of 12.50 per m2 .
Question 4)It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same?
l
2πr
Surface area of a Right Circular cone
l
2πr
V
A
B
Curved surface Area of the cone = Area of the sector VAB
= ½ * ( arc length * radius)
= ½ * 2 π r * l
= π r l
C.S.R = ½ * (circumference of base * slant height)
r
l=2r
Curved Surface Area of a Cone
=Π r ( l + r)
Examples…
Example 1)The diameter of a cone is 14cm and
its slant height is 9 cm. Find the area of its
curved surface .
Solution) S = Π r l
Here , r = 14/2 cm and l = 9 cm
s = 22/7 * 7 * 9 cm2 = 198 cm2
Example 2) Find the total surface area of a cone , if
its slant height is 9 m and the radius of its base is12m.
Solution) S = Π r l + Π r2
S=Π r ( l + r)
S = ( 22/7 * 12 * (12 + 9)) m2
= 792 m2
Example 3) The radius of a cone is 3 cm and vertical height is 4 cm. Find the area of the curved surface.
Solution)
we have r = 3 cm and h = 4 cm
l2 = r2 + h2
l2 = 3*3+ 4*4
l2 = 5 cm
Area of the curved surface = S = Π r l
=22/7 * 3 * 5
= 47.14cm2
4cm
3cm
Questions for practices…
Question 1) Find the curved surface area of a cone , if its slant height is 60 cm and the radius of its base is 21 cm.
Question 2) The radius of a cone is 5cm and vertical height is 12cm. Find the area of the curved surface.
Question 3)The radius of a cone is 7cm. And area of curved surface is 176cm2 .Find the slant height.
Surface Area of a Sphere
2r
2r
r
Surface Area of a Sphere = 4 Π r2
Curved Surface Area of a Hemi Sphere = 2 Π r2
Total Surface Area of a Hemi Sphere = 2 Π r2 + Π r2
= 3 Π r2
Examples…
Example 1) Find the surface area of a sphere of radius 7cm.
Solution) s = 4 Π r2
Here , r = 7cm
s = 4 * 22/7 * 7 * 7 cm2 = 616 cm2
Example 2) Find the surface area and total surface area of a hemisphere of radius 21cm.
Solution) S = 2 Π r 2 and s1 = 3Π r2
Here, r = 21
s= 2 * 22/7 * 21 * 21 cm2
s= 2772 cm2 s1= 4158 cm2
Questions for practices…
Question1)Find the surface area of a sphere of radius
(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm
Question 2) Find the surface area of a sphere of diameter.
(i)14 cm (ii) 21cm (iii)3.5cm
Continue…
Question 3) Find the total surface area of a hemi sphere and a solid hemisphere each of radius 10cm .
Question 4) The surface area of a sphere is 5544cm2 , find its diameter.
Volume of a Cuboid
b
Volume of a Cuboid = Base area * Height
= length * breadth * Height
Examples…
Example 1) The volume of a cuboid is 440 cm3 and the area of its base is 88cm2. Find the breadth of the tank if its length and the depth are respectively 2.5m and 10m.
Solution)
Volume = 440 cm3 and Area of the base = 88cm2
Height = Volume / Area of the base
= 440/ 88 cm = 5 cm
Questions for practices…
Question 1) A cuboidal water tank is 6m long , 5m wide and 4.5 m deep. How many litre of water can it hold?
Question 2) A cubical vessel is 10 m long and 8m wide. How high must it be made to hold 380 cubic metres of a liquid ?
a
a
a
VOLUME OF CUBE
Area of base (square) = a2
Height of cube = a
Volume of cube = Area of base x height
= a2 x a = a3
(unit)3
Examples….
Example 1) How many 3 metre cubes can be cut from a cuboid measuring 18 m * 12 m * 9 m.
Solution) Edge of each cube = 3m
Volume of each cube = (edge)3 = 3*3*3 m3
= 27m3
Volume of the cuboid = (18 * 12 * 9) m3
= 1944 m3
Number of cube
= volume of the cuboid/volume of the cube
= 1944 / 27
= 72
r
h
Volume of cylinder
Volume of cylinder = Area of base x vertical height
= π r2* h
Examples…
Example 1) Find the volume of a right circular cylinder , if the radius (r) of its base and height(h) are 7cm and 15cm respectively.
Solution)
volume of cylinder = π r2 h
Here r = 7cm and h = 15cm
Volume of the cylinder = 22/7 * 7 * 7 * 15 cm3
= 2310 cm3
Example 2) The area of the base of a right circular cylinder is 154cm2 and its height is 15cm . Find the volume of the cylinder.
Solution)
volume of a cylinder
= area of the base * height
Here , area of the base = 154cm and height= 15cm
= (154 * 15) cm3
Questions for practices…
Question 1 ) Find the volume of a right circular cylinder , if the radius (r) of its base and height (h) are 7cm and 15cm respectively.
Question 2) The volume of a cylinder is 448 π cm3. Find its lateral surface area and total surface area.
Question 3)A well with 10m inside diameter is dug 14m deep. Earth taken out of it is spread all around to a width of 5m to form an embankment. Find the height of embankment.
l = Slant height
h
Base
r
Cone
Volume of a Cone
h
h
Here the vertical height and radius of cylinder & cone are same.
r
r
3( volume of cone) = volume of cylinder
3( V )= π r2h
V = 1/3 π r2h
Volume = 3V
Volume =V
If both cylinder and cone have same height and radius then volume of a cylinder is three times the volume of a cone
Mr. Mohan has only a little jar of juice he wants to distribute it to his three friends. This time he choose the cone shaped glass so that quantity of juice seem to appreciable.
Examples
Example 1) A conical tank is 3m deep and its circular top has radius 1.75m. Find the capacity of the tank in kilolitres.
Solution)
we have , r= 1.75 m , h= 3m
Capacity of the tank = 1/3 π r2h
Capacity of the tank
= 1/3 * 22/7 *1.75 * 1.75 * 3 m3 Capacity of the tank = 9.625 m3
= 9.625 kilolitre
( 1m3 = 1 kilolitre)
Example 2) The height and the slant height of a cone are 21 cm and 28 cm respectively. Find the volume of the cone.
Solution )
l2 = r2 + h2
r = 7√ 7 cm.
volume of the cone = 1/3 π r2h
=1/3 * 22/7 *7√7 *√7 *21 cm3
= 7546cm3
Questions for practices…
Question 1 ) Find the volume of a right circular cone 1.02m high, if the radius of its base is 28cm.
Question 2) The area of the base of a right circular cone is 314cm2 and its height is 15cm. Find the volume of the cone.
Question 3) A semi-circular sheet of metal of diameter 28cm is bent into an open conical cup. Find the depth and capacity of cup.
Comparison of Area and volume of different geometrical figures
Volume of a Sphere
r
r
h=r
Here the vertical height and radius of cone are same as radius of sphere.
volume of Sphere =4( volume of cone)
V = 4( 1/3πr2h ) = 4( 1/3πr3)
V =4/3 π r3
r
V1
If we make a cone having radius and height equal to the radius of sphere. Then a water filled cone can fill the sphere in 4 times.
r
r
V=1/3 πr2h
If h = r then
V=1/3 πr3
V1 = 4V = 4(1/3 πr3)
= 4/3 πr3
Examples….
Example 1) Find the volume of a sphere of radius 7cm.
Solution) V= 4/3 πr3
Here, r = 7cm
V = 4/3 * 22/7 * 7 * 7 * 7 cm3
= 1437.33 cm3
Example 2) A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?
Solution)
The volume of water the bowl can contain = 2/3 πr3
= 2/3 *22/7 * 3.5*3.5*3.5 cm3
= 89.8 cm3
Questions for practices…
Question 1) A hemispherical bowl is made of steel 0.5cm thick. The inside radius of the bowl is 4cm. Find the volume of steel used in making the bowl.
Question 2) The volume of the two sphere are in the ratio 64:27. Find the difference of their surface areas , if the sum of their radii is 7.
Question 3) A solid sphere of radius 3cm is melted and then cast into small spherical balls each of diameter 0.6 cm. Find the number of balls thus obtained.
Volume of a Cylinder= Π r2 h
Volume of a Right Circular Cone =1/3 π r2h
Volume of a sphere = 4/3 π r3
Volume of a Hemi sphere = 2/3 π r3
Thank you….