Surface Areas and Volumes for 9th class ppt

2. Wherever we look, usually we see solids. So far
, in all our study, we have been dealing with figures that can be easily drawn on our notebooks
Or blackboards . These are called plane figures.
We have understood what rectangles, squares,
Cylinders and circles are. What we mean by their
Perimeters, and areas and how we can find them
We have learnt these in earlier classes it would be I
Intresting to see what happens if we cut out many
Of these plane figure of the same shape and size
From cardboard sheet and stack them up in a
Vertical file. By this process , we shall obtain some

3. Solid figures such as a cuboid a cylinder a cube
e.t.c. u have shall now learn
To find the surface area and volume of
cuboiods and cylinders in detail and
Extend these studies to some other solids such
as cones and spheres.

5. As we know that if we have to make a
cuboid we
Want a bottom, four walls and a
top, therefore
Six rectangular pieces to cover the
complete outer
Surface of cuboid.
If we take the length of cuboid as ‘l’
breadth as ‘b’
And height as ‘h’, then the figure with these
Dimensions would be like as the shape.
So the sum of the area of six rectangles is:

6. Area of rectangle 1= (lxh)
+
Area of rectangle 2= (lxb)
+
Area of rectangle 3= (lxh)
+
Area of rectangle 4= (lxb)
+
Area of rectangle 5= (bxh)
+
Area of rectangle 6= (bxh)
= 2(lxb)+2(bxh)+2(lxh)
= 2(lxb)+(bxh)+(hxl)
= 2(lb+bh+hl)
This give us:
surface area of cuboid = 2(lb+bh+hl)

7. Cut out a neatly paper cone that does not have any overlapped paper.
Straight along its side and opening its out, to see the shape of the
paper
That forms a surface of cone. The line along with you cut the cone is the
Slant height of the cone which is represented by the ‘l’. It look like a
part of
A round cake.
If you now bring the side marked ‘a’ and ‘b; at the tip together, you can
see
That the curved portion will form the circular base of the cone.
If the paper line 1 is now cut into hundred of little pieces along the line
Drawn from point ‘o’, each cut portion is almost a small triangle, whose
Height is the slant height ‘l’ of the cone.

8. Now the area of each triangle
= 1/2xbase of each triangle
So, the area of the entire piece of paper= sum of all the area
= 1/2b1l+1/2b2l+1/2b3l+……
= 1/2xlxlenghth of entire curved boundary
(as b1+b2 +b3 +…. Makes up the curved portion)
But the curved portion of the figure make up the perimeter of the base of the
Cone and the circumference of the base of the cone = 2x22/7xR
So the curved surface area of the cone
= 1/2xlx2x22/7xR

9. •Surface area of the cuboid= 2(lb+bh+hl)
•Surface area of the cube = 6a2
•Curved surface area of the cylinder= 2x22/7xRH
•Total surface area of the cylinder= 2x22/7xR(h+r)
•Curved surface area of the cone= 22/7xRL
•Volume of the cuboid= LxBxH
•Volume of the cube= a3
•Volume of cylinder= 22/7xR2H

10. •Volume of the cone= 1/3x22/7xR2H
•Volume of the sphere= 4/3x22/7xR3
•Volume of the hemisphere= 2/3x22/7xR3
•Curved surface area of the cuboid= 2(l+b)h
•Curved surface area of the cube= 4a2