Surface area and volume of different Geometrical Figures

Cube ParallelopipedCylinder Cone

faceface

face

Total faces = 6 ( Here three faces are visible)

1

2 3

Dice (Pasa)

Faces of cube

Faces of Parallelopiped

Brick

Book

Fac

e

Face

Face

Total faces = 6 ( Here only three faces are visible.)

Cores

Total cores = 12 ( Here only 9 cores are visible)

Cores

Note Same is in the case in parallelopiped.

Surface area = Area of all six faces

= 6a2

ab

Surface areaCube Parallelopiped

Surface area = Area of all six faces

= 2(axb + bxc +cxa)

c

a

a

a

Click to see the faces of parallelopiped.

(Here all the faces are square) (Here all the faces are rectangular)

Area of base (square) = a x b

a

Height of cube = c

Volume of cube = Area of base x height

= (a x b) x c

b

c

b

Volume of Parallelopiped Click to animate

Volume of Cube

a

a

Area of base (square) = a2

Height of cube = a

Volume of cube = Area of base x height

= a2 x a = a3

Click to see

a

(unit)3

Circumference of circle = 2 π r

Area covered by cylinder = Surface area of of cylinder = (2 π r) x( h)

r h

Outer Curved Surface area of cylinder

Activity -: Keep bangles of same radius one over another. It will form a cylinder.

It is the area covered by the outer surface of a cylinder.

Formation of Cylinder by bangles

Circumference of circle = 2 π r

r

Click to animate

Total Surface area of a solid cylinder

=(2 π r) x( h) + 2 π r2

Curved surface

Area of curved surface + area of two circular surfaces=

circular surfaces

= 2 π r( h+ r)

2πr

h

r

h

Surface area of cylinder = Area of rectangle= 2 πrh

Other method of Finding Surface area of cylinder with the help of paper

Volume of cylinder

Volume of cylinder = Area of base x vertical height

= π r2 xh

r

h

Cone

Baser

h

l = Slant height

3( V ) = π r2h

r

h h

r

Volume of a Cone Click to See the experiment

Here the vertical height and radius of cylinder & cone are same.

3( volume of cone) = volume of cylinder

V = 1/3 π r2h

if both cylinder and cone have same height and radius then volume of a cylinder is three times the volume of a cone ,

Volume = 3V Volume =V

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.

l

2πr

l

2πr

l

Area of a circle having sector (circumference) 2π l = π l 2

Area of circle having circumference 1 = π l 2/ 2 π l

So area of sector having sector 2 π r = (π l 2/ 2 π l )x 2 π r = π rl

Surface area of cone

Surface area

6a2 2π rh π r l 4 π r2

Volume a3 π r2h 1/3π r2h 4/3 π r3

Comparison of Area and volume of different geometrical figures

Surface area

6r2

=2 π r2

(about)

2π r2 2π r2 2 π r2

Volume r3 3.14 r3 0.57π r3 0.47π r3

Area and volume of different geometrical figures

r/√2

r

l=2rr

r

r

Total Surface area

4π r2 4π r2 4π r2 4 π r2

Volume 2.99r3 3.14 r3 2.95 r3 4.18 r3

Total surface Area and volume of different geometrical figures and nature

r

r

l=3r r

r

1.44r

So for a given total surface area the volume of sphere is maximum. Generally most of the fruits in the nature are spherical in nature because it enables them to occupy less space but contains big amount of eating material.

22r

Think :- Which shape (cone or cylindrical) is better for collecting resin from the tree

Click the next

r

3r

V= 1/3π r2(3r)

V= π r3

Long but Light in weight

Small niddle will require to stick it in the tree,so little harm in tree

V= π r2 (3r)

V= 3 π r3

Long but Heavy in

weight

Long niddle will require to stick it in the tree,so much harm in tree

r

Cone shape

Cylindrical shape

Bottle

V1

r

V=1/3 πr2h

If h = r thenV=1/3 πr3

rr

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.

V1 = 4V = 4(1/3 πr3)

= 4/3 πr3

4( 1/3πr2h ) = 4( 1/3πr3 ) = V

h=rr

Volume of a Sphere Click to See the experiment

Here the vertical height and radius of cone are same as radius of sphere.

4( volume of cone) = volume of Sphere

V = 4/3 π r3

r

Thanks

U.C. Pandey R.C.Rauthan, G.C.Kandpal