# Surface area and volume of different Geometrical Figures - PowerPoint PPT Presentation

Surface area and volume of different Geometrical Figures

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Surface area and volume of different Geometrical Figures

## Surface area and volume of different Geometrical Figures

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1. Cube Parallelopiped Cylinder Cone Surface area and volume of different Geometrical Figures

2. Faces of cube face face face 1 Dice (Pasa) 3 2 Total faces = 6 ( Here three faces are visible)

3. Faces of Parallelopiped Face Face Face Book Brick Total faces = 6 ( Here only three faces are visible.)

4. Cores Cores Total cores = 12 ( Here only 9 cores are visible) Note Same is in the case in parallelopiped.

5. Surface area Cube Parallelopiped c a b a a Click to see the faces of parallelopiped. a (Here all the faces are rectangular) (Here all the faces are square) Surface area = Area of all six faces = 6a2 Surface area = Area of all six faces = 2(axb + bxc +cxa)

6. Volume of Parallelopiped Click to animate c b b a Area of base (square) = a x b Height of cube = c Volume of cube = Area of base x height = (a x b) x c

7. Volume of Cube Click to see a a a Area of base (square) = a2 Height of cube = a Volume of cube = Area of base x height = a2 x a = a3 (unit)3

8. Outer Curved Surface area of cylinder r r h Click to animate Activity -: Keep bangles of same radius one over another. It will form a cylinder. Circumference of circle = 2 π r Formation of Cylinder by bangles 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 of cylinder = (2 π r) x( h)

9. Total Surface area of a solid cylinder Curved surface circular surfaces = Area of curved surface + area of two circular surfaces =(2 π r) x( h) + 2 π r2 = 2 π r( h+ r)

10. Other method of Finding Surface area of cylinder with the help of paper r h h 2πr Surface area of cylinder = Area of rectangle= 2 πrh

11. Volume of cylinder r h Volume of cylinder = Area of base x vertical height = π r2 xh

12. Cone l = Slant height h Base r

13. Volume of a Cone Click to See the experiment 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

14. 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

15. 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.

16. Surface area of cone l 2πr l l 2πr 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

17. Comparison of Area and volume of different geometrical figures

18. Area and volume of different geometrical figures r r r r/√2 l=2r r

19. Total surface Area and volume of different geometrical figures and nature r r r r l=3r 1.44r 22r 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.

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

21. 3r r r 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

22. Bottle Cone shape Cylindrical shape

23. 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

24. Volume of a Sphere Click to See the experiment r r h=r Here the vertical height and radius of cone are same as radius of sphere. 4( volume of cone) = volume of Sphere 4( 1/3πr2h) = 4( 1/3πr3 ) = V V = 4/3 π r3

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