Frustum

Deals with curved surface area, total surface area and volume of the frustum of a cone

1. M E N S U R A T I O N- sk STD X M A H A R A S H T R A S T A T E B O A R D O F E D U C A T I O N , M U M B A I F R U S T U M

2. MENSURATION Mensuration  is a branch of  mathematics  which deals with the surface area and volume of solid, plane and geometrical figures.

3. MENSURATION The  area  of a figure is the number of unit squares that cover the surface of a closed figure.  Area  is measured in square units such as square centimetres, square feet, square inches, etc.

4. MENSURATION In  math ,  volume  can be defined as the 3-dimensional space enclosed by a boundary or occupied by an object. ... Here, for example, the  volume  of the cuboid or rectangular prism, with unit cubes has been determined in cubic units.

5. MENSURATION

6. MENSURATION F R U S T U M In geometry, a frustum (plural: frusta or frustums) is the portion of a solid (normally a cone or pyramid) that lies between one or two parallel planes cutting it.

7. MENSURATION- surface area FRUSTUM OF A CONE Curved surface area = l Total surface area = (r +l) Remember : l 2 = r 2 + h 2   Curved surface area = l (r 1 + r 2 ) Total surface area = l(r 1 + r 2 ) + r 1 2 + r 2 2 = [l (r 1 + r 2 ) + r 1 2 + r 2 2 ] Remember: l 2 = (r 1 - r 2 ) 2 + h 2  

8. MENSURATION-volume cone The volume of a cone means the third part of the volume of a cylinder having the same base and the same height. It takes three  cones  to fill up a  cylinder .

9. MENSURATION- volume FRUSTUM OF A CONE Volume = h ( r 1 2 + r 2 2 + r 1 x r 2 )   r2 h r1

10. MENSURATION- surface area - volume FRUSTUM OF A CONE Volume = h ( r 1 2 + r 2 2 + r 1 x r 2 )   r2 r1 h l Curved surface area = l (r 1 + r 2 ) Total surface area = l(r 1 + r 2 ) + r 1 2 + r 2 2 = [l (r 1 + r 2 ) + r 1 2 + r 2 2 ] Remember: l 2 = (r 1 - r 2 ) 2 + h 2  

11. APPLICATION MENSURATION- surface area - volume Find how many litres of water a bucket can hold ( 1 lit = 1000 cm 3 ) R1 = 14 cm, r2 = 7 cm, h=30 cm - h 10 Volume: h ( r 1 2 + r 2 2 + r 1 x r 2 ) x 30(14 x 14 + 7 x 7 +14 x 7) x x 30 x 7 x 7( 4 +1 +2) 1540(7) 10780 cubic cm 10.780 litres     10

12. APPLICATION MENSURATION- surface area - volume The circumference of the circular faces of a bucket is 132 cm and 88 cm. if the height is 24 cm then find its curved surface area. Circumference 1 2 r 1 = 2 r 1 = r 1 21cm= r1   Circumference 2 2 r 1 = 2 r 2 = r 2 14 cm= r 2     Slant height: l 2 = (r 1 - r 2 ) 2 + h 2 L 2 = (21-14) 2 + 24 2 = 49 + 576 = 625 l = 25 Curved surface area = l (r 1 + r 2 ) = x 25 x (21+14) = x 25 x 35 =22 x 25 x 5 = 2750 cm 2  

13. APPLICATION MENSURATION- surface area - volume Given: r 1 = 14 cm, r 2 = 6 cm, h = 6 cm. Find: curved surface area, total surface area and its volume   Slant height: l 2 = (r 1 - r 2 ) 2 + h 2 =(14-6) 2 + 6 2 = 64 +36 =100 I = 10 cm Curved surface area: = l (r 1 + r 2 ) = 3.14 x 10 (14+6) =3.14 x 10 x 20 = 314 x 2 =628 cm 2       Total surface area = [l(r 1 +r 2 ) 2 + r 1 2 + r 2 2 ] 3.14 [10(14+6) + 14x 14 + 6 x6 ] 3;14 x 4 (50 +49 + 9) 3.14 x 4 (108) 3.14 x 432 = 1356.48 cm 2  

14. Curved surface area = l (r 1 + r 2 ) Total surface area = l(r 1 + r 2 ) + r 1 2 + r 2 2 = [l (r 1 + r 2 ) + r 1 2 + r 2 2 ] Remember: l 2 = (r 1 - r 2 ) 2 + h 2   Volume = h ( r 1 2 + r 2 2 + r 1 x r 2 )   T H A N K Y O U