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# Mensuration ppt

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very nice ppt on indias mensuration

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• nice 1 really usefull. ;-)

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### Mensuration ppt

1. 1. CUBE Surface Area We will need to find the surface area of the top, base and a sides. Area of the top and bottom is 2a2a a Area of sides (CSA) is 4a2 Therefore the formula is: 6a2 a Volume V = a 3
2. 2. Cuboid Surface Area We will have to calculate the area of sides, top and base. Area of sides = (CSA) is 2(lh+bh) Area of top and base is 2(lb)Therefore the formula is: 2(lb+lh+bh) Volume v = lbh
3. 3. TSA = ‫ת‬r(s+r)
4. 4. 2CSA = 2πr
5. 5. Surface Area We need to find theouter surface area and the area of the base. Outer surface area (CSA) is 2 π r^2 Area of the base is π r^2 Therefore the formula is 3 π r^2 Volume V = πr^3
6. 6. Surface Area of the model = CSA of cone + CSA of cylinder = πrl + 2πrh = π r ( l + 2h )
7. 7. Surface Area of the model = CSA of thecylinder + CSA Of the cone + Area of the cylinder’s base = 2πrh + πrl + πr2 = πr ( 2h + l + r )
8. 8. Surface Area of the model =CSA of hemisphere + CSA of the cone = 2πr2 + πrl = πr ( 2r + l )
9. 9. Surface area of the model = CSA of the cylinder + Surface area of the sphere – Area of the cylinder’s base = 2 πrh + 4 πr2 – πr2 = πr ( 2h – r ) + 4πr2
10. 10. Surface area of the model = TSA of the the largercylinder + TSA of the smaller cylinder – Area of the smaller cylinder’s base = 2πRH + 2πR2 + 2πrh + 2πr2 – πr2 = 2πR ( H + R ) + 2πr ( h + r )
11. 11. Surface area of themodel = CSA of the cylinder + CSA of both the cones = 2πrh + 2πrl = 2πr ( h + l )
12. 12. Surface area of the model = CSA of the cylinder + CSA of the two hemispheres = 2πrh + 2 ( 2πr2 ) = 2πrh + 4πr2 = 2πr ( h + 2r )
13. 13. When two cubes are joined, they form a cuboid. In the given model when the two cubes of side ‘a’ are joined, we get a cuboid of dimension > l = a + a b = a h = a So the model’s surface area is 2 ( lb + bh +hl )= 2 { ( a + a ) a + a2 + a ( a + a) }= 2 ( 2a2 + a2 + 2a2 )= 2 ( 5a2)= 10a2
14. 14. Surface area of the model = CSA of the cylinder + CSA of the two hemispheres of the same dimension = 2πrh + 2πr2 + 2πr2 = 2πrh + 4πr2 = 2πr ( h + 2r )
15. 15. Surface area of the model =TSA of the cuboid + CSA of thehemisphere - CSA of the top of thehemisphere = 2 ( lb + hl + hb ) + 2πr2 – πr2 = 2 ( lb + hl + hb ) + πr2
16. 16. Surface area of the model =TSA of the cuboid + CSA of thehemisphere - CSA of the top of thehemisphere = 2 ( lb + hl + hb ) + 2πr2 – πr2 = 2 ( lb + hl + hb ) + πr2
17. 17. Surface area of the model = CSA of the cylinder + CSA of the hemisphere = 2πrh + 2πr2 = 2πr ( h + r )
18. 18. • Thank you - Dhruv Sahdev
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