1.

The diameters of top and bottom portions of a milk can are 56 cm and 14 cm respectively. The height of the can is 72 cm . Find the (a) area of metal sheet required to make the can (without lid). (b) volume of milk which the container can hold.

Answer» The milk can is in the shape of a frustum with R = 28 cm, r = 7 cm and h = 72 cm.
(a) Area of metal sheet required =Curved surface area + Area of bottom base
`= pi l (R +r)+pi r^(2)`
Now, `l=sqrt((R-r^(2))+h^(2))=sqrt((28-7)^(2)+72^(2))=sqrt(21^(2)+72^(2))=sqrt(9(7^(2)+24^(2)))`
` =3 sqrt(49+576)=3xxsqrt(625)=3xx25=75 cm `
` :. "Area of metal sheet"=(22)/(7)xx75(28+7)+(22)/(7)xx 7^(2)=22 xx 75 xx 5 +22 xx 7`
`=22(375+7)`
`=22(382)=8404 cm^(2)`.
(b) Ammount of milk which the container can hold `=(1)/(3) pi h(R^(2)+Rr+r^(2))`
`=(1)/(3)xx (22)/(7) xx 72 (28^(2)+7 xx 28 +7^(2))`
`=(22)/(7) xx 24 (7xx 4 xx28+7xx 28+7xx 7)`
`=(22)/(7) xx 24xx 7(112+28+7)`
` =22 xx 24 xx (147)=77616 cm^(3)`.


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