A closed cylindrical tank, made of thin iron-sheet, has diameter $= 8.4 m$ and height $5.4 m .$ How much metal sheet, to the nearest $m^{2},$ is used in making this tank, if $\frac{1}{15}$ of the sheet actually used was wasted in making the tank ?

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Solution

Diameter of cylinder $= 8.4 m$
Radius $= 4.2$ m
Height $= 5.4$ m
Total surface area of cylinder $= 2 π r (r + h) = 253.44 m^{2}$
Wasted metal sheet $= \frac{1}{15} \times 253.44 = 16.896 m$
Total metal sheet required $= 253.44 + 16.896 = 270$ m

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A closed cylindrical tank, made of thin iron-sheet, has diameter =8.4m and height 5.4m. How much metal sheet, to the nearest m2, is used in making this tank, if 115 of the sheet actually used was wasted in making the tank ?

Diameter of cylinder =8.4mRadius =4.2 mHeight =5.4 mTotal surface area of cylinder =2πr(r+h)=253.44m2Wasted metal sheet =115×253.44=16.896mTotal metal sheet required =253.44+16.896=270 m

A closed cylindrical tank, made of thin iron-sheet, has diameter $= 8.4 m$ and height $5.4 m .$ How much metal sheet, to the nearest $m^{2},$ is used in making this tank, if $\frac{1}{15}$ of the sheet actually used was wasted in making the tank ?

Diameter of cylinder $= 8.4 m$
Radius $= 4.2$ m
Height $= 5.4$ m
Total surface area of cylinder $= 2 π r (r + h) = 253.44 m^{2}$
Wasted metal sheet $= \frac{1}{15} \times 253.44 = 16.896 m$
Total metal sheet required $= 253.44 + 16.896 = 270$ m