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Example 14 - An open metal bucket is in shape of a frustum - Examples

  1. Chapter 13 Class 10 Surface Areas and Volumes
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Example 14 An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet (see Fig. 13.23). The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket, where we do not take into account the handle of the bucket. Also, find the volume of water the bucket can hold.(Take π = 22/7) Area of metallic sheet used = curved surface area of frustum + curved surface area of cylinder + area of circular base Curved surface area of frustum Diameter of bigger circular end = 45 cm Radius = r1 = 45/2 cm Diameter of smaller circular end = 25 cm Radius = r2 = 25/2 cm Now, Height of frustum = Total height of bucket – Height of cylinder = 40 – 6 = 34 cm Also, Slant height of frustum (l) = √(h2+(r1−r2)2) = √((34)2+( 45/2−25/2)^2 ) = √((34)2+( (40 − 25)/2)^2 ) = √((34)2+(20/2)2) = √((34)2+(10)2) = √(1156+10) = √1256 = 35.44 cm Curved surface area of frustum = π(r1+r2)𝑙 = 𝜋×35.44(45/2+25/2) = 𝜋×35.44×(70/2) = 𝜋×35.44×35 = 1240.4 𝜋 cm2 Area of circular base Base is a circle with radius = r2 = 25/2 Area of circular base = πr22 = π×(25/2)^2 = π×(12.5)2 = 156.25 π cm2 Curved Surface Area of cylinder Radius of cylinder = r2 = 25/2 Height of cylinder = 6 cm Curved Surface Area of cylinder = 2𝜋𝑟ℎ = 2×𝜋×25/2×6 = 150𝜋 cm2 Now, Area of metallic sheet used = curved surface area of frustum + area of circular base + curved surface area of cylinder = 1240.4𝜋+156.25𝜋+150𝜋 = 𝜋(1240.4+156.26+150) = 22/7×(1546.66) = 22×220.95 = 4860.9 cm2 Now, We need to find volume of water in bucket Volume of water in bucket = Volume of frustum = 1/3 𝜋ℎ(𝑟12+𝑟22+𝑟2𝑟2) = 1/3×22/7×34 ((45/2)^2+(25/2)^2+45/2×25/2) = 22/7×34/3 (( 22.5)2+(12.5)2+22.5×12.5) = 22/7×34/3 (506.25 + 156.25 + 281.25) = 22/7×34/3(943.75) = 22×34×44.94 = 33615.48 cm3 = 33615.48/1000 litres = 33.62 litres (approx.) Hence, area of metallic shed used = 4860.9 cm2 Volume of water that bucket can hold = 33.62 litres (approx.)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
  • Rohit Kumar's image

    Hey in example 14 of chapter 13 of class 10  why you do not take vol of cylinder on which a frustrum is mounted the cylinder is hollow and it has height that means some water goes to the cylindrical base so why you do not take the volume of cylinder you only take the volume of frustrum sir plzz answer my question as soon as possible

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