NCERT Class 9 Solutions: Surface Areas and Volumes (Chapter 13) Exercise 13.7 – Part 1

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volume of cone=1/3 πr^2 h

Volume of Cone

volume of cone=1/3 πr^2 h

Relation between the height, radius and slant height of a cone.

Slant Height Radius and Height of Cone

Relation between the height, radius and slant height of a cone.

By the Pythagoras theorem, (Slantheight)2=(Radius)2+(Height)2

Q-1 Find the volume of the right circular cone with

  1. Radius 6cm , height 7cm

  2. Radius 3.5cm , height 12cm

Solution:

  1. Radius (r)=6cm

    Height (h)=7cm

    Therefore, volume of the cone

    • =13πr2h

    • =(13×227×6×6×7)cm3

    • =264cm3

  2. Radius (r)=3.5cm Height (h)=12cm

Volume of the cone

  • =13πr2h

  • =(13×227×3.5×3.5×12)cm3

  • =154cm3

Q-2 Find the capacity in liters of a conical vessel with

  1. Radius 7cm , slant height 25cm

  2. Height 12cm , slant height 13cm

Solution:

  1. Radius (r)=7cm Slant height (l)=25cm

    Consider the height of the conical vessel h . Then by applying Pythagoras theorem,

    • h=l2r2

    • h=25272

    • h=62549

    • h=576

    • h=24cm

    Now, volume of the cone

    • =13πr2h

    • =(13×227×7×7×24)cm3

    • =1232cm3

    Therefore, capacity of the vessel =(12321000)l=1.232l

  2. Height (h)=12cm Slant height (l)=13cm

    Consider the radius of the conical vessel r . Again using Pythagoras theorem,

  • r=l2h2

  • r=132122

  • r=169144

  • r=25

  • r=5cm

Volume of the cone

  • =13πr2h

  • =(13×227×5×5×12)cm3

  • =(22007)cm3

Therefore, capacity of the vessel =(22007000)l=1135l , 1l=1000cm3

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