NCERT Class 9 Solutions: Surface Areas and Volumes (Chapter 13) Exercise 13.3 – Part 3

Surface area and volume of a cone

Surface Area and Volume of a Cone

Surface area and volume of a cone

Q-6 The slant height and base diameter of a conical tomb are 25m and 14m respectively. Find the cost of white-washing its curved surface at the rate of Rs210 per 100m2 Solution:

We know,

  • Base diameter = 14m

  • Radius (r) =142m=7m

  • Slant height of tomb (l) =25m

Therefore curved surface area = πrlm2 =(227×25×7)m2=550m2

Rate of white- washing 100 m2 =Rs210

Rate of white- washing 1 m2 =210100

Therefore, total cost of white-washing the tomb with area 550 =Rs(550×210100)=Rs.1155

Q-7 A joker’s cap is in the form of a right circular cone of base radius 7cm and height 24cm . Find the area of the sheet required to make 10 such caps.Solution:

  • The cone radius (r) =7cm

  • The cone height (h) =24cm

Now, l be the slant height

  • l=h2+r2

  • l=242+72

  • l=625

  • l=25m

Sheet required for one cap = Curved surface of the cone

  • πrlcm2

  • (227×7×25)cm2

  • 550cm2

Sheet required for 10caps=550×10cm2=5500cm2

Q-8 A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs12perm2 , what will be the cost of painting all these cones? (Use π=3.14 and take 1.04=1.02 )


We are given,

  • Radius of the cone (r) =402cm=20cm=0.2m

  • Height of the cone (h) =1m

Now, the slant height of a cone l is given by

  • l=h2+r2

  • l=1+(0.2)2

  • l=1.04

  • l=1.02m

We know that, rate of painting Rs12perm2

Curved surface area of 1 cone πrlm2 = (3.14×0.2×1.02)m2=0.64056m2

Therefore, curved surface of such 50 cones = (50×0.64056)m2=32.028m2

Finally, cost of painting all these cones is Rs(32.028×12)=Rs384.34 (since, the cost of painting is Rs12perm2 )

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