NCERT Class 9 Solutions: Surface Areas and Volumes (Chapter 13) Exercise 13.3 – Part 2

Surface area of cone=πrl

Lateral Surface Area of Cone

Surface area of cone=πrl

Slant height l or s = h2+r2

Q-4 A conical tent is 10m high and the radius of its base is 24m . Find

  1. Slant height of the tent.

  2. Cost of the canvas required to make the tent, if the cost of 1m2 canvas is Rs70


  • Radius of the base (r) =24m

  • Height of the conical tent (h) =10m

Therefore, the slant height of the cone is l .

  • l2=h2+r2

  • l=h2+r2

  • l=102+242

  • l=100+576

  • l=676

  • l=26m


Canvas required to make the conical tent is the curved or lateral surface of the cone. Note that the tent does not have a base. Also, cost of 1m2 canvas =Rs70

Area of canvas = πrl=(227×24×26)m2 ( π=227 ) =137287m2

Therefore, cost of canvas = 137287×70=137280 Rs.

5. What length of tarpaulin 3m wide will be required to make conical tent of height 8m and base radius 6m ? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20cm (Use π=3.14 ).Solution:

  • Radius of the base (r) =6m

  • Height of the conical tent (h) =8m

Now, l is the slant height of the cone =h2+r2=82+62=100=10

Curved surface area of conical tent = πrl=(3.14×6×10)m2=188.4m2

We know that breadth of tarpaulin =3m . Let, x is length of tarpaulin sheet required of which we know from the problem that 20cmor0.2m will be wasted in cutting. So, the actual length used will be (x0.2m) . Now area of this sheet with length (x0.2m) and breadth of 3m must match the lateral surface area of the sheet,

That is, [(x0.2)×3]=188.4m2

  • x0.2=62.8

  • x=63m

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