NCERT Class 9 Solutions: Surface Areas and Volumes (Chapter 13) Exercise 13.5 – Part 1

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Formulas for volume of common solids

Volumes of Common Solids

Formulas for volume of common solids

Q-1 A matchbox measures 4cm×2.5cm×1.5cm . What will be the volume of a packet containing 12 such boxes?


  • Each matchbox measures =4cm×2.5cm×1.5cm

  • Therefore, l=4cm , b=2.5cm and h=1.5cm

Volume of one matchbox

  • (l×b×h)

  • (4×2.5×1.5)cm3

  • 15cm3

Volume of a packet containing 12 such boxes

  • (12×15)cm3

  • 180cm3

Q-2 A cuboidal water tank is 6m long, 5m wide and 4.5m deep. How many litres of water can it hold? ( 1m3=1000l )


  • Dimensions of water tank =6m×5m×4.5m

  • l=6m,b=5mandh=4.5m

Therefore Volume of the tank

  • lbhm3

  • (6×5×4.5)m3

  • 135m3

Therefore, the tank can hold

  • 135×1000liters [Note that, 1 m3=1000litres ]

  • 135000 Liters of water.

Q-3 A cuboidal vessel is 10m long and 8m wide. How high must it be made to hold 380 cubic metres of a liquid?Solution:

We know that,

  • Length =10m ,

  • Breadth =8m

  • Volume =380m3

Volume of cuboid =Length×Breadth×Height

Therefore, Height =VolumeofcuboidLength×Breadth = 38010×8=4.75m

Q-4 Find the cost of digging a cuboidal pit 8m long, 6m broad and 3m deep at the rate of Rs30perm3.


We know,

  • l=8m,

  • b=6m

  • h=3m

Volume of the pit

  • lbhm3

  • (8×6×3)m3

  • 144m3

Rate of digging is Rs30perm3 . Therefore, total cost of digging the pit = Rs.(144×30)=Rs.4320

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