Volume Of A Combination Of Solids

A solid which is bounded by six rectangular faces is known as cuboid and if the length, breadth and height of the cuboid are equal, then it is a cube.

8 vertices, 6 faces and 12 edges are there in both cube as well as cuboid. Base of the cuboid is any face of the cuboid.

For a cuboid which has length (l), breadth (b) and height (h) has:

  • Volume = l×b×h
  • Total surface area = 2(lb+bh+lh)

For a cube with length x,

  • Volume = x3 (because l = b = h = x)
  • Total surface area = 6x2

Volume of combination of solids Examples:

Q1. For a room of dimension 10m×8m×9m.Find out the longest pole that can be put in this room.

Solution: Longest pole is the longest diagonal of the room=Combination Of Solids

Combination Of Solids

=15.652m

The longest pole that can be put inside the room has length=15.652m.

Q2. A cube has a volume 343cm3.Find the surface area of the cube.

Solution: The volume of the cube=a3=323

Combination Of Solids

Total surface area=6a2

=6×72=294cm2

Total surface are of the cube=294cm2

Q3. Two cubes are joined end to end and has the volume 81cm3. Find the total surface area of the cube which is formed now.

Solution:

Combination Of Solids

When 2 cubes are joined end to end, it becomes a cuboid.

Volume of the cuboid=81cm3 = 2 x volume of each cube.

Let x be side of each cube.

2x3=81

x=9/ √2 cm

Length of the resulting cuboid=2x = 2 x 9/ √2 cm = 9√2 cm.

Breadth=9/√2 cm Height=9/√2 cm

Total surface area of the cuboid=2(lb+bh+lh)

= 2(9 √2×9/√2 + 9/√2 ×9√2 +9/√2 ×9 √2)

= 2( 81 + 81/2 + 81/2) = 2 x 162

=324 sq. cm.

Q4. A cuboidal water tank is aluminium steel sheet which is 4.5m thick. The outer dimensions are 1.5m×2.5m×3m. Find the internal dimensions and total surface area of the tank.

Solution: External dimensions of the cube are:

L=150cm, b=250cm and h=300cm.

As we know that the sheet is 4.5m thick, the internal dimensions are:

L=(150-9)cm=141cm

B=(250-9)=241cm

H= (300-9) =291cm

Total surface area of the tank=2(lb+bh+lh)

=2(1.5×2.5+2.5×3+3×1.5)

=31.5 m3.

Q5. How many tissue boxes of size 10cm×8cm×9cm can be adjusted inside a cupboard box of size 36cm×40cm×100cm.

Solution: Volume of the tissue box = 10 x 8 x 9 cm3

=720cm3

Volume of the cupboard = 36 x 40 x 100 cm3.

=144,000cm3

Combination Of Solids

=200 boxes

Therefore we can say that 200 tissue boxes can be adjusted in the cupboard box. For more information contact byju’s mentors.

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