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:
- Total surface area=2(lb+bh+lh)
For a cube with length x,
- Volume=x3 (because l = b = h = x)
- Total surface area=6x2
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=
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
Total surface area=6a2
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.
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.
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:
H= (300-9) =291cm
Total surface area of the tank=2(lb+bh+lh)
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
Volume of the cupboard = 36 x 40 x 100 cm3.
Therefore we can say that 200 tissue boxes can be adjusted in the cupboard box.