In volume of combination of solids concept, we will come across the different solids and their respective volumes. As we know, all three-dimensional shapes have surface areas and volumes, based on their dimensions.

Now, if we see in our real-life, there are many shapes which are formed by combining different shapes. For example, a tent is formed by the combination of a cone and cylinder shape. Similarly, ice cream cone is a combination of a cone and a hemi-sphere. Hence, the volume of this combination of solids will be equal to the sum of the volume of individual solid.

A solid which is bounded by six rectangular faces is known as a cuboid and if the length, breadth and height of the cuboid are equal, then it is a cube. Both cube and cuboid have 8 vertices, 6 faces and 12 edges. The faces of cuboid are in rectangular shape and faces of cube are in square shape. Let us learn here the volumes for such shapes and their combination.

## Volume of Solids

Before we solve the problems based on combination of solids, let us have a look at the volumes of all the three-dimensional solid shapes.

For a cuboid which has length (l), breadth (b) and height (h), the formula for volume and surface area is given by:

- Volume = l×b×h
- Total surface area = 2(lb+bh+lh)
- Length of diagonal of cuboid = √(l
^{2}+b^{2}+h^{2})

For a cube having edge length equal to x, the formula for volume and surface area is given by:

- Volume = x
^{3}(because l = b = h = x) - Total surface area = 6x
^{2} - Length of diagonal of cube = √3x

Similarly, for other shapes such as sphere, cone and cylinder, the formulas for volumes are:

- Volume of Sphere = (4/3)πr
^{3} - Volume of Cone = (1/3) πr
^{2}h - Volume of Cylinder = πr
^{2}h - Volume of Hemi-sphere = (2/3)πr
^{3}

## Solved Examples

**Q1**. **A cylinder of volume 150 cu.cm is placed with a cone, whose height is 4cm. If the height of cone and cylinder is equal, then find the total volume of shape formed by the combination of cylinder and cone.**

**Solution**: Given, Volume of cylinder = 150 cu.cm

Height of cylinder = Height of cone = 4cm

By the formula of volume of cylinder we know,

V1 = πr^{2}h

150 = πr^{2}(4)

r^{2}= 150/4π …..(1)

Now, volume of cone is given by:

V2 = 1/3 πr^{2}h

By putting the value of r^{2} from eq. 1 we get;

Therefore, V2 = 1/3 π (150/4π) (4)

On solving the above, we get;

V2 = 50 cu.cm.

Therefore, the total volume of the combined solids, V = V1 + V2 = 150 + 50 = 200 cu.cm.

**Note:** Volume of Cone = 1/3 (Volume of Cylinder)

**Q2. A cube has a volume of 343 cm ^{3}. Find its surface area.**

**Solution**: Given, volume of 343 cm^{3}

As we know,

volume of the cube = side^{3}

Therefore,

side^{3} = 343

side = 7 cm

Hence, surface area of the cube = 6(side)^{2}

= 6 (7)^{2}

= 294 sq.cm.

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

**Solution**: External dimensions of the cube are:

l = 150 cm, b = 250 cm and h = 300 cm

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

L = (150-9) = 141 cm

B = (250-9) = 241 cm

H = (300-9) = 291 cm

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

= 2(1.5×2.5+2.5×3+3×1.5)

= 31.5 m^{3}.

**Q.4**. **How many tissue boxes of size 10 cm × 8 cm × 9 cm can be adjusted inside a cupboard box of size 36 cm × 40 cm × 100 cm.**

**Solution**: Volume of the tissue box = 10 x 8 x 9 cm^{3}

= 720 cm^{3}

Volume of the cupboard = 36 x 40 x 100 cm^{3}.

= 144,000 cm^{3}

=200 boxes

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