NCERT Solution Class 10th Maths Chapter – 12 Surface Areas and Volumes Examples

Solution: This top is exactly like the object we have discussed in Fig. 12.5. So, we can conveniently use the result we have arrived at there. That is-
TSA of the toy = CSA of hemisphere + CSA of cone

Now, the curved surface area of the hemisphere = 1/2 (4πr²) = 2πr²
= (2 × 22/7 × 35/2 × 35/2) cm²

Also, the height of the cone = height of the top-height (radius) of the hemispherical part
= (5 – 3.5/2) cm = 3.25 cm

So, the slant height of the cone (l) = √r²+h² = √(3.5/2)² +(3.25)² cm = 3.7 cm (approx)
Therefore, CSA of cone = πrl = (22/7 × 3.5/2 × 3.7) cm²
This gives the surface area of the top as
= (2 × 22/7 × 3.5/2 × 3.5/2) cm² + (22/7 × 3.5/2 × 3.7) cm²
= 22/7 × 3.5/2 (3.5 + 3.7) cm² = 11/2 × (3.5 + 3.7) cm² = 39.6 cm² (approx)

You may note that ‘total surface area of the top’ is not the sum of the total surface areas of the cone and hemisphere.

Example 2. The decorative block shown in Fig. 12.7 is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. (Take л = 22/7)

Solution: The total surface area of the cube = 6 × (edge)² = 6 × 5 × 5 cm² = 150 cm². Note that the part of the cube where the hemisphere is attached is not included in the surface area.

So, the surface area of the block = TSA of cube – base area of hemisphere + CSA of hemisphere
= 150 – πr² + 2 m² = (150 + πr²) cm²
= 150 cm² + (22/7 × 4.2/2 × 4.2/2) cm²
= (150 + 13.86) cm² = 163.86 cm²

Example 3. A wooden toy rocket is in the shape of a cone mounted on a cylinder, as shown in Fig. 12.8. The height of the entire rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Take π = 3.14)

Solution: Denote radius of cone by r, slant height of cone by l, height of cone by h, radius of cylinder by r’ and height of cylinder by h’.

Then r = 2.5 cm, h = 6 cm, r’ = 1.5 cm, h’ = 26-6 = 20 cm and l = √r² + h² = √2.5² + 6² cm = 6.5 cm

Here, the conical portion has its circular base resting on the base of the cylinder, but the base of the cone is larger than the base of the cylinder. So, a part of the base of the cone (a ring) is to be painted.

So, the area to be painted orange = CSA of the cone + base area of the cone
– base area of the cylinder
= πrl + πr² – π(r’)²
= π[(2.5 × 6.5) + (2.5)² – (1.5)²] cm²
= π[20.25] cm² = 3.14 × 20.25 cm²
= 63.585 cm²

Now, the area to be painted yellow = CSA of the cylinder
+ area of one base of the cylinder
= 2πr’h’ + π(r’)²
= πr’ (2h’ + r’)
= (3.14 × 1.5) (2 × 20 × 1.5) cm²
= 4.71 × 41.5 cm²
= 195.465 cm²

Example 5. Shanta runs an industry in a shed which is in the shape of a cuboid surmounted by a half cylinder (see Fig. 12.12). If the base of the shed is of dimension 7 m x 15 m, and the height of the cuboidal portion is 8 m, find the volume of air that the shed can hold. Further, suppose the machinery in the shed occupies a total space of 300 m³, and there are 20 workers, each of whom occupy about 0.08 m³ space on an average. Then, how much air is in the shed? (Take л = 22/7)

Solution: The volume of air inside the shed (when there are no people or machinery) is given by the volume of air inside the cuboid and inside the half cylinder, taken together.

Now, the length, breadth and height of the cuboid are 15 m, 7 m and 8 m, respectively.
Also, the diameter of the half cylinder is 7 m and its height is 15 m.
So, the required volume = volume of the cuboid + 1/2 Volume of the cylinder
= [15 × 7 × 8 + 1/2 x 22/7 x 7/2 x 7/2 x 15] m³ = 1128.75 m³

Next, the total space occupied by the machinery = 300 m³
And the total space occupied by the workers = 20 × 0.08 m³ = 1.6 m³
Therefore, the volume of the air, when there are machinery and workers
= 1128.75 – (300.00 + 1.60) = 827.15 m³

Example 6. A juice seller was serving his customers using glasses as shown in Fig. 12.13. The inner diameter of the cylindrical glass was 5 cm, but the bottom of the glass had hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 10 cm, find the apparent capacity of the glass and its actual capacity. (Use π = 3.14.)

Solution: Since the inner diameter of the glass = 5 cm and height = 10 cm, the apparent capacity of the glass = πr²h
= 3.14 × 2.5 × 2.5 × 10 cm³ = 196.25 cm³

But the actual capacity of the glass is less by the volume of the hemisphere at the base of the glass.
i.e., it is less by 2/3 πr³ = 2 × 3.14 × 2.5 × 2.5 × 2.5 cm³ = 32.71 cm³

So, the actual capacity of the glass = apparent capacity of glass – volume of the hemisphere
= (196.25 – 32.71) cm³
= 163.54 cm³

Example 7. A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (Take л = 3.14)

Solution: Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere (see Fig. 12.14). The radius BO of the hemisphere (as well as of the cone)= 1/2 × 4 cm = 2 cm.

So, volume of the toy = 2/3πr³ + 1/3πr²h
= [2/3 × 3.14 × (2)³ + 3.14 × (2)² × 2) cm³ = 25.12 cm³

Now, let the right circular cylinder EFGH circumscribe the given solid. The radius of the base of the right circular cylinder = HP = BO=2 cm, and its height is EH = AO + OP = (2+2) cm = 4 cm

So, the volume required = volume of the right circular cylinder – volume of the toy
= (3.14 × 224 – 25.12) cm³
= 25.12 cm³

Hence, the required difference of the two volumes = 25.12 cm³.