CBSE Class 10 Mathematics Surface Area and Volume Notes Set 05

Formulae of Area and Volume of Solids

Let us first revise the formulae of the area and volume of the basic solids.

Combination of Solids

Two types of questions arise from a combination of solids: finding the surface area of the combination and finding its volume.

The total surface area of a combination of solids is the sum of the surface areas of the individual solids, leaving out the parts that overlap (the joined faces). The volume of a combination of solids is the sum of the volumes of the solids.

Knowledge Hub

The term capacity is used for how much a container can hold and is commonly measured in litres, while volume is used for how much space an object occupies and is commonly measured in cubic metres. Only hollow objects have capacity, while both hollow and solid objects have volume.

Examples

Example 1. A solid is in the form of a right circular cylinder mounted on a solid hemisphere of the same radius having a diameter of 21 cm and the total height of the solid is 24.5 cm. If both the cylinder and hemisphere are made up of same material, then determine the weight of the solid where the weight of 1 cm³ material is 5 g.
Answer:

Radius of the hemispherical part \( r = \frac{21}{2} = 10.5 \) cm. Height of the cylindrical part \( = 24.5 - 10.5 = 14 \) cm.
Volume of the hemisphere \( = \frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (10.5)^3 = 2425.5 \) cm³.
Volume of the cylinder \( = \pi r^2 h = \frac{22}{7} \times (10.5)^2 \times 14 = 4851 \) cm³.
Total volume \( = 2425.5 + 4851 = 7276.5 \) cm³. Weight \( = 7276.5 \times 5 = 36382.5 \) g \( = 36.3825 \) kg.
Hence, the weight of the solid is 36.3825 kg.

Example 2. A pencil is in the form of a cone mounted on a cylinder with the same radius. The total length of the pencil is 15 cm of which 1.5 cm is the conical portion. The diameter of the pencil is 0.7 cm. Find the total surface area of the pencil.
Answer:

Radius \( r = \frac{0.7}{2} = 0.35 \) cm. Height of the cylinder \( H = 15 - 1.5 = 13.5 \) cm and height of the cone \( h = 1.5 \) cm.
Slant height of the cone \( l = \sqrt{r^2+h^2} = \sqrt{0.1225+2.25} = \sqrt{2.3725} \approx 1.54 \) cm.
Total surface area \( = \) curved surface of cone \( + \) curved surface of cylinder \( + \) base of cylinder \( = \pi rl + 2\pi rH + \pi r^2 \).
\( = \frac{22}{7} \times 0.35 \times 1.54 + 2 \times \frac{22}{7} \times 0.35 \times 13.5 + \frac{22}{7} \times 0.35 \times 0.35 \approx 1.69 + 29.7 + 0.385 \approx 31.78 \) cm².
Hence, the total surface area of the pencil is about 31.78 cm².

Example 3. The length of a room is twice its height and its breadth is \( \frac{3}{2} \) times its height. The cost of white washing the walls at the rate of Rs. 2.40 per m² is Rs. 268.80. Find the cost of tiling the floor at the rate of Rs. 9.35 per m².
Answer:
Let the height of the room be \( h \) metres. Then length \( = 2h \) m and breadth \( = \frac{3}{2}h \) m.
Area of the four walls \( = 2(l+b)h = 2\left(2h+\frac{3}{2}h\right)h = 7h^2 \) m².
Cost of white washing \( = 2.40 \times 7h^2 = 16.80h^2 = 268.80 \Rightarrow h^2 = 16 \Rightarrow h = 4 \) m.
So length \( = 8 \) m and breadth \( = 6 \) m. Area of the floor \( = 8 \times 6 = 48 \) m².
Cost of tiling \( = 9.35 \times 48 = \) Rs. 448.80.
Hence, the cost of tiling the floor is Rs. 448.80.

Example 4. A conical cavity of height 24 cm and of base radius 7 cm is drilled out from a solid cylinder of height 30 cm and radius 7 cm. Find the volume and the total surface area of the remaining solid.
Answer:

Slant height of the cone \( l = \sqrt{24^2+7^2} = 25 \) cm.
Volume of the remaining solid \( = \) volume of cylinder \( - \) volume of cone \( = \pi r^2\left(H - \frac{h}{3}\right) = \frac{22}{7} \times 7 \times 7 \times (30 - 8) = 3388 \) cm³.
The cone's base coincides with the whole top face of the cylinder, so the top face is removed. The surface of the remaining solid consists of the curved surface of the cylinder, the curved surface of the cone and the lower circular base.
T.S.A. \( = 2\pi rH + \pi rl + \pi r^2 = \frac{22}{7} \times 7 \times (60 + 25 + 7) = 2024 \) cm².
Hence, the volume is 3388 cm³ and the total surface area is 2024 cm².