SURFACE AREAS AND VOLUMES-Notes
Maths - Notes
Solid Figure
Surface Area
Surface area is of two types:
- Curved Surface Area (CSA):
The area of only the curved portion of a solid.
- Total Surface Area (TSA):
The sum of the curved surface area and the areas of all plane faces (bases, top, etc.).
Volume
Right Circular Solid
Cuboid
A cuboid is a solid bounded by six rectangular faces, where opposite faces are equal and parallel.
Let:
\(\text{Length }=l\)
\(\text{Breadth }=b\)
\(\text{Height }=h\)
Formulae
- Curved Surface Area
(CSA):
\[\boxed{\;\boldsymbol{2h(l+b)}\;}\] - Total Surface Area
(TSA):
\[\boxed{\;\boldsymbol{2(lb+bh+hl)}\;}\] - Volume:
\[\boxed{\;\boldsymbol{l\times b\times h}\;}\]
Cube
A cube is a special cuboid in which all edges are equal.
Let side=\(a\)
Formulae
- Curved Surface Area
(CSA):
\[\boxed{\;\boldsymbol{6a^2}\;}\] - Total Surface Area
(TSA):
\[\boxed{\;\boldsymbol{2(lb+bh+hl)}\;}\] - Volume:
\[\boxed{\boldsymbol{\quad a^3\quad}}\]
Right Circular Cylinder
A right circular cylinder is formed by revolving a rectangle about one of its sides.
Let
Radius of base =\(r\)
Height=\(h\)
Formulae
- Curved Surface Area
(CSA):
\[\boxed{\;\boldsymbol{2\pi r h}\;}\] - Total Surface Area
(TSA):
\[\boxed{\boldsymbol{2\pi r(r+h)}}\] - Volume:
\[\boxed{\;\boldsymbol{\pi r^2h}\;}\]
Right Circular Cone
A right circular cone is obtained by revolving a right-angled triangle about one of its perpendicular
sides.
Let
Radius of the base: \(=r\)
Height: \(=h\)
Slant Height: \(l=\sqrt{r^2=h^2}\)
Formulae
- Curved Surface Area
(CSA):
\[\boxed{\;\boldsymbol{\pi r l}\;}\] - Total Surface Area
(TSA):
\[\boxed{\;\boldsymbol{\pi r(r+l)}\;}\] - Volume:
\[\boxed{\;\boldsymbol{\dfrac{1}{3}\pi r^2h}\;}\]
Sphere
A sphere is a perfectly round solid in which every point on the surface is at the same distance from the
centre.
Let Radius: \(=r\)
Formulae
- Surface Area
(TSA):
\[\boxed{\;\boldsymbol{4\pi r^2}\;}\] - Volume:
\[\boxed{\;\boldsymbol{\dfrac{4}{3}\pi r^3}\;}\]
Hemisphere
A hemisphere is exactly half of a sphere.
Let Radius: \(=r\)
Formulae
- Curved Surface Area
(CSA):
\[\boxed{\;\boldsymbol{2\pi r^2}\;}\] - Total Surface Area
(TSA):
\[\boxed{\;\boldsymbol{3\pi r^2}\;}\] - Volume:
\[\boxed{\;\boldsymbol{\dfrac{2}{3}\pi r^3}\;}\]
Important Aspects and Key Observations
Units of Measurement
- Surface area is measured in square units (\(cm^2,\ m^2\)).
- Volume is measured in cubic units (\(cm^3,\ m^3\)).
- Capacity may be expressed in litres, where \[\boxed{\;\boldsymbol{1\ cm^3=1\ mL}\;}\]
Slant Height Concept
For cones, the slant height plays a crucial role in surface area calculations and must be computed using the Pythagorean theorem.
Conversion of Solids
Many problems involve melting, reshaping, or recasting solids.
In such cases:
Composite Solids
Some objects are formed by combining two or more standard solids (e.g., cone on a hemisphere).
For such figures:
- Surface area is calculated by adding only exposed surfaces
- Volume is calculated by adding volumes of all parts
Example-1
Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere (see Fig 12.6). The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour. (Take \(\pi = \frac{22}{7}\))
Solution
TSA of the top = CSA of hemisphere + CSA of
cone
Diameter of hemisphere = 3.5cm
therefore, Radius = 3.5/2 = 1.75cm
Radius of base of core = 1.75
Height of core = Total Height of toy - Radius of hemisphere
h = 5-1.75 = 3.25
Curved surface Area of cone \((CSA_{c})\)
Total Surface Area of toy
$$\begin{aligned}TSA&=CSA_{hs}+CSA_{c}\\ &=19.25+20.33\\ &=39.58\\ &\approx 39.6\ cm^{2}\end{aligned}$$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 \(\pi = \frac{22}{7}\))
Solution
TSA of block = CSA of cube + CSA of hemisphere -
base area of hemisphere
Edge of cube = 5cm
Diameter of hemisphere = 4.2cm therefore Radius of-hemisphere = 4.2/2=2.1
CSA of Cube \((CSA_{c})\)
CSA of hemisphere \((CSA_{h})\)
$$\begin{aligned}CSA_{h}&=2\pi r^{2}\\ &=2\times \dfrac{22}{7}\times 2\cdot 1\times 2\cdot 1\\ &=2\times 22\times 0.3\times 2.1\\ &=13.2\times 2.1\\ &=27.72\end{aligned}$$Base Area of hemisphere \((B_{A})\)
$$\begin{aligned}B_{A}&=\pi r^{2}\\ &=\dfrac{22}{7}\times 2\cdot 1\times 2\cdot 1\\ &=22\times 0.3\times 2.1\\ &=6.6\times 2.1\\ &=13.86\end{aligned}$$TSA of Decorative Block
$$\begin{aligned}TSA&=CSA_{h}+CSA_{c}-B_{A}\\ &=27.72+150-13.86\\ &=163.86\ cm^{2}\end{aligned}$$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 \(\pi = 3.14\))
Solution
Total Height of Rocket = 26cm
Height of come = 6cm
therefore, Height of cylindrical part = 26-6 = 20 cm
Diameter of conical portion = 5 cm
therefore, Radius of conical part = 5/2 = 2.5 cm
Diameter of cylindrical part = 3 cm
Radius of cylindrical part =3/2=1.5 cm
Total Surface Area of come to be painted orange
TSA = CSA of cone + Base Area of cone - Base
Area
of cylinder
Radius of cone = 2.5 cm
height of cone = 6 cm
CSA of Cone \((CSA_{c})\)
Base Area of cone \((B_{A})\)
$$\begin{aligned}B_{A}&=\pi r^{2}\\ &=\dfrac{22}{7}\times 2.5\times 2.5\\ &=\dfrac{22}{7}\times 6.25\\ &=19.642\end{aligned}$$Radius of cylinder = 1.5cm
Base of cylinder \((B_{c})\)
TSA to be painted orange
$$\begin{aligned}TSA=&CSA_{c}+B_{A}-B_{c}\\ &=51.071+19.462-7.071\\ &\approx 63.46\end{aligned}$$
Area to be painted in yellow= CSA of
Cylinder + Area of one
base of cylinder
Radius of cylinder = 1. 5 cm
Height of cylinder = 20cm
CSA of Cylinder
Base Area of Cylinder \((B_{A})\)
$$\begin{aligned}B_{A}&=\pi r^{2}\\ &=3.14\times 1.5\times 1.5\\ &=7.065cm^{2}\end{aligned}$$TSA of Cylinder to be painted in yellow
$$\begin{aligned}TSA&=CSA+BA\\ &=188.4+7.065\\ &=195.465cm^{2}\end{aligned}$$Example-4
Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end (see Fig. 12.9). The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath. (Take \(\pi = \dfrac{22}{7}\))
Solution
Height of Cylinder = 1. 45m = 145cm
Radius of cylinder = 30cm
Total surface Area (TSA) of the bird bath =
CSA of Cylinder + CSA of hemisphere
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 × 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 \(\pi = \dfrac{22}{7}\))
Solution
Base of the shed = \(7\times 15\)
Height of the Cuboidal part= 8m
Diameter of cylindrical part = 7m Radius = 7/2 = 3.5 m
Height of Cylinder = 15m
Volume of Air that shed occupies =
volume of Cuboidal part + volume of half cylinder
Volume of Air is shed is 1128.75 m³
Volume occupied by Machinery = 300m³ and
volume occupied by 20 workers
\(= 20 \times 0.08\ m^3 = 1.6\ m^3\)
Total occupation = \(30071.6 = 301.6\)
Thus Air is present in the shed
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 a 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. (Take \(\pi = \dfrac{22}{7}\))
Solution
Inner diameter of the glass = 5cm
Inner Radius of the glass = 5/2 = 2.5cm
Height of Glass = 10cm
Apparent Capacity of glass = volume of the Cylinder
Volume of Cylinder \(V\)
Actual capacity of Glass = Volume of
cylinder - Volume of hemisphere
Radius of hemisphere = 2.5cm
Volume of hemisphere \(V_{h}\)
Hence, Actual Capacity
$$196.25-32.71=163.54\ cm^{3}$$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 \(\pi = 3.14\))
Solution
Volume of toy = Volume of cone + Volume
of hemisphere
Height of cone = 2cm
Diameter of base = 4cm
Radius of base = 4/2=2 cm
Volume of cone \(V_{c}\)
Volume of hemisphere \(V_{h}\)
$$\begin{aligned}V_{h}&=\dfrac{2}{3}\pi r^{3}\\ &=\dfrac{2}{3}\times 3.14\times 2^{3}\\ &=\dfrac{2}{3}\times 3.14\times 8\\ &=16.746\ cm^{3}\end{aligned}$$Total volume of toy
$$\begin{aligned}&8.373+16.746\\ &=25.12cm^{3}\end{aligned}$$
