TOPICS

Unit-13(Examples)

Surface Area and Volume

**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.). 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 π = 22/7).

Given that:- Diameter of Toy (Hemisphere and Cone) = 3.5cm Radius(r) = Diameter/2 = 3.5/2 = 35/20 = 7/4cm = 1.75cm Height of Toy (Hemisphere + Cone)= 5cm Total Surface Area(TSA) of Toy = Curved surface area(CSA) of hemisphere + Curved surface area(CSA) of cone Now, the CSA of hemisphere = 1/2 x (4πr^{2}) = 1/2 x 4 x 22/7 x 7/4 x 7/4 = 2 x 22/4 x 7/4 = 77/4cm^{2}= 19.25cm^{2}Also, height of the cone(h) = height of the top - height(radius) of hemispherical part = (5 - 7/4)cm = 3.25cm So, slant height of the cone(l) = √r² + h² = √(1.75)² + (3.25)² = √3.06 + 10.56 = √13.62 = 3.7cm (approx.) Therefore, CSA of the cone = πrl = 22/7 x 7/4 x 3.7 = 11/2 x 3.7 = 40.7/2 = 20.35cm^{2}Total Surface Area(TSA) of Toy = 19.25^{2}+ 20.35cm^{2}= 39.60cm^{2}(approx.)

**Example-2 :-** The decorative block shown in Fig. 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. (Use π = 22/7)

Given that:- Edge or side of cube = 5cm Diameter of hemisphere = 4.2cm Radius of hemisphere = 4.2/2 = 2.1cm Note that the part of the cube where the hemisphere is attached is not included in the surface area. Total surface area of the block = TSA of cube - base area of hemisphere + CSA of hemishere Now TSA of cube = 6 x (edge)^{2}= 6 x 5^{2}= 6 x 25 = 150cm^{2}Also, Base area of hemisphere = πr^{2}= 22/7 x 2.1 x 2.1 = 22 x 0.3 x 2.1 = 22 x 0.63 = 13.86cm^{2}Also, CSA of hemisphere = 2πr^{2}= 2 x 13.86 = 27.72cm^{2}TSA of the block = 150 - 13.86 + 27.72 = 150 + 13.86 = 163.86cm^{2}

**Example-3 :-** A wooden toy rocket is in the shape of a cone mounted on a cylinder, as shown in Fig. 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. (Use π = 3.14)

Let radius of cone = r slant height of cone = l height of cone = h radius of cylinder = r′ height of cylinder = 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 = √6.25 + 36 = √42.25 = 6.5cm 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 + πr^{2}- π(r′)^{2}= π(rl + r^{2}- r′^{2}) = 3.14{2.5 x 6.5 + (2.5)^{2}- (1.5)^{2}} = 3.14(16.25 + 6.25 - 2.25) = 3.14 x 20.25 = 63.585cm^{2}Now, area painted yellow = CSA of the cylinder + area of one base of the cylinder = 2πr′h′ + π(r′)^{2}= πr′(2h′ + r′) = 3.14 x 1.5(2 x 20 + 1.5) = 4.710 x 41.5 = 195.465cm^{2}

**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.). The height of the cylinder is 1.45 m and its radius is 30 cm. Find the toal surface area of the bird-bath. (Use π = 22/7)

Let h be height of the cylinder, and r the common radius of the cylinder and hemisphere. Given that:- the common radius of the cylinder and hemisphere(r) = 30cm height of the cylinder = 1.45m = 145cm Then, the total surface area of the bird-bath = CSA of cylinder + CSA of hemisphere = 2πrh + 2πr^{2}= 2πr(h + r) = 2 x 22/7 x 30(145 + 30) = 44/7 x 30 x 175 = 33000cm^{2}= 3.3m^{2}(in meter)

**Example-5 :-** Shanta runs an industry in a shed which is in the shape of a cuboid surmounted by a half cylinder (see Fig.). 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 m^{3} , and there are 20 workers, each of whom occupy about 0.08 m^{3} space on an average. Then, how much air is in the shed? (Use π = 22/7)

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, Length of cuboid(l) = 15m bredth of cuboid(b) = 7m height of cuboid(h) = 8m Also, the diameter of half cylinder = 7m and radius(r) = 7/2 m height(h) = 15m So, the required volume of shed = volume of cuboid + 1/2 volume of the cylinder = l x b x h + 1/2 x πr^{2}h = 15 x 7 x 8 + 1/2 x 22/7 x 7/2 x 7/2 x 15 = 15(56 + 77/4) = 15 x 301/4 = 4515/4 = 1128.75m^{3}The total space occupied by the machinery = 300m^{3}And the total space occupied by the workers = 20 x 0.08m^{3}= 1.6m^{3}Therefore, the volume of the air, when there are machinery and workers = 1128.75 - (300.00 + 1.60) = 827.15m^{3}

**Example-6 :-** A juice seller was serving his customers using glasses as shown in Fig. 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. (Use π = 3.14)

Given that:- Inner diameter of glass = 5cm and radius(r) = 5/2 = 2.5cm height of glass = 10 cm The apparent capacity(or volume) of the glass = πr^{2}h = 3.14 x 2.5 x 2.5 x 10 = 196.25cm^{3}But the actual capacity of the glass is less by the volume of the hemisphere at the base of the glass. Now, volume of the hemisphere = 2/3 x πr^{3}= 2/3 x 3.14 x 2.5 x 2.5 = 32.71 cm^{3}So, the actual capacity(or volume) of the glass = apparent capacity(or volume) of the glass - volume of the hemisphere = 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 π = 3.14)

Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere. The diameter BC of base = 4cm The radius BO of the hemisphere(as well as of the cone) = 4/2 = 2cm Height of the cone = 2cm So, volume of toy = volume of hemisphere + volume of cone = 2/3 x πr^{3}+ 1/3 x πr^{2}h = 1/3 x πr^{2}(2r + h) = 1/3 x 3.14 x 4 x [(2 x 2) + 2] = 1/3 x 3.14 x 4 x 6 = 75.36/3 = 25.12cm^{3}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 height is EH = AO + OP = (2 + 2) cm = 4 cm So, the volume required = volume of the right circular cylinder – volume of the toy = πr^{2}h - (2/3 x πr^{3}+ 1/3 x πr^{2}h) = 3.14 x 2 x 2 x 4 - 25.12 = 50.24 - 25.12 = 25.12cm^{3}Hence, the required difference of the two volumes = 25.12cm^{3}

**Example-8 :-** A cone of height 24 cm and radius of base 6 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.

Given that:- Height of the cone(h) = 24cm radius of base(r) = 6cm Now, child reshapes it in the form of a sphere. Since, the volume of clay in the form of the cone and the sphere remains the same, we have Volume of sphere = Volume of cone 4/3 x πr^{3}= 1/3 x πr^{2}h 4 x r = h 4 x r = 24 r = 24/4 r = 6 cm Therefore, the radius of the sphere is 6 cm.

**Example-9 :-** Selvi’s house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (an underground tank) which is in the shape of a cuboid. The sump has dimensions 1.57 m × 1.44 m × 95cm. The overhead tank has its radius 60 cm and height 95 cm. Find the height of the water left in the sump after the overhead tank has been completely filled with water from the sump which had been full. Compare the capacity of the tank with that of the sump. (Use π = 3.14)

Given that: The sump(cuboid) dimensions = l x b x h = 1.57 m × 1.44 m × 95cm = 157cm x 144cm x 95cm Radius of overhead tank (cylinder) = 60cm Height of overhead tank (cylinder) = 95cm According to question, The volume of water in the overhead tank (cylinder) = The volume of the water removed from the sump (cuboid). Now, the volume of water in the overhead tank (cylinder) = πr^{2}h = 3.14 x 60 x 60 x 95 cm^{3}The volume of water in the sump (cuboid) when full = l x b x h = 157 x 144 x 95 cm^{3}Therefore, The volume of water left in the sump after filling the tank = The volume of the water removed from the sump (cuboid) - The volume of water in the overhead tank (cylinder) = [(157 x 144 x 95) - (3.14 x 60 x 60 x 95)] = 2147760 - 1073880 = 1073880cm^{3}= 1.073m^{3}So, the height of the water left in the sump = volume of water left in the sump/(l x b) = 1.073/(1.57 x 1.44) = 1.073/2.260 = 0.475m Also, capacity of tank/capacity of sump = (3.14 x 0.6 x 0.6 x 0.95)/(1.57 × 1.44 × 0.95) = 1.073880/2.147760 = 1/2 (approx.) Therefore, the capacity of the tank is half the capacity of the sump.

**Example-10 :-** A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.

Given that: Diameter of copper rod = 1cm radius of copper rod = 1/2 cm length of copper rod = 8cm The volume of the rod = πr^{2}h = 22/7 x 1/2 x 1/2 x 8 = 22/7 x 2 cm^{3}The length of the new wire of the same volume = 18m = 1800cm If r_{1}is the radius (in cm) of cross-section of the wire, its volume = π x r_{1}^{2}x 1800 = 22/7 x r_{1}^{2}x 1800 Therefore, 22/7 x r_{1}^{2}x 1800 = 22/7 x 2 r_{1}^{2}x 1800 = 2 r_{1}^{2}= 2/1800 r_{1}^{2}= 1/900 r_{1}= 1/30 So, the diameter of the cross section, i.e., the thickness of the wire is 1/15 cm, i.e., 0.67mm (approx.).

**Example-11 :-** A hemispherical tank full of water is emptied by a pipe at the rate of 4 by 3/7 litres per second. How much time will it take to empty half the tank, if it is 3m in diameter? (Take π = 22/7)

Given that: Diameter of hemispherical tank = 3m Radius of hemispherical tank = 3/2 m Volume of hemispherical tank = 2/3 x πr^{3}= 2/3 x 22/7 x 3/2 x 3/2 x 3/2 = 99/14 m^{3}So, the volume of the water to be emptied = 1/2 x 99/14 = 99/28 x 1000 = 99000/28 litres Since, 25/7 litres of water is emptied in = 1 second 99000/28 litres of water is emptied in = 99000/28 x 7/25 = 16.5 minutes

**Example-12 :-** The radii of the ends of a frustum of a cone 45 cm high are 28 cm and 7 cm (see Fig.). Find its volume, the curved surface area and the total suface area. (Take π = 22/7)

We have : r_{1}= 28 cm, r_{2}= 7 cm and the height of frustum (h) = 45 cm. (i) Volume of frustum = 1/3 x πh(r_{1}^{2}+ r_{2}^{2}+ r_{1}r_{2}) = 1/3 x 22/7 x 45 x [(28)^{2}+ (7)^{2}+ (28 x 7)] = 22/7 x 15 x [784 + 49 + 196] = 330/7 x 1029 = 330 x 147 = 48510 cm^{3}

(ii) We have l = √h² + (r_{1}+ r_{2})² = √(45)² + (28 - 7)² = √(45)² + (21)² = √2025 + 441 = √2466 = 49.65 cm So, the curved surface area of the frustum = πl(r_{1}+ r_{2}) = 22/7 x 49.65 x (28 + 7) = 22/7 x 49.64 x 35 = 22 x 49.64 x 5 = 5461.5 cm^{2}

(iii) Total curved surface area of the frustum = πl(r_{1}+ r_{2}) + π(r_{1}^{2}+ r_{2}^{2}) = 5461.5 + [22/7 x (28)^{2}] + [22/7 x (7)^{2}] = 5461.5 + (22/7 x 784) + (22/7 x 49) = 5461.5 + (22 x 112) + (22 x 7) = 5461.5 + 2464 + 154 = 8079.5 cm^{2}

**Example-13 :-** Hanumappa and his wife Gangamma are busy making jaggery out of sugarcane juice. They have processed the sugarcane juice to make the molasses, which is poured into moulds in the shape of a frustum of a cone having the diameters of its two circular faces as 30 cm and 35 cm and the vertical height of the mould is 14 cm (see Fig.). If each cm^{3} of molasses has mass about 1.2 g, find the mass of the molasses that can be poured into each mould. (Take π = 22/7)

We have the diameters of a frustum of a cone two circular faces as 30 cm and 35 cm. Here, The radius of the smaller base (r_{1}) = 30/2 = 15cm, The radius of the larger base (r_{2}) = 35/2 cm, Vertical height of the mould (h) = 14cm The mould is in the shape of a frustum of a cone, the quantity (volume) of molasses that can be poured into it = 1/3 x πh x (r_{1}^{2}+ r_{2}^{2}+ r_{1}r_{2}) = 1/3 x 22/7 x 14 x [(15)^{2}+ (35/2)^{2}+ (15 x 35/2)] = 1/3 x 22 x 2 x [225 + 306.25 + 262.5] = 44/3 x 793.75 = 11641.7 cm^{3}Given that:- 1cm^{3}of molasses has mass 1.2g. So, the mass of the molasses that can be poured into each mould = (11641.7 × 1.2) g = 13970.04g = 13.97kg = 14kg (approx.)

**Example-14 :-** An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet (see Fig.). The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket, where we do not take into account the handle of the bucket. Also, find the volume of water the bucket can hold. (Take π = 22/7)

We have, An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. Here, The diameters of two circular ends of the bucket are 45cm and 25cm. The radius of the larger base (r_{1}) = 45/2 cm The radius of the smaller base (r_{2}) = 25/2 cm The total height of the bucket 40cm which includes the height of the base. So, the height of the frustum of the cone (h) = 40 - 6 = 34cm the cylindrical base is (h_{2}) = 6cm Therefore, the slant height of the frustum, l = √h² + (r_{1}- r_{2})² = √(34)^{2}+ (45/2 - 25/2)² = √1156 + 100 = √1256 = 35.44 cm The area of metallic sheet used = curved surface area of frustum of cone + area of circular base + curved surface area of cylinder = πl(r_{1}+ r_{2}) + πr_{2}^{2}+ 2πr_{2}h_{2}= π[l(r_{1}+ r_{2}) + r_{2}^{2}+ 2r_{2}h_{2}] = 22/7 x [{35.44 x (45/2 + 25/2)} + (12.5)^{2}+ (2 x 12.5 x 6)] = 22/7 x [1240.4 + 156.25 + 150] = 22/7 x 1545.25 = 4860.9 cm^{3}

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