Chapter 11: Surface Areas and Volumes
Exercise 11.1
Q1
Diameter of the base of a cone is \( 10.5 \text{ cm} \) and its slant height is \( 10 \text{ cm} \). Find its curved surface area. \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
Radius (\( r \)) of the base of cone \( = \frac{10.5}{2} = 5.25 \text{ cm} \)
Slant height (\( l \)) of cone \( = 10 \text{ cm} \)
CSA of cone \( = \pi rl \)
\[ = \left(\frac{22}{7} \times 5.25 \times 10\right) \text{ cm}^{2} = 165 \text{ cm}^{2} \]
Therefore, the curved surface area of the cone is \( 165 \text{ cm}^{2} \).
Q2
Find the total surface area of a cone, if its slant height is \( 21 \text{ m} \) and diameter of its base is \( 24 \text{ m} \). \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
Radius (\( r \)) of the base of cone \( = \frac{24}{2} = 12 \text{ m} \)
Slant height (\( l \)) of cone \( = 21 \text{ m} \)
Total surface area of cone \( = \pi r(r+l) \)
\[ = \left[\frac{22}{7} \times 12 \times (12+21)\right] \text{ m}^{2} \]
\[ = \left(\frac{22}{7} \times 12 \times 33\right) \text{ m}^{2} = 1244.57 \text{ m}^{2} \]
Therefore, the total surface area of the cone is \( 1244.57 \text{ m}^{2} \).
Q3
Curved surface area of a cone is \( 308 \text{ cm}^{2} \) and its slant height is \( 14 \text{ cm} \). Find (i) radius of the base and (ii) total surface area of the cone. \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
(i) Slant height (\( l \)) \( = 14 \text{ cm} \). Let radius be \( r \).
CSA \( = \pi rl \Rightarrow 308 = \frac{22}{7} \times r \times 14 \Rightarrow 308 = 44r \)
\[ \Rightarrow r = \frac{308}{44} = 7 \text{ cm} \]
(ii) Total Surface Area \( = \pi rl + \pi r^2 = 308 + \frac{22}{7} \times (7)^2 = 308 + 154 = 462 \text{ cm}^{2} \).
Q4
A conical tent is \( 10 \text{ m} \) high and the radius of its base is \( 24 \text{ m} \). Find (i) slant height of the tent (ii) cost of the canvas required to make the tent, if the cost of \( 1 \text{ m}^{2} \) canvas is Rs 70. \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
(i) \( h=10 \text{ m}, r=24 \text{ m} \).
\[ l = \sqrt{r^2 + h^2} = \sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26 \text{ m} \]
(ii) CSA \( = \pi rl = \frac{22}{7} \times 24 \times 26 = \frac{13728}{7} \text{ m}^{2} \).
Cost \( = \text{Area} \times \text{Rate} = \frac{13728}{7} \times 70 = 13728 \times 10 = \text{Rs } 137280 \).
Q5
What length of tarpaulin \( 3 \text{ m} \) wide will be required to make conical tent of height \( 8 \text{ m} \) and base radius \( 6 \text{ m} \)? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately \( 20 \text{ cm} \). \( [\text{Use } \pi=3.14] \)
Solution:
\( h=8 \text{ m}, r=6 \text{ m} \). Slant height \( l = \sqrt{6^2 + 8^2} = \sqrt{100} = 10 \text{ m} \).
CSA \( = \pi rl = 3.14 \times 6 \times 10 = 188.4 \text{ m}^{2} \).
Let length be \( L \). Effective area \( = (L - 0.2) \times 3 = 188.4 \).
\[ L - 0.2 = \frac{188.4}{3} = 62.8 \Rightarrow L = 63 \text{ m} \]
Q6
The slant height and base diameter of a conical tomb are \( 25 \text{ m} \) and \( 14 \text{ m} \) respectively. Find the cost of white-washing its curved surface at the rate of Rs 210 per \( 100 \text{ m}^{2} \). \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
\( l=25 \text{ m}, r=7 \text{ m} \).
CSA \( = \pi rl = \frac{22}{7} \times 7 \times 25 = 550 \text{ m}^{2} \).
Cost \( = \frac{550}{100} \times 210 = 5.5 \times 210 = \text{Rs } 1155 \).
Q7
A joker's cap is in the form of right circular cone of base radius \( 7 \text{ cm} \) and height \( 24 \text{ cm} \). Find the area of the sheet required to make 10 such caps. \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
\( r=7 \text{ cm}, h=24 \text{ cm} \). \( l = \sqrt{7^2 + 24^2} = \sqrt{625} = 25 \text{ cm} \).
CSA of 1 cap \( = \pi rl = \frac{22}{7} \times 7 \times 25 = 550 \text{ cm}^{2} \).
Area for 10 caps \( = 10 \times 550 = 5500 \text{ cm}^{2} \).
Q8
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of \( 40 \text{ cm} \) and height \( 1 \text{ m} \). If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per \( \text{m}^{2} \), what will be the cost of painting all these cones? \( (\text{Use } \pi=3.14 \text{ and take } \sqrt{1.04}=1.02) \)
Solution:
\( r=20 \text{ cm} = 0.2 \text{ m}, h=1 \text{ m} \).
\( l = \sqrt{0.2^2 + 1^2} = \sqrt{1.04} = 1.02 \text{ m} \).
CSA of 1 cone \( = \pi rl = 3.14 \times 0.2 \times 1.02 = 0.64056 \text{ m}^{2} \).
CSA of 50 cones \( = 50 \times 0.64056 = 32.028 \text{ m}^{2} \).
Cost \( = 32.028 \times 12 \approx \text{Rs } 384.34 \).
Exercise 11.2
Q1
Find the surface area of a sphere of radius: (i) \( 10.5 \text{ cm} \) (ii) \( 5.6 \text{ cm} \) (iii) \( 14 \text{ cm} \). \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
(i) \( 4\pi r^2 = 4 \times \frac{22}{7} \times (10.5)^2 = 1386 \text{ cm}^{2} \).
(ii) \( 4\pi r^2 = 4 \times \frac{22}{7} \times (5.6)^2 = 394.24 \text{ cm}^{2} \).
(iii) \( 4\pi r^2 = 4 \times \frac{22}{7} \times (14)^2 = 2464 \text{ cm}^{2} \).
Q2
Find the surface area of a sphere of diameter: (i) \( 14 \text{ cm} \) (ii) \( 21 \text{ cm} \) (iii) \( 3.5 \text{ m} \). \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
(i) \( r=7 \text{ cm} \). SA \( = 4 \times \frac{22}{7} \times 7^2 = 616 \text{ cm}^{2} \).
(ii) \( r=10.5 \text{ cm} \). SA \( = 4 \times \frac{22}{7} \times (10.5)^2 = 1386 \text{ cm}^{2} \).
(iii) \( r=1.75 \text{ m} \). SA \( = 4 \times \frac{22}{7} \times (1.75)^2 = 38.5 \text{ m}^{2} \).
Q3
Find the total surface area of a hemisphere of radius \( 10 \text{ cm} \). \( [\text{Use } \pi=3.14] \)
Solution:
TSA of hemisphere \( = 3\pi r^2 = 3 \times 3.14 \times 10^2 = 3 \times 314 = 942 \text{ cm}^{2} \).
Q4
The radius of a spherical balloon increases from \( 7 \text{ cm} \) to \( 14 \text{ cm} \) as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Solution:
Ratio \( = \frac{4\pi (7)^2}{4\pi (14)^2} = \left(\frac{7}{14}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \). Ratio is 1:4.
Q5
A hemispherical bowl made of brass has inner diameter \( 10.5 \text{ cm} \). Find the cost of tin-plating it on the inside at the rate of Rs 16 per \( 100 \text{ cm}^{2} \). \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
\( r=5.25 \text{ cm} \). Inner CSA \( = 2\pi r^2 = 2 \times \frac{22}{7} \times (5.25)^2 = 173.25 \text{ cm}^{2} \).
Cost \( = \frac{173.25}{100} \times 16 = \text{Rs } 27.72 \).
Q6
Find the radius of a sphere whose surface area is \( 154 \text{ cm}^{2} \). \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
\( 4\pi r^2 = 154 \Rightarrow 4 \times \frac{22}{7} \times r^2 = 154 \).
\( r^2 = \frac{154 \times 7}{88} = \frac{49}{4} \Rightarrow r = 3.5 \text{ cm} \).
Q7
The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface area.
Solution:
\( r_{moon} = \frac{1}{4} r_{earth} \).
Ratio \( = \frac{4\pi (r_m)^2}{4\pi (r_e)^2} = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \). Ratio is 1:16.
Q8
A hemispherical bowl is made of steel, \( 0.25 \text{ cm} \) thick. The inner radius of the bowl is \( 5 \text{ cm} \). Find the outer curved surface area of the bowl. \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
Inner radius \( r=5 \). Thickness \( = 0.25 \). Outer radius \( R = 5.25 \text{ cm} \).
Outer CSA \( = 2\pi R^2 = 2 \times \frac{22}{7} \times (5.25)^2 = 173.25 \text{ cm}^{2} \).
Q9
A right circular cylinder just encloses a sphere of radius \( r \). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder, (iii) ratio of the areas obtained in (i) and (ii).
Solution:
(i) SA of sphere \( = 4\pi r^2 \).
(ii) Cylinder radius \( = r \), height \( h = 2r \). CSA \( = 2\pi rh = 2\pi r(2r) = 4\pi r^2 \).
(iii) Ratio \( = 1:1 \).
Exercise 11.3
Q1
Find the volume of the right circular cone with (i) radius \( 6 \text{ cm} \), height \( 7 \text{ cm} \) (ii) radius \( 3.5 \text{ cm} \), height \( 12 \text{ cm} \). \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
(i) \( V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 36 \times 7 = 264 \text{ cm}^{3} \).
(ii) \( V = \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times 12 = 154 \text{ cm}^{3} \).
Q2
Find the capacity in litres of a conical vessel with (i) radius \( 7 \text{ cm} \), slant height \( 25 \text{ cm} \) (ii) height \( 12 \text{ cm} \), slant height \( 13 \text{ cm} \). \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
(i) \( h = \sqrt{25^2 - 7^2} = 24 \text{ cm} \). \( V = \frac{1}{3}\pi (7)^2(24) = 1232 \text{ cm}^{3} = 1.232 \text{ litres} \).
(ii) \( r = \sqrt{13^2 - 12^2} = 5 \text{ cm} \). \( V = \frac{1}{3}\pi (5)^2(12) = \frac{2200}{7} \text{ cm}^{3} = \frac{11}{35} \text{ litres} \).
Q3
The height of a cone is \( 15 \text{ cm} \). If its volume is \( 1570 \text{ cm}^{3} \), find the diameter of its base. \( [\text{Use } \pi=3.14] \)
Solution:
\( \frac{1}{3} \times 3.14 \times r^2 \times 15 = 1570 \Rightarrow 15.7r^2 = 1570 \Rightarrow r^2=100 \Rightarrow r=10 \).
Diameter \( = 20 \text{ cm} \).
Q4
If the volume of a right circular cone of height \( 9 \text{ cm} \) is \( 48\pi \text{ cm}^{3} \), find the diameter of its base.
Solution:
\( \frac{1}{3}\pi r^2(9) = 48\pi \Rightarrow 3r^2 = 48 \Rightarrow r^2=16 \Rightarrow r=4 \).
Diameter \( = 8 \text{ cm} \).
Q5
A conical pit of top diameter \( 3.5 \text{ m} \) is \( 12 \text{ m} \) deep. What is its capacity in kilolitres? \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
\( r=1.75 \text{ m}, h=12 \text{ m} \).
\( V = \frac{1}{3} \times \frac{22}{7} \times (1.75)^2 \times 12 = 38.5 \text{ m}^{3} = 38.5 \text{ kL} \).
Q6
The volume of a right circular cone is \( 9856 \text{ cm}^{3} \). If the diameter of the base is \( 28 \text{ cm} \), find (i) height of the cone (ii) slant height of the cone (iii) curved surface area of the cone. \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
(i) \( r=14 \). \( \frac{1}{3}\pi (14)^2 h = 9856 \Rightarrow h=48 \text{ cm} \).
(ii) \( l = \sqrt{14^2 + 48^2} = 50 \text{ cm} \).
(iii) \( \text{CSA} = \frac{22}{7} \times 14 \times 50 = 2200 \text{ cm}^{2} \).
Q7
A right triangle ABC with sides \( 5 \text{ cm} \), \( 12 \text{ cm} \) and \( 13 \text{ cm} \) is revolved about the side \( 12 \text{ cm} \). Find the volume of the solid so obtained.
Solution:
Revolving about 12 cm means height \( h=12 \) and radius \( r=5 \).
\( V = \frac{1}{3}\pi (5)^2(12) = 100\pi \text{ cm}^{3} \).
Q8
If the triangle ABC in the Question 7 above is revolved about the side \( 5 \text{ cm} \), then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
Solution:
Revolving about 5 cm means height \( h=5 \) and radius \( r=12 \).
\( V = \frac{1}{3}\pi (12)^2(5) = 240\pi \text{ cm}^{3} \).
Ratio \( = \frac{100\pi}{240\pi} = 5:12 \).
Q9
A heap of wheat is in the form of a cone whose diameter is \( 10.5 \text{ m} \) and height is \( 3 \text{ m} \). Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required. \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
\( r=5.25 \text{ m}, h=3 \text{ m} \).
\( V = 86.625 \text{ m}^{3} \).
\( l = 6.05 \text{ m} \). Canvas Area \( = \pi rl = 99.825 \text{ m}^{2} \).
Exercise 11.4
Q1
Find the volume of a sphere whose radius is (i) \( 7 \text{ cm} \) (ii) \( 0.63 \text{ m} \). \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
(i) \( V = \frac{4}{3}\pi (7)^3 = 1437.33 \text{ cm}^{3} \).
(ii) \( V = \frac{4}{3}\pi (0.63)^3 \approx 1.05 \text{ m}^{3} \).
Q2
Find the amount of water displaced by a solid spherical ball of diameter (i) \( 28 \text{ cm} \) (ii) \( 0.21 \text{ m} \). \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
(i) \( r=14 \text{ cm} \). \( V = 11498.66 \text{ cm}^{3} \).
(ii) \( r=0.105 \text{ m} \). \( V = 0.004851 \text{ m}^{3} \).
Q3
The diameter of a metallic ball is \( 4.2 \text{ cm} \). What is the mass of the ball, if the density of the metal is \( 8.9 \text{ g per cm}^{3} \)? \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
\( r=2.1 \text{ cm} \). \( V = 38.808 \text{ cm}^{3} \).
Mass \( = 38.808 \times 8.9 \approx 345.39 \text{ g} \).
Q4
The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Solution:
Ratio of volumes \( = (\frac{1}{4})^3 = \frac{1}{64} \).
Q5
How many litres of milk can a hemispherical bowl of diameter \( 10.5 \text{ cm} \) hold? \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
\( r=5.25 \text{ cm} \). \( V = \frac{2}{3}\pi (5.25)^3 \approx 303 \text{ cm}^{3} = 0.303 \text{ litres} \).
Q6
A hemispherical tank is made up of an iron sheet \( 1 \text{ cm} \) thick. If the inner radius is \( 1 \text{ m} \), then find the volume of the iron used to make the tank. \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
\( r_1=1 \text{ m}, r_2=1.01 \text{ m} \).
\( V = \frac{2}{3}\pi (1.01^3 - 1^3) \approx 0.06348 \text{ m}^{3} \).
Q7
Find the volume of a sphere whose surface area is \( 154 \text{ cm}^{2} \). \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
\( 4\pi r^2 = 154 \Rightarrow r=3.5 \text{ cm} \).
\( V = \frac{4}{3}\pi (3.5)^3 \approx 179.67 \text{ cm}^{3} \).
Q8
A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs 498.96. If the cost of white-washing is Rs 2.00 per square meter, find the (i) inside surface area of the dome, (ii) volume of the air inside the dome. \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
(i) Area \( = 498.96 / 2 = 249.48 \text{ m}^{2} \).
(ii) \( 2\pi r^2 = 249.48 \Rightarrow r=6.3 \text{ m} \). \( V = \frac{2}{3}\pi (6.3)^3 \approx 523.9 \text{ m}^{3} \).
Q9
Twenty seven solid iron spheres, each of radius \( r \) and surface area \( S \) are melted to form a sphere with surface area \( S' \). Find the (i) radius \( r' \) of the new sphere, (ii) ratio of \( S \) and \( S' \).
Solution:
(i) \( \text{Vol}_{new} = 27 \times \text{Vol}_{old} \Rightarrow (r')^3 = 27r^3 \Rightarrow r' = 3r \).
(ii) \( S/S' = r^2 / (3r)^2 = 1/9 \). Ratio is 1:9.
Q10
A capsule of medicine is in the shape of a sphere of diameter \( 3.5 \text{ mm} \). How much medicine (in \( \text{mm}^{3} \)) is needed to fill this capsule? \( [\text{Assume } \pi=\frac{22}{7}] \)
Solution:
\( r=1.75 \text{ mm} \). \( V = \frac{4}{3}\pi (1.75)^3 \approx 22.46 \text{ mm}^{3} \).