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Chapter 12

Surface Areas and Volumes

NCERT solutions and explanations for Class 10 Mathematics Chapter 12 Surface Areas and Volumes — covering surface area and volume of combinations of solids (cone, cylinder, hemisphere, sphere, cube, cuboid), conversion of solids via volume equivalence, frustum of a cone, and representative worked examples from the NCERT textbook.

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Questions

7
Q1

A toy is in the form of a cone of radius 3.53.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.515.5 cm. Find the total surface area of the toy. (Take π=227\pi = \frac{22}{7})

Let the common radius be r=3.5r = 3.5 cm. The height of the hemispherical part is equal to its radius, so height of the hemisphere =r=3.5= r = 3.5 cm. Height of the conical part h=Total height−r=15.5−3.5=12h = \text{Total height} - r = 15.5 - 3.5 = 12 cm. Slant height of the cone l=r2+h2=(3.5)2+122=12.25+144=156.25=12.5l = \sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5 cm. Total surface area of the toy =CSA of cone+CSA of hemisphere=πrl+2πr2=πr(l+2r)=227×3.5×(12.5+2×3.5)=227×72×(12.5+7)=11×19.5=214.5= \text{CSA of cone} + \text{CSA of hemisphere} = \pi r l + 2\pi r^2 = \pi r(l + 2r) = \frac{22}{7} \times 3.5 \times (12.5 + 2 \times 3.5) = \frac{22}{7} \times \frac{7}{2} \times (12.5 + 7) = 11 \times 19.5 = 214.5 cm2^2.
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Q2

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. The entire top is 55 cm in height and the diameter of the top is 3.53.5 cm. Find the area he has to colour. (Take π=227\pi = \frac{22}{7})

Diameter =3.5= 3.5 cm, so radius r=3.52=1.75r = \frac{3.5}{2} = 1.75 cm. The top consists of a hemisphere surmounted on a cone. Height of hemispherical part =r=1.75= r = 1.75 cm. Height of cone h=5−1.75=3.25h = 5 - 1.75 = 3.25 cm. Slant height of cone l=r2+h2=(1.75)2+(3.25)2=3.0625+10.5625=13.625=3.7l = \sqrt{r^2 + h^2} = \sqrt{(1.75)^2 + (3.25)^2} = \sqrt{3.0625 + 10.5625} = \sqrt{13.625} = 3.7 cm (approx.). Area to be coloured =CSA of hemisphere+CSA of cone=2πr2+πrl=πr(2r+l)=227×3.52(2×3.52+3.7)=112×(3.5+3.7)=112×7.2=39.6= \text{CSA of hemisphere} + \text{CSA of cone} = 2\pi r^2 + \pi r l = \pi r(2r + l) = \frac{22}{7} \times \frac{3.5}{2} \left(2 \times \frac{3.5}{2} + 3.7\right) = \frac{11}{2} \times (3.5 + 3.7) = \frac{11}{2} \times 7.2 = 39.6 cm2^2 (approx.).
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Q3

A wooden toy rocket is in the shape of a cone mounted on a cylinder. The height of the entire rocket is 2626 cm, while the height of the conical part is 66 cm. The base of the conical portion has a diameter of 55 cm, while the base diameter of the cylindrical portion is 33 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\pi = 3.14)

For the cone: radius r=52=2.5r = \frac{5}{2} = 2.5 cm, height h=6h = 6 cm. Slant height l=r2+h2=2.52+62=6.25+36=42.25=6.5l = \sqrt{r^2 + h^2} = \sqrt{2.5^2 + 6^2} = \sqrt{6.25 + 36} = \sqrt{42.25} = 6.5 cm. For the cylinder: radius r′=32=1.5r' = \frac{3}{2} = 1.5 cm, height h′=26−6=20h' = 26 - 6 = 20 cm. Orange area: The cone's base overhangs the cylinder, so a ring of area πr2−π(r′)2\pi r^2 - \pi (r')^2 also needs orange. Orange area =CSA of cone+πr2−π(r′)2=π[rl+r2−(r′)2]=3.14×[2.5×6.5+2.52−1.52]=3.14×[16.25+6.25−2.25]=3.14×20.25=63.585= \text{CSA of cone} + \pi r^2 - \pi (r')^2 = \pi [r l + r^2 - (r')^2] = 3.14 \times [2.5 \times 6.5 + 2.5^2 - 1.5^2] = 3.14 \times [16.25 + 6.25 - 2.25] = 3.14 \times 20.25 = 63.585 cm2^2. Yellow area: The cylinder is open at the top (mounted by the cone), so only the CSA and the bottom base are painted yellow. Yellow area =CSA of cylinder+π(r′)2=2πr′h′+π(r′)2=πr′(2h′+r′)=3.14×1.5×(2×20+1.5)=4.71×41.5=195.465= \text{CSA of cylinder} + \pi (r')^2 = 2\pi r' h' + \pi (r')^2 = \pi r'(2h' + r') = 3.14 \times 1.5 \times (2 \times 20 + 1.5) = 4.71 \times 41.5 = 195.465 cm2^2.
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Q4

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 22 cm and the diameter of the base is 44 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\pi = 3.14)

The base diameter is 44 cm, so the common radius r=2r = 2 cm. Height of cone h=2h = 2 cm. Volume of the toy =Volume of hemisphere+Volume of cone=23πr3+13πr2h=13πr2(2r+h)=13×3.14×22×(2×2+2)=13×3.14×4×6=13×75.36=25.12= \text{Volume of hemisphere} + \text{Volume of cone} = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (2r + h) = \frac{1}{3} \times 3.14 \times 2^2 \times (2 \times 2 + 2) = \frac{1}{3} \times 3.14 \times 4 \times 6 = \frac{1}{3} \times 75.36 = 25.12 cm3^3. For the circumscribing cylinder: radius =r=2= r = 2 cm, height =r+h=2+2=4= r + h = 2 + 2 = 4 cm. Volume of cylinder =πr2H=3.14×4×4=50.24= \pi r^2 H = 3.14 \times 4 \times 4 = 50.24 cm3^3. Difference in volumes =50.24−25.12=25.12= 50.24 - 25.12 = 25.12 cm3^3.
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Q5

A vessel is in the form of an inverted cone. Its height is 88 cm and the radius of its top, which is open, is 55 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.50.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

The vessel is a cone with radius r=5r = 5 cm and height h=8h = 8 cm, filled completely with water. Volume of water in the cone =13πr2h=13π×52×8=13π×25×8=2003π= \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \times 5^2 \times 8 = \frac{1}{3}\pi \times 25 \times 8 = \frac{200}{3}\pi cm3^3. When lead shots are dropped, one-fourth of the water overflows: Volume of water that flows out =14×2003π=503π= \frac{1}{4} \times \frac{200}{3}\pi = \frac{50}{3}\pi cm3^3. This outflow volume equals the total volume of all lead shot spheres. Volume of one lead shot (sphere) =43πR3= \frac{4}{3}\pi R^3, where R=0.5R = 0.5 cm. So volume of one shot =43π×(0.5)3=43π×0.125=0.53π=16π= \frac{4}{3}\pi \times (0.5)^3 = \frac{4}{3}\pi \times 0.125 = \frac{0.5}{3}\pi = \frac{1}{6}\pi cm3^3. Let nn be the number of shots. Then: n×16π=503πn \times \frac{1}{6}\pi = \frac{50}{3}\pi. Cancel π\pi: n×16=503  ⟹  n=503×6=100n \times \frac{1}{6} = \frac{50}{3} \implies n = \frac{50}{3} \times 6 = 100. Thus, 100100 lead shots were dropped.
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Q6

A juice seller was serving his customers using glasses. The inner diameter of the cylindrical glass was 55 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 1010 cm, find the apparent capacity of the glass and its actual capacity. (Use π=3.14\pi = 3.14)

The glass is a cylinder with a hemispherical bump at the bottom. Radius r=52=2.5r = \frac{5}{2} = 2.5 cm, height of cylinder h=10h = 10 cm. Apparent capacity (volume of the full cylinder) =πr2h=3.14×(2.5)2×10=3.14×6.25×10=196.25= \pi r^2 h = 3.14 \times (2.5)^2 \times 10 = 3.14 \times 6.25 \times 10 = 196.25 cm3^3. Volume of the hemispherical raised portion =23πr3=23×3.14×(2.5)3=23×3.14×15.625=98.1253=32.71= \frac{2}{3}\pi r^3 = \frac{2}{3} \times 3.14 \times (2.5)^3 = \frac{2}{3} \times 3.14 \times 15.625 = \frac{98.125}{3} = 32.71 cm3^3 (approx.). Actual capacity =Apparent capacity−Volume of hemispherical portion=196.25−32.71=163.54= \text{Apparent capacity} - \text{Volume of hemispherical portion} = 196.25 - 32.71 = 163.54 cm3^3.
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Q7

A metallic bucket, open at the top, is in the shape of a frustum of a cone. The height of the bucket is 2424 cm, the radius of the upper circular end is 1414 cm, and the radius of the lower circular end is 77 cm. Find: (i) the area of the metallic sheet used to make the bucket (ignoring the thickness of the metal), and (ii) the volume of water the bucket can hold. (Take π=227\pi = \frac{22}{7})

(i) The bucket is a frustum of a cone, open at the top. Let upper radius r1=14r_1 = 14 cm, lower radius r2=7r_2 = 7 cm, and height h=24h = 24 cm. Slant height of the frustum: l=h2+(r1−r2)2=242+(14−7)2=576+49=625=25l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{24^2 + (14 - 7)^2} = \sqrt{576 + 49} = \sqrt{625} = 25 cm. The metallic sheet area consists of the curved surface area of the frustum plus the area of the circular base (the lower end): Area of sheet =CSA of frustum+Area of circular base=π(r1+r2)l+πr22=π×(14+7)×25+π×72=π×21×25+π×49=525π+49π=574π= \text{CSA of frustum} + \text{Area of circular base} = \pi (r_1 + r_2) l + \pi r_2^2 = \pi \times (14 + 7) \times 25 + \pi \times 7^2 = \pi \times 21 \times 25 + \pi \times 49 = 525\pi + 49\pi = 574\pi cm2^2. Using π=227\pi = \frac{22}{7}: Area =574×227=82×22=1804= 574 \times \frac{22}{7} = 82 \times 22 = 1804 cm2^2. (ii) Volume of the bucket (frustum) =πh3(r12+r22+r1r2)=π×243(142+72+14×7)=8π×(196+49+98)=8π×343=2744π= \frac{\pi h}{3}(r_1^2 + r_2^2 + r_1 r_2) = \frac{\pi \times 24}{3}(14^2 + 7^2 + 14 \times 7) = 8\pi \times (196 + 49 + 98) = 8\pi \times 343 = 2744\pi cm3^3. Using π=227\pi = \frac{22}{7}: Volume =2744×227=392×22=8624= 2744 \times \frac{22}{7} = 392 \times 22 = 8624 cm3^3. Since 11 litre =1000= 1000 cm3^3, the bucket can hold 8.6248.624 litres.
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