Surface Areas and Volumes: Class 10 Maths · Chapter 12

Unless a question says otherwise, take $\pi =\frac{22}{7}\mathrm{.}$

1. 1
   NumericalEasyNCERT Exercise 12.1, Q1

   Three wooden blocks shaped like cubes, each of volume $27\mathrm{c}{\mathrm{m}}^3,$ are glued in a row to make a cuboid.

   HintAsk AI

   Find the edge of a cube from its volume first.

   1. (a)

      Find the surface area of the cuboid.

      cm²

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      Show solution

      Edge of each cube $=\sqrt[3]{27}=3\mathrm{c}\mathrm{m},$ so the cuboid is 9 cm by 3 cm by 3 cm.

      $2(27+9+27)=126\mathrm{c}{\mathrm{m}}^2\mathrm{.}$

   2. (b)

      How much less is this than the total surface area of the three separate cubes?

      cm²

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      Three cubes: $3\times 6\times 9=162\mathrm{c}{\mathrm{m}}^2\mathrm{.}$

      $162-126=36\mathrm{c}{\mathrm{m}}^2:$ four hidden faces of $9\mathrm{c}{\mathrm{m}}^2$ each.

2. 2
   NumericalMediumNCERT Exercise 12.1, Q2

   A dairy's milk vat has a hemispherical bottom and straight cylindrical sides of the same radius, and no lid. It is 1.4 m across and 1.9 m deep from the rim to the lowest point. Its inside is to be coated with tin.

   HintAsk AI

   The depth includes the hemisphere's radius. The top is open, so there is no lid.

   1. (a)

      Find the inner surface area.

      m²

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      Show solution

      $r=0.7\mathrm{m};$ cylinder height $h=1.9-0.7=1.2\mathrm{m}\mathrm{.}$

      Inside $=2\pi r^2+2\pi rh=2\pi r(r+h)\mathrm{.}$

      $2\times \frac{22}{7}\times 0.7\times 1.9=8.36{\mathrm{m}}^2\mathrm{.}$

   2. (b)

      Tinning costs ₹ 250 per m². Find the cost.

      ₹

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      $8.36\times 250=2090,$ so ₹ 2090.

3. 3
   NumericalMediumNCERT Exercise 12.1, Q3

   A fishing float is a cone of radius 1.4 cm fixed on a hemisphere of the same radius. The float is 6.2 cm tall in all. Find its total surface area.

   cm²

   Check

   HintAsk AI

   Subtract the hemisphere's radius to get the cone's height.

   Show solution

   Cone height $=6.2-1.4=4.8\mathrm{c}\mathrm{m};$ $l=\sqrt{1.96+23.04}=5\mathrm{c}\mathrm{m}\mathrm{.}$

   TSA $=2\pi r^2+\pi rl=\pi r(2r+l)\mathrm{.}$

   $\frac{22}{7}\times 1.4\times 7.8=4.4\times 7.8=34.32\mathrm{c}{\mathrm{m}}^2\mathrm{.}$

4. 4
   NumericalMediumNCERT Exercise 12.1, Q4

   A block of sandalwood, 14 cm × 10 cm × 8 cm, rests on its 14 cm × 10 cm face. A hemisphere is to be fixed on its top face.

   HintAsk AI

   A circle fits on a rectangle only if its diameter is at most the shorter side.

   1. (a)

      How wide can the hemisphere be, at most?

      cm

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      The top face is 14 cm by 10 cm, so the largest circle on it has diameter 10 cm.

   2. (b)

      With that hemisphere, find the total surface area of the solid.

      cm²

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      Show solution

      Cuboid: $2(140+80+112)=664\mathrm{c}{\mathrm{m}}^2\mathrm{.}$

      The hemisphere adds $\pi r^2=\frac{22}{7}\times 25\approx 78.57\mathrm{c}{\mathrm{m}}^2\mathrm{.}$

      Total $\approx 742.57\mathrm{c}{\mathrm{m}}^2\mathrm{.}$

5. 5
   Short answerHardNCERT Exercise 12.1, Q5

   A wooden sindoor box is a cube of edge $a\mathrm{.}$ A hemispherical hollow of radius $\frac{a}{4}$ is scooped out of the centre of its top face.

   HintAsk AI

   The hollow removes a circle and adds a curved inside.

   1. (a)

      Find the surface area of the box in terms of $a\mathrm{.}$

      Show solution

      Answer$\frac{a^2}{16}(96+\pi )$

      Cube: $6a^2\mathrm{.}$ The hollow removes $\pi (\frac{a}{4}{)}^2$ and adds $2\pi (\frac{a}{4}{)}^2,$ a net $+\frac{\pi a^2}{16}\mathrm{.}$

      Surface area $=6a^2+\frac{\pi a^2}{16}=\frac{a^2}{16}(96+\pi )\mathrm{.}$

   2. (b)

      Evaluate it for $a=8\mathrm{c}\mathrm{m}\mathrm{.}$

      cm²

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      $\frac{64}{16}(96+\frac{22}{7})=4\times 99.14\approx 396.57\mathrm{c}{\mathrm{m}}^2\mathrm{.}$

6. 6
   NumericalEasyNCERT Exercise 12.1, Q6

   A domestic gas cylinder is roughly a cylinder with a hemisphere at each end. It is 70 cm long in all and 28 cm across. Find the area of its outer surface, ignoring the valve and the collar.

   cm²

   Check

   HintAsk AI

   It is a capsule: the two ends together make one sphere.

   Show solution

   $r=14\mathrm{c}\mathrm{m};$ the cylindrical part is $70-28=42\mathrm{c}\mathrm{m}$ long.

   Area $=2\pi r(2r+h)=2\times \frac{22}{7}\times 14\times 70.$

   $=88\times 70=6160\mathrm{c}{\mathrm{m}}^2\mathrm{.}$

7. 7
   NumericalMediumNCERT Exercise 12.1, Q7

   A desert-camp tent is a cylinder 2.5 m high and 7 m across, topped by a cone of slant height 4.5 m. No canvas is used for the floor.

   HintAsk AI

   Count the cylinder's side and the cone's curved surface only.

   1. (a)

      Find the area of canvas needed.

      m²

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      Show solution

      $r=3.5\mathrm{m}\mathrm{.}$ Canvas $=2\pi rh+\pi rl=\pi r(2h+l)\mathrm{.}$

      $\frac{22}{7}\times 3.5\times 9.5=11\times 9.5=104.5{\mathrm{m}}^2\mathrm{.}$

   2. (b)

      Canvas costs ₹ 240 per m². Find the cost.

      ₹

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      $104.5\times 240=25080,$ so ₹ 25 080.

8. 8
   NumericalMediumNCERT Exercise 12.1, Q8

   An incense-stick stand is a solid wooden cylinder of radius 3 cm and height 10 cm. A conical hollow of radius 3 cm and depth 4 cm is cut into its top. Find the total surface area of the stand.

   cm²

   Check

   HintAsk AI

   Count the side, the bottom circle and the sloping inside of the hollow. The top circle has gone.

   Show solution

   Slant height of the hollow: $l=\sqrt{9+16}=5\mathrm{c}\mathrm{m}\mathrm{.}$

   TSA $=2\pi rh+\pi r^2+\pi rl=\pi (60+9+15)=84\pi \mathrm{.}$

   $84\times \frac{22}{7}=264\mathrm{c}{\mathrm{m}}^2\mathrm{.}$

9. 9
   NumericalHardNCERT Exercise 12.1, Q9

   A carpenter turns a solid wooden cylinder 14 cm across and 20 cm tall. From the centre of each flat end he scoops out a hemisphere 7 cm across. What is the total surface area of the finished article?

   cm²

   Check

   HintAsk AI

   Each end now has a flat ring and a curved hollow.

   Show solution

   $R=7\mathrm{c}\mathrm{m},$ $r=3.5\mathrm{c}\mathrm{m}\mathrm{.}$ Side: $2\pi Rh=280\pi \mathrm{.}$

   Two end rings: $2\pi (R^2-r^2)=2\pi (49-12.25)=73.5\pi \mathrm{.}$

   Two hollows: $2\times 2\pi r^2=49\pi \mathrm{.}$

   Total $=402.5\pi =402.5\times \frac{22}{7}=1265\mathrm{c}{\mathrm{m}}^2\mathrm{.}$

10. 10
    Quick answerEasy

    A hemisphere of radius $r$ is fixed on one face of a cube whose edge is more than $2r\mathrm{.}$ The TSA of the solid is the cube's TSA

    Aplus $2\pi r^2$Bplus $\pi r^2$Cminus $\pi r^2$Dplus $3\pi r^2$

    Check

    Show answer

    Answer (B) plus $\pi r^2$

    It hides $\pi r^2$ of the face and adds $2\pi r^2,$ a net $+\pi r^2\mathrm{.}$ Adding $3\pi r^2$ wrongly uses the hemisphere's whole TSA.

    HintAsk AI

    One circle is covered and one curved surface is added.

Unless a question says otherwise, take $\pi =\frac{22}{7}\mathrm{.}$

1. 1
   Short answerMediumNCERT Exercise 12.2, Q1

   A wooden spinning toy has a hemispherical base of radius $r$ and a cone of height $2r$ fixed on its flat face. Take $r=3\mathrm{c}\mathrm{m}\mathrm{.}$

   HintAsk AI

   Write both volumes with $r$ before putting in numbers.

   1. (a)

      Find the volume of the toy, leaving $\pi$ as a symbol.

      Show solution

      Answer$36\pi \mathrm{c}{\mathrm{m}}^3$

      $\frac{2}{3}\pi r^3+\frac{1}{3}\pi r^2(2r)=\frac{4}{3}\pi r^3\mathrm{.}$

      With $r=3:$ $\frac{4}{3}\pi \times 27=36\pi \mathrm{c}{\mathrm{m}}^3\mathrm{.}$

   2. (b)

      Show that this equals the volume of a sphere of radius $r\mathrm{.}$

      Show solution

      $\frac{2}{3}\pi r^3+\frac{2}{3}\pi r^3=\frac{4}{3}\pi r^3,$ the sphere's volume. A cone of height $2r$ has exactly the volume of the missing upper hemisphere.

2. 2
   NumericalMediumNCERT Exercise 12.2, Q2

   A paper Diwali lantern (आकाश कंदील) is a cylinder capped by a cone at each end. It is 28 cm across and 60 cm long in all, and each cone is 12 cm high. Find the volume of air inside it.

   cm³

   Check

   HintAsk AI

   The cylinder's length is what is left after both cones.

   Show solution

   $r=14\mathrm{c}\mathrm{m};$ the cylinder is $60-24=36\mathrm{c}\mathrm{m}$ long.

   Cylinder: $\pi r^2\times 36.$ Two cones: $2\times \frac{1}{3}\pi r^2\times 12=\pi r^2\times 8.$

   $V=\frac{22}{7}\times 196\times 44=27104\mathrm{c}{\mathrm{m}}^3\mathrm{.}$

3. 3
   NumericalMediumNCERT Exercise 12.2, Q3

   Thin Diwali candles are cast in moulds. Each candle is a cylinder of diameter 1.4 cm and length 10 cm with a cone of height 3 cm on top. About 10% extra wax is melted to allow for spills. How much wax must be melted for 50 candles?

   cm³

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   HintAsk AI

   Find one candle's volume, multiply by 50, then add 10%.

   Show solution

   $r=0.7\mathrm{c}\mathrm{m}\mathrm{.}$ One candle: $\pi r^2(10+\frac{3}{3})=\frac{22}{7}\times 0.49\times 11=16.94\mathrm{c}{\mathrm{m}}^3\mathrm{.}$

   50 candles: $847\mathrm{c}{\mathrm{m}}^3\mathrm{.}$ With 10% extra: $847\times 1.1=931.7\mathrm{c}{\mathrm{m}}^3\mathrm{.}$

4. 4
   NumericalMediumNCERT Exercise 12.2, Q4

   A wooden tray for festival laddoos is a cuboid 21 cm × 14 cm × 4 cm. Six hemispherical hollows, each of radius 2.1 cm, are carved in its top. How much wood is left in the tray?

   cm³

   Check

   HintAsk AI

   Subtract six hemispheres from the cuboid.

   Show solution

   Cuboid: $21\times 14\times 4=1176\mathrm{c}{\mathrm{m}}^3\mathrm{.}$

   One hollow: $\frac{2}{3}\times \frac{22}{7}\times {2.1}^3=19.404\mathrm{c}{\mathrm{m}}^3;$ six: $116.424\mathrm{c}{\mathrm{m}}^3\mathrm{.}$

   Wood $=1176-116.424\approx 1059.58\mathrm{c}{\mathrm{m}}^3\mathrm{.}$

5. 5
   NumericalMediumNCERT Exercise 12.2, Q5

   Like the thirsty crow in the old story, Meena drops spherical pebbles of radius 1 cm into a cylindrical pot of radius 4 cm that is partly full of water. All the pebbles sink. How many pebbles make the water rise by 2.5 cm?

   Check

   HintAsk AI

   The water pushed up has the same volume as the pebbles.

   Show solution

   Water pushed up $=\pi \times 4^2\times 2.5=40\pi \mathrm{c}{\mathrm{m}}^3\mathrm{.}$

   One pebble $=\frac{4}{3}\pi \times 1^3=\frac{4}{3}\pi \mathrm{c}{\mathrm{m}}^3\mathrm{.}$

   Number $=40\pi ÷\frac{4}{3}\pi =30.$

6. 6
   NumericalMediumNCERT Exercise 12.2, Q6

   A gym dumbbell is solid iron: two equal discs (cylinders) of radius 6 cm and thickness 3 cm joined by a cylindrical handle of radius 1.5 cm and length 12 cm. The density of iron is about 7.8 g per cm³. Use $\pi =\mathrm{3.14.}$

   HintAsk AI

   Three cylinders: add their volumes, then multiply by the density.

   1. (a)

      Find the volume of iron.

      cm³

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      Show solution

      Discs: $2\times \pi \times 36\times 3=216\pi \mathrm{.}$ Handle: $\pi \times 2.25\times 12=27\pi \mathrm{.}$

      Total $=243\pi =763.02\mathrm{c}{\mathrm{m}}^3\mathrm{.}$

   2. (b)

      Find its mass in kg, to two decimal places.

      kg

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      Show solution

      $763.02\times 7.8=5951.556\mathrm{g}\approx 5.95\mathrm{k}\mathrm{g}\mathrm{.}$

7. 7
   NumericalHardNCERT Exercise 12.2, Q7

   A cylindrical drum of radius 35 cm and height 100 cm is filled to the brim with water. A solid made of a cylinder (radius 21 cm, height 50 cm) on a hemisphere of the same radius is lowered in, rounded end down, until it touches the bottom, and water spills over. How many litres stay in the drum?

   L

   Check

   HintAsk AI

   The water that spills out has the volume of the solid, if the solid is fully under water.

   Show solution

   Drum: $\frac{22}{7}\times {35}^2\times 100=385000\mathrm{c}{\mathrm{m}}^3\mathrm{.}$

   Solid: $\pi r^2(50+\frac{2}{3}\times 21)=\frac{22}{7}\times 441\times 64=88704\mathrm{c}{\mathrm{m}}^3\mathrm{.}$

   It is 71 cm tall, so it is fully under water.

   Left $=296296\mathrm{c}{\mathrm{m}}^3\approx 296.3\mathrm{L}\mathrm{.}$

8. 8
   NumericalMediumNCERT Exercise 12.2, Q8

   A clay surahi has a spherical body of inner diameter 18 cm and a cylindrical neck of inner diameter 4 cm and length 10 cm. The seller says it holds 3.5 litres. Find its capacity in litres, to two decimal places, treating the neck as a separate cylinder on top of the sphere. Is the seller right? Use $\pi =\mathrm{3.14.}$

   L

   Check

   HintAsk AI

   Add a sphere and a cylinder, then divide by 1000.

   Show solution

   Body: $\frac{4}{3}\times 3.14\times 9^3=3052.08\mathrm{c}{\mathrm{m}}^3\mathrm{.}$

   Neck: $3.14\times 2^2\times 10=125.6\mathrm{c}{\mathrm{m}}^3\mathrm{.}$

   Total $=3177.68\mathrm{c}{\mathrm{m}}^3\approx 3.18\mathrm{L}\mathrm{.}$ The claim of 3.5 L is too high.

9. 9
   Quick answerEasy

   A hemispherical hollow of volume $60\mathrm{c}{\mathrm{m}}^3$ is scooped out of a solid cylinder of volume $500\mathrm{c}{\mathrm{m}}^3\mathrm{.}$ The volume of the remaining solid is

   A$560\mathrm{c}{\mathrm{m}}^3$B$500\mathrm{c}{\mathrm{m}}^3$C$440\mathrm{c}{\mathrm{m}}^3$D$380\mathrm{c}{\mathrm{m}}^3$

   Check

   Show answer

   Answer (C) $440\mathrm{c}{\mathrm{m}}^3$

   A hollow subtracts: $500-60=440\mathrm{c}{\mathrm{m}}^3\mathrm{.}$ (The surface area, by contrast, goes up.)

   HintAsk AI

   Does a hollow add space to the solid or take it away?

Unless a question says otherwise, take $\pi =\frac{22}{7}\mathrm{.}$

1. 1
   Quick answerEasy

   Two cubes, each of edge 4 cm, are joined face to face. The total surface area of the cuboid formed is

   A$192\mathrm{c}{\mathrm{m}}^2$B$160\mathrm{c}{\mathrm{m}}^2$C$128\mathrm{c}{\mathrm{m}}^2$D$176\mathrm{c}{\mathrm{m}}^2$

   Check

   Show answer

   Answer (B) $160\mathrm{c}{\mathrm{m}}^2$

   Each cube has $96\mathrm{c}{\mathrm{m}}^2;$ joining hides two faces: $192-32=160\mathrm{c}{\mathrm{m}}^2\mathrm{.}$ 192 forgets the hidden faces.

   HintAsk AI

   Two faces of $16\mathrm{c}{\mathrm{m}}^2$ go inside.

2. 2
   Quick answerEasy

   A cone of radius $r$ and height $k$ stands on a cylinder of the same radius and height $h\mathrm{.}$ The volume of the solid is

   A$\pi r^2(h+k)$B$\pi r^2(h+3k)$C$\pi r^2(h+\frac{k}{3})$D$\frac{1}{3}\pi r^2(h+k)$

   Check

   Show answer

   Answer (C) $\pi r^2(h+\frac{k}{3})$

   $\pi r^{2}h+\frac{1}{3}\pi r^{2}k=\pi r^2(h+\frac{k}{3})\mathrm{.}$

   HintAsk AI

   Add the two volumes and take out $\pi r^2\mathrm{.}$

3. 3
   Quick answerMedium

   From a solid cylinder of radius $r$ and height $2r,$ a hemisphere of radius $r$ is scooped out of each end. The volume left is

   A$\frac{2}{3}\pi r^3$B$\frac{4}{3}\pi r^3$C$2\pi r^3$D$\frac{1}{3}\pi r^3$

   Check

   Show answer

   Answer (A) $\frac{2}{3}\pi r^3$

   $2\pi r^3-\frac{4}{3}\pi r^3=\frac{2}{3}\pi r^3\mathrm{.}$

   HintAsk AI

   Two hemispheres make one sphere.

4. 4
   Quick answerEasy

   Assertion (A): A cone is fixed on a hemisphere of the same radius. The TSA of the solid is the CSA of the cone plus the CSA of the hemisphere. Reason (R): The two flat circular faces that are joined are not on the outside of the solid.

   ABoth A and R are true, and R is the correct explanation of A.BBoth A and R are true, but R is not the correct explanation of A.CA is true but R is false.DA is false but R is true.

   Check

   Show answer

   Answer (A) Both A and R are true, and R is the correct explanation of A.

   Both are true, and the hidden flat faces are exactly why only the two curved surfaces are counted.

   HintAsk AI

   Is R true, and does it explain A?

5. 5
   Quick answerMedium

   Assertion (A): When two solids are joined, the volume of the new solid is less than the sum of their volumes. Reason (R): When two solids are joined, part of their surfaces gets hidden.

   ABoth A and R are true, and R is the correct explanation of A.BBoth A and R are true, but R is not the correct explanation of A.CA is true but R is false.DA is false but R is true.

   Check

   Show answer

   Answer (D) A is false but R is true.

   Volumes add exactly, so A is false. R is true: it is why surface areas do not simply add.

   HintAsk AI

   Does hiding a face remove any space?

6. 6
   Short answerMedium

   A solid is a cylinder of radius 7 cm and height 10 cm with a cone of the same radius and height 24 cm on top.

   HintAsk AI

   The cylinder's bottom circle still shows; its top circle does not.

   1. (a)

      Find its total surface area.

      cm²

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      Show solution

      $l=\sqrt{7^2+{24}^2}=25\mathrm{c}\mathrm{m}\mathrm{.}$

      TSA $=2\pi rh+\pi r^2+\pi rl=\pi (140+49+175)=364\pi \mathrm{.}$

      $364\times \frac{22}{7}=1144\mathrm{c}{\mathrm{m}}^2\mathrm{.}$

   2. (b)

      Find its volume.

      cm³

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      Show solution

      $\pi r^2(h+\frac{k}{3})=\pi \times 49\times 18=882\pi \mathrm{.}$

      $882\times \frac{22}{7}=2772\mathrm{c}{\mathrm{m}}^3\mathrm{.}$

7. 7
   Long answerHard

   A village water tank is a cylinder of radius 3 m and height 7 m with a hemispherical bottom of the same radius and a flat lid. Use $\pi =\mathrm{3.14.}$

   HintAsk AI

   Capacity: cylinder + hemisphere. Paint: the two curved surfaces only.

   1. (a)

      Find its capacity in litres.

      L

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      $V=\pi r^{2}h+\frac{2}{3}\pi r^3=\pi (63+18)=81\pi \mathrm{.}$

      $81\times 3.14=254.34{\mathrm{m}}^3=254340\mathrm{L}\mathrm{.}$

   2. (b)

      The outside of its curved surfaces is painted at ₹ 45 per m². Find the cost.

      ₹

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      Show solution

      Area $=2\pi rh+2\pi r^2=\pi (42+18)=60\pi =188.4{\mathrm{m}}^2\mathrm{.}$

      $188.4\times 45=8478,$ so ₹ 8478.

   3. (c)

      Each household uses 450 L a day. How many households can one full tank supply for a day?

      Check

      Show solution

      $254340÷450=565.2,$ so 565 households.

8. 8
   Long answerMedium

   Case study. A sweet shop serves two round gulab jamuns in a hemispherical steel bowl (कटोरी) of inner radius 4.2 cm and fills the bowl to the brim with syrup. Each gulab jamun is a sphere of radius 1.4 cm.

   HintAsk AI

   Bowl: half a sphere. Syrup: bowl minus the two spheres.

   1. (a)

      Find the capacity of the bowl.

      cm³

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      Show solution

      $\frac{2}{3}\pi r^3=\frac{2}{3}\times \frac{22}{7}\times {4.2}^3=155.232\mathrm{c}{\mathrm{m}}^3\mathrm{.}$

   2. (b)

      How much syrup is in one serving?

      cm³

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      Show solution

      Two gulab jamuns: $2\times \frac{4}{3}\pi (1.4{)}^3\approx 23.00\mathrm{c}{\mathrm{m}}^3\mathrm{.}$

      Syrup $\approx 155.23-23.00=132.23\mathrm{c}{\mathrm{m}}^3\mathrm{.}$

   3. (c)

      The outside of the bowl is engraved. Taking its outer radius as 4.2 cm too, find the area engraved.

      cm²

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      Show solution

      $2\pi r^2=2\times \frac{22}{7}\times 17.64=110.88\mathrm{c}{\mathrm{m}}^2\mathrm{.}$

   4. (d)

      The syrup comes from a cylindrical tin of radius 7 cm and height 20 cm. How many full servings can one tin fill?

      Check

      Show solution

      Tin $=\frac{22}{7}\times 49\times 20=3080\mathrm{c}{\mathrm{m}}^3\mathrm{.}$

      $3080÷132.23\approx 23.3,$ so 23 full servings.