Maths · Mensuration · chapter 12 of 14

Class 10 Maths Chapter 12: Surface Areas and Volumes worksheet with answer key

A 20-mark worksheet on Surface Areas and Volumes for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

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Question 1
A solid is a cylinder of radius r and height h with a hemisphere of radius r fixed at each end. Its total surface area is
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Answer: (b) πr(2h + 4r)

TSA = CSA of cylinder + CSA of two hemispheres = 2πrh + 2(2πr²) = πr(2h + 4r). The flat ends are covered, so they are not counted.

Question 2
A toy is made by fixing a cone on a hemisphere so that the join is smooth. The base radius of the cone must be
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Answer: (d) equal to the radius of the hemisphere

For a smooth join the cone's base must exactly cover the hemisphere's flat face, so the two radii are equal.

Question 3
Asha works out the paint needed to cover a toy and writes her answer as 550 cm³. What does the unit tell you?
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Answer: (c) She has found a volume, but paint covers a surface, so the answer must be in cm².

cm³ belongs to volume and cm² to surface area. Paint covers a surface, so the unit shows a volume formula was used by mistake.

Question 4
The volume of a solid hemisphere of radius 21 cm is (Take π = 22/7)
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Answer: (d) 19404 cm³

V = (2/3)πr³ = (2/3) × (22/7) × 9261 = 19404 cm³. 38808 cm³ is the full sphere; 2772 cm² is the curved surface, not a volume.

Question 5
The curved surface area of a hemisphere of radius 14 cm is (Take π = 22/7)
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Answer: (c) 1232 cm²

CSA of a hemisphere = 2πr² = 2 × (22/7) × 196 = 1232 cm². 2464 cm² is the whole sphere; 1848 cm² adds the flat face.

Question 6 · 2 marks
On one face of a cube of edge 21 cm is fixed a hemisphere of diameter 14 cm. Take π = 22/7. Find the total surface area of the solid.
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r = 7 cm. Surface = 6a² − πr² + 2πr² = 6a² + πr² = 2646 + 154 = 2800 cm².

  • subtracts the covered circle and adds the hemisphere's curved surface (1)
  • gets 2800 cm² (1)
Question 7 · 2 marks
A solid made of a cone of radius 3 cm and height 4 cm on a hemisphere of radius 3 cm is lowered fully into a cylindrical vessel of radius 5 cm containing water. By how much does the water level rise?
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Volume of solid = 12π + 18π = 30π cm³. Rise x: π(25)x = 30π, so x = 1.2 cm.

  • finds the solid's volume as 30π cm³ (1)
  • equates it to π(5²)x and gets 1.2 cm (1)
Question 8 · 2 marks
A solid toy is a cone of radius 7 cm and height 7 cm fixed on a hemisphere of the same radius. Take π = 22/7. Find the volume of the toy.
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Volume = (1/3)πr²h + (2/3)πr³ = (1/3)(22/7)(49)(7) + (2/3)(22/7)(343) = 1078/3 + 2156/3 = 1078 cm³.

  • writes volume of toy = volume of cone + volume of hemisphere with both formulas (1)
  • computes 1078 cm³ (1)
Question 9 · 3 marks
A lane-marker float for a swimming pool is a hemisphere of radius 9 cm with a cone on its flat face. The float is 21 cm tall. The cone is painted red and the hemisphere white. Take π = 3.14. Which colour covers the larger area, and by how much?
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Cone height = 21 − 9 = 12 cm, l = √(81 + 144) = 15 cm. Red (cone) = πrl = 3.14 × 9 × 15 = 423.9 cm². White (hemisphere) = 2πr² = 2 × 3.14 × 81 = 508.68 cm². The joined circle is inside and is painted neither colour. White covers more, by 508.68 − 423.9 = 84.78 cm².

  • cone height 12 cm and l = 15 cm (1)
  • red 423.9 cm² and white 508.68 cm², the join not counted (1)
  • white, larger by 84.78 cm² (1)
Question 10 · 3 marks
From a solid cube of edge 14 cm, a cone of radius 7 cm and depth 6 cm is hollowed out of the top face. Take π = 22/7. Find the volume of wood left.
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Cube = 14³ = 2744 cm³. Cone = (1/3)(22/7)(49)(6) = 308 cm³. Left = 2744 − 308 = 2436 cm³.

  • cube volume 2744 cm³ (1)
  • cone volume 308 cm³ (1)
  • subtracts: 2436 cm³ (1)
Question 11 · 3 marks
A paperweight is a cuboid 10 cm × 8 cm × 6 cm with a hemisphere of radius 3.5 cm fixed on its top face. Take π = 22/7. Find the total surface area of the paperweight.
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Cuboid TSA = 2(80 + 48 + 60) = 376 cm². Covered circle πr² = 38.5 cm²; hemisphere CSA 2πr² = 77 cm². Total = 376 − 38.5 + 77 = 414.5 cm².

  • cuboid TSA 376 cm² (1)
  • subtracts πr² and adds 2πr² (1)
  • 414.5 cm² (1)
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