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About Surface Areas and Volumes — Class 10 Olympiad

Find surface area and volume of combined solids, frustum of a cone, and solve conversion problems. This topic is part of the Olympiad Class 10 mathematics syllabus (chapter: Module 10). On this page you can practice 50 questions across three difficulty levels — 20 easy, 20 medium, and 10 hard — each with a visual step-by-step solution, plus a timed 27-question mastery test with a live leaderboard. Worked examples and sample questions from the topic are below.

Surface Areas and Volumes — solved examples for Class 10 Olympiad

Example 1easy

A decorative toy is made by surmounting a conical top on a hemispherical base. The radius of the hemisphere is 3 cm and the height of the cone is 4 cm. Find the total surface area of the toy. (Use π = 3.14)
  1. A)103.62 cm²
  2. B)94.20 cm²
  3. C)113.04 cm²
  4. D)84.78 cm²

Step-by-step solution

  1. Calculate the slant height (l) of the cone:
    l=√(r2+h2)=√(32+42)=√(9+16)=√25=5cml = √(r² + h²) = √(3² + 4²) = √(9 + 16) = √25 = 5 cm
  2. The total surface area (TSA) of the toy is the sum of the curved surface area (CSA) of the cone and the CSA of the hemisphere.
    CSAcone=πrl=3.14×3×5=47.1cm2CSA_cone = πrl = 3.14 × 3 × 5 = 47.1 cm²
  3. CSA of hemisphere:
    CSAhemisphere=2πr2=2×3.14×32=2×3.14×9=56.52cm2CSA_hemisphere = 2πr² = 2 × 3.14 × 3² = 2 × 3.14 × 9 = 56.52 cm²
  4. TSA of toy:
    TSA=CSAcone+CSAhemisphere=47.1+56.52=103.62cm2TSA = CSA_cone + CSA_hemisphere = 47.1 + 56.52 = 103.62 cm²

Answer: 103.62 cm²

Example 2medium

A solid metallic cuboid of dimensions 20 cm × 10 cm × 5 cm has a cylindrical hole of radius 2 cm and height 5 cm drilled through its center, perpendicular to the 20 cm × 10 cm face. What is the total surface area of the remaining solid?
  1. A)(700 - 8π) cm²
  2. B)(700 + 4π) cm²
  3. C)(700 + 12π) cm²
  4. D)(700 + 20π) cm²

Step-by-step solution

  1. The original surface area of the cuboid is 2(20×10 + 10×5 + 20×5) = 2(200 + 50 + 100) = 2(350) = 700 cm².
  2. When the cylindrical hole is drilled, two circular areas (each with radius 2 cm) are removed from the cuboid's surface. The area removed is 2 × π × (2 cm)² = 8π cm².
  3. Simultaneously, the inner curved surface of the cylinder is exposed. The curved surface area of the cylinder (height 5 cm, radius 2 cm) is 2 × π × 2 cm × 5 cm = 20π cm².
  4. The total surface area of the remaining solid is the original cuboid surface area minus the two circular areas, plus the curved surface area of the cylinder: 700 - 8π + 20π = (700 + 12π) cm².

Answer: (700 + 12π) cm²

Example 3hard

A frustum of a cone has top radius 'r' and bottom radius '3r'. If its slant height is twice its height 'h', what is the volume of the frustum?
  1. A)πh(13r²/3)
  2. B)πh(13r²/2)
  3. C)πh(7r²/3)
  4. D)πh(7r²/2)

Step-by-step solution

  1. Let the height of the frustum be 'h', top radius r₁ = r, and bottom radius r₂ = 3r. The slant height l = 2h.
  2. Using the Pythagorean theorem for the frustum: l² = h² + (r₂ - r₁)².
  3. Substitute the given values: (2h)² = h² + (3r - r)². This simplifies to 4h² = h² + (2r)² => 4h² = h² + 4r² => 3h² = 4r².
  4. The volume of a frustum is V = (1/3)πh(r₁² + r₂² + r₁r₂). Substitute r₁ = r and r₂ = 3r: V = (1/3)πh(r² + (3r)² + r(3r)) = (1/3)πh(r² + 9r² + 3r²) = (1/3)πh(13r²) = 13πhr²/3.

Answer: πh(13r²/3)

Practice questions on Surface Areas and Volumes

  1. Q1.easy

    A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its total volume. (Use π = 22/7)
    1. A)205.17 mm³
    2. B)242.00 mm³
    3. C)183.33 mm³
    4. D)291.67 mm³
    Show answer

    Answer: 242.00 mm³

    Hint: First, determine the radius and the height of the cylindrical part by accounting for the hemispherical ends.

  2. Q2.easy

    A solid metallic sphere of radius 12 cm is melted and recast into 'n' identical solid cones, each of radius 3 cm and height 4 cm. What is the value of 'n'?
    1. A)96
    2. B)144
    3. C)192
    4. D)216
    Show answer

    Answer: 192

    Hint: The total volume of the metallic sphere must be equal to the sum of the volumes of 'n' cones.

  3. Q3.easy

    If the radius of a spherical balloon is increased from 7 cm to 14 cm, what is the ratio of its initial surface area to its new surface area?
    1. A)1:1
    2. B)1:2
    3. C)1:3
    4. D)1:4
    Show answer

    Answer: 1:4

    Hint: Recall that the surface area of a sphere is directly proportional to the square of its radius.

  4. Q4.medium

    A large solid metallic sphere of radius R is melted down and recast into N identical smaller solid spheres. If the sum of the total surface areas of these N smaller spheres is 3 times the total surface area of the original large sphere, find the value of N.
    1. A)27
    2. B)9
    3. C)64
    4. D)8
    Show answer

    Answer: 27

    Hint: Relate the radii of the large and small spheres using volume conservation, then use this relationship with the given surface area condition.

  5. Q5.medium

    A solid right circular cone of height 24 cm and base radius 6 cm is cut into two parts by a plane parallel to its base. If the plane is 12 cm above the base, what is the ratio of the volume of the frustum (lower part) to the volume of the smaller cone (upper part)?
    1. A)1:1
    2. B)7:1
    3. C)3:1
    4. D)1:7
    Show answer

    Answer: 7:1

    Hint: Use similar triangles to find the dimensions of the smaller cone, then relate its volume to the original cone's volume.

  6. Q6.medium

    A cylindrical water tank of height 2.5 m and radius 2 m is initially full. Water is drained from the tank through a pipe of diameter 10 cm at a speed of 5 m/s. How long will it take for the water level in the tank to drop by 1 meter?
    1. A)4 minutes
    2. B)4 minutes 30 seconds
    3. C)5 minutes
    4. D)5 minutes 20 seconds
    Show answer

    Answer: 5 minutes 20 seconds

    Hint: Calculate the volume of water to be drained, then determine the flow rate through the pipe to find the time.

  7. Q7.hard

    A solid metallic sphere of radius R is melted and recast into N identical cubes. If the total surface area of the N cubes is equal to the surface area of the original sphere, find the value of N.
    1. A)π/6
    2. B)π
    3. C)6/π
    4. D)3/π
    Show answer

    Answer: π/6

    Hint: Equate the total volume before and after recasting, and then equate the total surface areas. Express the side of the cube in terms of R.

  8. Q8.hard

    A solid toy is in the form of a hemisphere surmounted by a right circular cone. If the radius of the hemisphere is equal to the radius of the cone (r), and the volume of the hemisphere is equal to the volume of the cone, find the ratio of the total surface area of the toy to the surface area of the hemisphere.
    1. A) (1 + √5)/2
    2. B) (2 + √5)/2
    3. C) (1 + √3)/2
    4. D) (2 + √3)/2
    Show answer

    Answer: (2 + √5)/2

    Hint: First, use the volume equality to find a relationship between the cone's height and its radius. Then, calculate the total surface area of the toy and the curved surface area of the hemisphere.

  9. Q9.hard

    Water flows at a rate of 10 m³/minute from a pipe into a conical tank whose base diameter is 20 m and depth is 24 m. How long will it take to fill the tank to a depth of 18 m?
    1. A)16.875 minutes
    2. B)18.75 minutes
    3. C)22.5 minutes
    4. D)25 minutes
    Show answer

    Answer: 16.875 minutes

    Hint: Use similar triangles to find the radius of the water surface at a given depth. The volume of water is proportional to the cube of the height. Note: For numerical options, assume the given rate is a 'π-adjusted' rate if needed.

These are 9 of the 50 questions available for Surface Areas and Volumes. Start practicing above to unlock visual solutions, AI coaching, and the topic leaderboard — completely free.