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Surface Areas and Volumes Class 9 Maths Worksheet with Answers Chapter 11

Updated:6 Mins Read

We are close to the end of the book. Here we are in the 11th chapter, Surface Areas and Volumes from the NCERT book. The worksheet below is divided into 4 parts – Basic, Standard, Advance, and HOTS. The questions are selected carefully so that you can go from direct formulas to mixed application problems. Complete the chapter first, then attempt the worksheet in sequence to avoid mixing formulas and units. Do attempt only after you have finished the chapter and revised the formula list properly.

  • Cone surface formulas
  • Sphere and hemisphere
  • Volumes of solids
  • Unit conversions practice
  • Real-life applications

Class 9 Maths Worksheet – Chapter 11: Surface Areas and Volumes

Note: Use π = 22/7 unless stated otherwise. Write all answers with correct units (cm², m², cm³, m³).

Basic

  1. Find the curved surface area (CSA) of a cone with radius 7 cm and slant height 10 cm.
  2. Find the total surface area (TSA) of the same cone (radius 7 cm, slant height 10 cm).
  3. A cone has radius 7 cm and height 24 cm. Find its slant height.
  4. Find the surface area of a sphere of radius 7 cm.
  5. Find the curved surface area of a hemisphere of radius 7 cm.
  6. Find the volume of a cone with radius 7 cm and height 12 cm.

Standard

  1. Find the volume of a sphere of radius 10.5 cm.
  2. Find the volume of a hemisphere of radius 21 cm.
  3. A conical tent has radius 7 m and slant height 25 m. Find (i) height of the tent (ii) volume of the tent.
  4. Find the surface area of a sphere of diameter 14 cm.
  5. A right circular cylinder just encloses a sphere of radius 7 cm. Find (i) surface area of the sphere (ii) curved surface area of the cylinder (iii) their ratio.
  6. Find the volume of a cone with radius 3.5 cm and height 12 cm.

Advance

  1. CSA of a cone is 308 cm² and its slant height is 14 cm. Find (i) radius (ii) TSA of the cone.
  2. A conical pit has top diameter 3.5 m and depth 12 m. Find its capacity in kilolitres (kL).
  3. Find the radius of a sphere whose surface area is 154 cm².
  4. A hemispherical bowl has inner diameter 10.5 cm. Find the cost of tin-plating the inside at ₹16 per 100 cm².
  5. A conical tomb has slant height 25 m and base diameter 14 m. Find the cost of white-washing its curved surface at ₹210 per 100 m².
  6. A hemispherical dome has base circumference 17.6 m. Find the cost of painting its curved surface at ₹5 per 100 cm².

HOTS

  1. The radius of a spherical balloon increases from 7 cm to 14 cm. Find the ratio of their surface areas.
  2. 27 solid spheres of radius r are melted to form one sphere of radius r′. Find (i) r′ in terms of r (ii) ratio of their surface areas S : S′.
  3. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold? (Take 1 litre = 1000 cm³)
  4. A medicine capsule is a sphere of diameter 3.5 mm. Find its volume in mm³.
  5. A cylinder and cone have the same base radius and height. If the cylinder volume is 3960 cm³, find the cone volume.
  6. A hemispherical dome is white-washed from inside at a cost of ₹4989.60. If the rate is ₹20 per m², find:
    • (i) inside curved surface area
    • (ii) radius of the dome (use π = 3.14)
    • (iii) volume of air inside the dome (use π = 3.14)

Answer Key (Answers + Hints)

Basic – Answers

  1. Ans: CSA = πrl = (22/7)×7×10 = 220 cm². Hint: Use CSA(cone)=πrl.
  2. Ans: TSA = πr(l+r) = (22/7)×7×(10+7) = 374 cm². Hint: TSA(cone)=πr(l+r).
  3. Ans: l = √(r²+h²) = √(7²+24²) = √(49+576) = √625 = 25 cm. Hint: Pythagoras in cone.
  4. Ans: SA = 4πr² = 4×(22/7)×7² = 616 cm². Hint: Sphere has only curved surface.
  5. Ans: CSA(hemisphere)=2πr² = 2×(22/7)×7² = 308 cm². Hint: Hemisphere curved area is half of sphere.
  6. Ans: V = (1/3)πr²h = (1/3)×(22/7)×7²×12 = 616 cm³. Hint: Cancel 3 with 12.

Standard – Answers

  1. Ans: V = (4/3)πr³ = (4/3)×(22/7)×(10.5)³ = 4851 cm³. Hint: 10.5 = 21/2 makes cancellation clean.
  2. Ans: V(hemisphere) = (2/3)πr³ = (2/3)×(22/7)×21³ = 19404 cm³. Hint: Hemisphere volume is half of sphere.
  3. Ans: l²=r²+h² ⇒ h = √(25²−7²)=√(625−49)=√576=24 m.
    V = (1/3)πr²h = (1/3)×(22/7)×7²×24 = 1232 m³.
    Hint: First find height using slant height.
  4. Ans: Diameter 14 ⇒ r=7, SA = 4πr² = 616 cm². Hint: Convert diameter to radius.
  5. Ans: Sphere SA = 616 cm².
    Cylinder: r=7, height=14 ⇒ CSA = 2πrh = 2×(22/7)×7×14 = 616 cm².
    Ratio = 616 : 616 = 1 : 1.
    Hint: “Just encloses” ⇒ height = diameter = 2r.
  6. Ans: V = (1/3)πr²h = (1/3)×(22/7)×(3.5)²×12 = 154 cm³. Hint: 3.5 = 7/2 helps cancellation.

Advance – Answers

  1. Ans: 308 = πrl = (22/7)×r×14 = 44r ⇒ r = 7 cm.
    TSA = πr(l+r) = (22/7)×7×(14+7) = 462 cm².
    Hint: First find r from CSA.
  2. Ans: r=1.75 m, h=12 m.
    V = (1/3)πr²h = (1/3)×(22/7)×(1.75)²×12 = 38.5 m³ = 38.5 kL.
    Hint: 1 m³ = 1 kL.
  3. Ans: 154 = 4πr² = 4×(22/7)×r² ⇒ r² = 12.25 ⇒ r = 3.5 cm.
    Hint: Solve for r² first.
  4. Ans: r=10.5/2 = 5.25 cm.
    Inside area = CSA(hemisphere)=2πr² = 2×(22/7)×(5.25)² = 173.25 cm².
    Cost = 173.25×(16/100) = ₹27.72.
    Hint: “Inside” means curved surface only.
  5. Ans: r=7 m, l=25 m ⇒ CSA = πrl = (22/7)×7×25 = 550 m².
    Cost = 550×(₹210/100) = ₹1155.
    Hint: Rate is per 100 m².
  6. Ans: 17.6 = 2πr ⇒ r = 17.6×7/44 = 2.8 m.
    CSA = 2πr² = 2×(22/7)×(2.8)² = 49.28 m².
    ₹5 per 100 cm² ⇒ ₹500 per m² ⇒ Cost = 49.28×500 = ₹24640.
    Hint: Convert cm² rate to m² rate.

HOTS – Answers

  1. Ans: SA ratio = 7² : 14² = 49 : 196 = 1 : 4. Hint: Surface area ∝ r².
  2. Ans: Volume conserved ⇒ 27r³ = (r′)³ ⇒ r′ = 3r.
    S : S′ = 4πr² : 4π(3r)² = 1 : 9.
    Hint: Surface area ∝ r², volume ∝ r³.
  3. Ans: r=5.25 cm.
    V = (2/3)πr³ = (2/3)×(22/7)×(5.25)³ = 303.1875 cm³.
    Litres = 303.1875/1000 = 0.303 L.
    Hint: Hemisphere volume is half of sphere volume.
  4. Ans: Diameter 3.5 mm ⇒ r=1.75 mm.
    V = (4/3)πr³ = (4/3)×(22/7)×(1.75)³ = 539/24 = 22.46 mm³.
    Hint: Keep units in mm³.
  5. Ans: Cone volume = (1/3) cylinder volume = 3960/3 = 1320 cm³. Hint: Same r and h ⇒ fixed 1:3 relation.
  6. Ans: (i) Inside curved area = 4989.60/20 = 249.48 m².
    (ii) 249.48 = 2πr² ⇒ r² = 249.48/(2×3.14) = 39.72 ⇒ r = 6.3 m.
    (iii) Volume = (2/3)πr³ = (2/3)×3.14×6.3³ = 523.43 m³.
    Hint: First convert cost to area using rate.

Worksheet for Other Class 9 Maths Chapters

Amit, a BE in Mechanical Engineering, is a math enthusiast dedicated to making math fun and accessible for kids in classes 1 to 10. With a knack for simplifying complex concepts, Amit offers easy-to-understand solutions, fostering a love for math in young minds across India.

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