Surface Areas and Volumes — Class 9 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Surface Areas and Volumes" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Surface Areas and Volumes" — 8 important questions with detailed answers for CBSE b…
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NCERT Solutions for Class 9 Mathematics — Surface Areas and Volumes. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Surface Areas and Volumes — All 8 Questions Solved
Q1. Find the total surface area and volume of a cuboid (rectangular prism) with length 12 cm, breadth 8 cm, and height 6 cm.
Step 1: Cuboid dimensions:
Length (l) = 12 cm
Breadth (b) = 8 cm
Height (h) = 6 cm
Step 2: Total Surface Area of Cuboid:
Formula: Total Surface Area = 2(lb + bh + hl)
Step 3: Calculate each product:
lb = 12 × 8 = 96 cm²
bh = 8 × 6 = 48 cm²
hl = 6 × 12 = 72 cm²
Step 4: Sum of products:
lb + bh + hl = 96 + 48 + 72 = 216 cm²
Step 5: Calculate surface area:
Total Surface Area = 2 × 216 = 432 cm²
Step 6: Volume of Cuboid:
Formula: Volume = l × b × h
Volume = 12 × 8 × 6
Volume = 96 × 6
Volume = 576 cm³
Step 7: Verification:
The cuboid has:
- 2 rectangular faces of dimensions 12 × 8 cm
- 2 rectangular faces of dimensions 8 × 6 cm
- 2 rectangular faces of dimensions 6 × 12 cm
Total surface area = 2(96) + 2(48) + 2(72) = 192 + 96 + 144 = 432 cm² ✓
Answer: Total Surface Area = 432 cm²; Volume = 576 cm³.
Q2. A cylinder has radius 7 cm and height 10 cm. Find its curved surface area, total surface area, and volume.
Step 1: Cylinder dimensions:
Radius (r) = 7 cm
Height (h) = 10 cm
Step 2: Curved Surface Area (Lateral Surface Area):
Formula: Curved Surface Area = 2πrh
Curved Surface Area = 2 × π × 7 × 10
Curved Surface Area = 140π cm²
Curved Surface Area ≈ 140 × 3.14159 ≈ 439.82 cm²
Step 3: Area of Two Circular Bases:
Area of one circle = πr²
Area of one circle = π × 7²
Area of one circle = 49π cm²
Area of two circles = 2 × 49π = 98π cm²
Area of two circles ≈ 98 × 3.14159 ≈ 307.88 cm²
Step 4: Total Surface Area:
Total Surface Area = Curved Surface Area + Area of two bases
Total Surface Area = 140π + 98π
Total Surface Area = 238π cm²
Total Surface Area ≈ 238 × 3.14159 ≈ 747.70 cm²
Step 5: Volume of Cylinder:
Formula: Volume = πr²h
Volume = π × 7² × 10
Volume = π × 49 × 10
Volume = 490π cm³
Volume ≈ 490 × 3.14159 ≈ 1539.38 cm³
Answer: Curved Surface Area = 140π cm² ≈ 439.82 cm²; Total Surface Area = 238π cm² ≈ 747.70 cm²; Volume = 490π cm³ ≈ 1539.38 cm³.
Q3. A cone has base radius 5 cm, height 12 cm. Find the slant height, curved surface area, total surface area, and volume.
Step 1: Cone dimensions:
Base radius (r) = 5 cm
Height (h) = 12 cm
Slant height (l) = ?
Step 2: Find slant height using Pythagoras theorem:
The slant height, radius, and height form a right triangle.
l² = r² + h²
l² = 5² + 12²
l² = 25 + 144
l² = 169
l = 13 cm
Step 3: Curved Surface Area (Lateral Surface Area):
Formula: Curved Surface Area = πrl
Curved Surface Area = π × 5 × 13
Curved Surface Area = 65π cm²
Curved Surface Area ≈ 65 × 3.14159 ≈ 204.20 cm²
Step 4: Area of Circular Base:
Area of base = πr²
Area of base = π × 5²
Area of base = 25π cm²
Area of base ≈ 25 × 3.14159 ≈ 78.54 cm²
Step 5: Total Surface Area:
Total Surface Area = Curved Surface Area + Base Area
Total Surface Area = 65π + 25π
Total Surface Area = 90π cm²
Total Surface Area ≈ 90 × 3.14159 ≈ 282.74 cm²
Step 6: Volume of Cone:
Formula: Volume = (1/3)πr²h
Volume = (1/3) × π × 5² × 12
Volume = (1/3) × π × 25 × 12
Volume = (1/3) × 300π
Volume = 100π cm³
Volume ≈ 100 × 3.14159 ≈ 314.16 cm³
Answer: Slant height = 13 cm; Curved Surface Area = 65π cm² ≈ 204.20 cm²; Total Surface Area = 90π cm² ≈ 282.74 cm²; Volume = 100π cm³ ≈ 314.16 cm³.
Q4. A sphere has radius 6 cm. Find its surface area and volume.
Step 1: Sphere radius:
r = 6 cm
Step 2: Surface Area of Sphere:
Formula: Surface Area = 4πr²
Surface Area = 4 × π × 6²
Surface Area = 4 × π × 36
Surface Area = 144π cm²
Surface Area ≈ 144 × 3.14159 ≈ 452.39 cm²
Step 3: Volume of Sphere:
Formula: Volume = (4/3)πr³
Volume = (4/3) × π × 6³
Volume = (4/3) × π × 216
Volume = (4/3 × 216) × π
Volume = (864/3) × π
Volume = 288π cm³
Volume ≈ 288 × 3.14159 ≈ 904.78 cm³
Step 4: Verification:
For r = 6:
4πr² = 4π(6)² = 144π ✓
(4/3)πr³ = (4/3)π(6)³ = (4/3)π(216) = 288π ✓
Answer: Surface Area = 144π cm² ≈ 452.39 cm²; Volume = 288π cm³ ≈ 904.78 cm³.
Q5. A hemisphere has radius 8 cm. Find its curved surface area, total surface area (curved + base), and volume.
Step 1: Hemisphere radius:
r = 8 cm
Step 2: Curved Surface Area (curved part only):
Formula: Curved Surface Area = 2πr²
Curved Surface Area = 2 × π × 8²
Curved Surface Area = 2 × π × 64
Curved Surface Area = 128π cm²
Curved Surface Area ≈ 128 × 3.14159 ≈ 402.12 cm²
Step 3: Area of Circular Base:
Area of base = πr²
Area of base = π × 8²
Area of base = 64π cm²
Area of base ≈ 64 × 3.14159 ≈ 201.06 cm²
Step 4: Total Surface Area:
Total Surface Area = Curved Surface Area + Base Area
Total Surface Area = 128π + 64π
Total Surface Area = 192π cm²
Total Surface Area ≈ 192 × 3.14159 ≈ 603.19 cm²
Step 5: Volume of Hemisphere:
Formula: Volume = (2/3)πr³
Volume = (2/3) × π × 8³
Volume = (2/3) × π × 512
Volume = (2/3 × 512) × π
Volume = (1024/3) × π
Volume ≈ 341.33π cm³
Volume ≈ 341.33 × 3.14159 ≈ 1072.33 cm³
Step 6: Relationship check:
A full sphere with radius 8 has volume (4/3)π(8)³ = (4/3)π(512) = (2048/3)π cm³
A hemisphere is half: (1/3) × (2048/3)π = (1024/3)π cm³ ✓
Answer: Curved Surface Area = 128π cm² ≈ 402.12 cm²; Total Surface Area = 192π cm² ≈ 603.19 cm²; Volume = (1024/3)π cm³ ≈ 1072.33 cm³.
Q6. A cube has side length 5 cm. Find its total surface area, lateral surface area, and volume.
Step 1: Cube side length:
a = 5 cm
Step 2: Total Surface Area of Cube:
A cube has 6 equal square faces.
Formula: Total Surface Area = 6a²
Total Surface Area = 6 × 5²
Total Surface Area = 6 × 25
Total Surface Area = 150 cm²
Step 3: Lateral Surface Area (area of 4 vertical faces):
Formula: Lateral Surface Area = 4a²
Lateral Surface Area = 4 × 5²
Lateral Surface Area = 4 × 25
Lateral Surface Area = 100 cm²
Step 4: Verification:
Total Surface Area = Lateral Surface Area + 2 × (Base Area)
150 = 100 + 2 × 25
150 = 100 + 50
150 = 150 ✓
Step 5: Volume of Cube:
Formula: Volume = a³
Volume = 5³
Volume = 125 cm³
Step 6: Diagonal of the cube:
Body diagonal = a√3 = 5√3 cm ≈ 8.66 cm
(This is the distance from one corner to the opposite corner through the interior.)
Answer: Total Surface Area = 150 cm²; Lateral Surface Area = 100 cm²; Volume = 125 cm³.
Q7. Compare the volumes of a cylinder and a cone with the same base radius 4 cm and height 9 cm. Show the relationship between them.
Step 1: Cylinder with radius 4 cm and height 9 cm:
Volume of cylinder = πr²h
Volume of cylinder = π × 4² × 9
Volume of cylinder = π × 16 × 9
Volume of cylinder = 144π cm³
Step 2: Cone with radius 4 cm and height 9 cm:
Volume of cone = (1/3)πr²h
Volume of cone = (1/3) × π × 4² × 9
Volume of cone = (1/3) × π × 16 × 9
Volume of cone = (1/3) × 144π
Volume of cone = 48π cm³
Step 3: Compare the volumes:
Ratio = Volume of cylinder / Volume of cone
Ratio = 144π / 48π
Ratio = 144 / 48
Ratio = 3
Step 4: Relationship:
Volume of cylinder = 3 × Volume of cone
Or: Volume of cone = (1/3) × Volume of cylinder
Step 5: Physical interpretation:
If you fill a cone with water and pour it into a cylinder with the same base radius and height, it takes exactly 3 cones to fill the cylinder.
Step 6: Numerical values:
Volume of cylinder ≈ 144 × 3.14159 ≈ 452.39 cm³
Volume of cone ≈ 48 × 3.14159 ≈ 150.80 cm³
Ratio ≈ 452.39 / 150.80 ≈ 3 ✓
Answer: Volume of cylinder = 144π cm³ ≈ 452.39 cm³; Volume of cone = 48π cm³ ≈ 150.80 cm³. The cylinder's volume is exactly 3 times the cone's volume.
Q8. A rectangular tank has dimensions 2 m × 1.5 m × 1 m. Find how many liters of water it can hold. (1 m³ = 1000 liters)
Step 1: Tank dimensions (cuboid):
Length (l) = 2 m
Breadth (b) = 1.5 m
Height (h) = 1 m
Step 2: Volume of the tank:
Volume = l × b × h
Volume = 2 × 1.5 × 1
Volume = 3 m³
Step 3: Convert to liters:
Given: 1 m³ = 1000 liters
Capacity of tank = 3 × 1000 = 3000 liters
Step 4: Additional calculations:
Surface area of tank (if we need to know surface area of material used):
Total Surface Area = 2(lb + bh + lh)
Total Surface Area = 2(2×1.5 + 1.5×1 + 1×2)
Total Surface Area = 2(3 + 1.5 + 2)
Total Surface Area = 2(6.5)
Total Surface Area = 13 m²
Step 5: If the tank is filled to 3/4 of its capacity:
Water volume = (3/4) × 3 = 2.25 m³ = 2250 liters
Answer: The tank can hold 3000 liters of water.
Surface Areas and Volumes — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such matchboxes?
- A. 18 cm³
- B. 180 cm³
- C. 36 cm³
- D. 90 cm³
Answer: B. 180 cm³ — The volume of one matchbox is length × breadth × height = 4 × 2.5 × 1.5 = 15 cm³. The volume of a packet containing 12 such matchboxes is 12 × 15 cm³ = 180 cm³.
MCQ 2. The total surface area of a cube is 294 cm². What is the length of its edge?
- A. 6 cm
- B. 7 cm
- C. 8 cm
- D. 9 cm
Answer: B. 7 cm — The total surface area (TSA) of a cube is given by 6a², where 'a' is the edge length. So, 6a² = 294 cm², which means a² = 49 cm², and thus a = 7 cm.
MCQ 3. The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder. (Use π = 22/7)
- A. 1 cm
- B. 2 cm
- C. 3 cm
- D. 4 cm
Answer: B. 2 cm — The curved surface area (CSA) of a cylinder is 2πrh. Given CSA = 88 cm² and h = 14 cm. So, 2 × (22/7) × r × 14 = 88. This simplifies to 88r = 88, so r = 1 cm. The diameter is 2r = 2 × 1 cm = 2 cm.
MCQ 4. A conical tent has a radius of 7 m and a height of 24 m. What is its slant height?
- A. 25 m
- B. 31 m
- C. 17 m
- D. 15 m
Answer: A. 25 m — The slant height (l) of a cone is given by the formula l = √(r² + h²). Substituting r = 7 m and h = 24 m, we get l = √(7² + 24²) = √(49 + 576) = √625 = 25 m.
MCQ 5. Calculate the volume of a sphere whose diameter is 6 cm. (Use π = 22/7)
- A. 36π cm³
- B. 108π cm³
- C. 27π cm³
- D. 18π cm³
Answer: A. 36π cm³ — The diameter is 6 cm, so the radius (r) is 3 cm. The volume of a sphere is (4/3)πr³. Substituting r = 3 cm, we get Volume = (4/3)π(3³) = (4/3)π(27) = 36π cm³.
MCQ 6. The total surface area of a hemisphere of radius 'r' is:
- A. 2πr²
- B. 3πr²
- C. 4πr²
- D. πr²
Answer: B. 3πr² — The total surface area (TSA) of a hemisphere is the sum of its curved surface area (2πr²) and the area of its circular base (πr²), which is 2πr² + πr² = 3πr².
MCQ 7. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of ₹7.50 per m².
- A. ₹555
- B. ₹405
- C. ₹300
- D. ₹450
Answer: A. ₹555 — Area of walls = 2h(l+b) = 2(3)(5+4) = 6(9) = 54 m². Area of ceiling = l×b = 5×4 = 20 m². Total area to whitewash = 54 + 20 = 74 m². Cost = 74 × ₹7.50 = ₹555.
MCQ 8. The capacity of a closed cylindrical tank is 1848 m³. If the diameter of its base is 14 m, what is its depth (height)? (Use π = 22/7)
- A. 12 m
- B. 18 m
- C. 24 m
- D. 36 m
Answer: A. 12 m — The volume of a cylinder is πr²h. Given V = 1848 m³ and diameter = 14 m, so radius (r) = 7 m. Thus, 1848 = (22/7) × 7² × h. This simplifies to 1848 = 22 × 7 × h = 154h. So, h = 1848/154 = 12 m.
MCQ 9. If each edge of a cube is doubled, how many times does its volume increase?
- A. 2 times
- B. 4 times
- C. 6 times
- D. 8 times
Answer: D. 8 times — Let the original edge be 'a'. Original volume = a³. If the edge is doubled, the new edge is 2a. The new volume = (2a)³ = 8a³. Thus, the volume increases 8 times.
MCQ 10. The dimensions of a cuboid are in the ratio 1:2:3 and its total surface area is 88 cm². Find its volume.
- A. 48 cm³
- B. 64 cm³
- C. 96 cm³
- D. 128 cm³
Answer: A. 48 cm³ — Let the dimensions be x, 2x, and 3x. TSA = 2(lb + bh + hl) = 2(x*2x + 2x*3x + 3x*x) = 2(2x² + 6x² + 3x²) = 2(11x²) = 22x². Given TSA = 88 cm², so 22x² = 88, which means x² = 4, and x = 2 cm. The dimensions are 2 cm, 4 cm, 6 cm. Volume = l×b×h = 2 × 4 × 6 = 48 cm³.
MCQ 11. The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)
- A. 5 cm
- B. 10 cm
- C. 15 cm
- D. 20 cm
Answer: B. 10 cm — The volume of a cone is (1/3)πr²h. Given V = 1570 cm³ and h = 15 cm. So, 1570 = (1/3) × 3.14 × r² × 15. This simplifies to 1570 = 3.14 × r² × 5, or 1570 = 15.7 r². Thus, r² = 1570/15.7 = 100, so r = 10 cm.
MCQ 12. Two spheres have radii in the ratio 2:3. What is the ratio of their volumes?
- A. 2:3
- B. 4:9
- C. 8:27
- D. 16:81
Answer: C. 8:27 — The volume of a sphere is proportional to the cube of its radius (V ∝ r³). If the radii are in the ratio r1:r2, then their volumes are in the ratio r1³:r2³. So, (2)³:(3)³ = 8:27.
MCQ 13. What is the curved surface area of a hemisphere with radius 'r'?
- A. πr²
- B. 2πr²
- C. 3πr²
- D. 4πr²
Answer: B. 2πr² — The curved surface area of a hemisphere is exactly half the surface area of a full sphere, which is (1/2) × 4πr² = 2πr².
MCQ 14. Find the total surface area of a cylinder if the radius of its base is 5 cm and its height is 10 cm. (Use π = 3.14)
- A. 314 cm²
- B. 471 cm²
- C. 628 cm²
- D. 157 cm²
Answer: B. 471 cm² — The total surface area (TSA) of a cylinder is 2πr(r+h). Substituting r = 5 cm and h = 10 cm, we get TSA = 2 × 3.14 × 5 × (5+10) = 2 × 3.14 × 5 × 15 = 150 × 3.14 = 471 cm².
MCQ 15. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres? (Use π = 22/7)
- A. 38.5 kl
- B. 77 kl
- C. 19.25 kl
- D. 9.625 kl
Answer: A. 38.5 kl — The radius (r) = diameter/2 = 3.5/2 = 1.75 m. Height (h) = 12 m. Volume of cone = (1/3)πr²h = (1/3) × (22/7) × (1.75)² × 12 = (1/3) × (22/7) × (3.0625) × 12 = 38.5 m³. Since 1 m³ = 1 kilolitre, the capacity is 38.5 kl.
MCQ 16. The volume of a cuboid is 384 cm³. If its length is 12 cm and height is 4 cm, what is its breadth?
- A. 6 cm
- B. 8 cm
- C. 10 cm
- D. 12 cm
Answer: B. 8 cm — The volume of a cuboid is length × breadth × height. Given V = 384 cm³, l = 12 cm, h = 4 cm. So, 384 = 12 × b × 4. This simplifies to 384 = 48b, so b = 384/48 = 8 cm.
MCQ 17. What is the lateral surface area of a cube whose edge is 5 cm?
- A. 25 cm²
- B. 50 cm²
- C. 100 cm²
- D. 125 cm²
Answer: C. 100 cm² — The lateral surface area (LSA) of a cube is the area of its four walls, which is 4a², where 'a' is the edge length. So, LSA = 4 × (5 cm)² = 4 × 25 cm² = 100 cm².
MCQ 18. If the radius of the base of a cylinder is halved, keeping the height same, what is the ratio of the volume of the new cylinder to the original cylinder?
- A. 1:2
- B. 1:4
- C. 2:1
- D. 4:1
Answer: B. 1:4 — Let the original radius be 'r' and height 'h'. Original volume V = πr²h. If the radius is halved, new radius r' = r/2. New volume V' = π(r/2)²h = π(r²/4)h = (1/4)πr²h. The ratio of new volume to original volume is (1/4)πr²h / πr²h = 1/4 or 1:4.
MCQ 19. The surface area of a sphere is 154 cm². Find its volume. (Use π = 22/7)
- A. 179.67 cm³
- B. 1437.33 cm³
- C. 308 cm³
- D. 616 cm³
Answer: A. 179.67 cm³ — Surface area of a sphere = 4πr². Given 4πr² = 154 cm². So, 4 × (22/7) × r² = 154, which gives r² = (154 × 7) / (4 × 22) = 49/4. Thus, r = 7/2 = 3.5 cm. Volume of sphere = (4/3)πr³ = (4/3) × (22/7) × (3.5)³ = (4/3) × (22/7) × 42.875 = 179.67 cm³ (approx).
MCQ 20. Find the curved surface area of a cone whose base radius is 5 cm and slant height is 13 cm. (Use π = 22/7)
- A. 157.14 cm²
- B. 204.28 cm²
- C. 204.14 cm²
- D. 163.57 cm²
Answer: B. 204.28 cm² — The curved surface area (CSA) of a cone is πrl. Substituting r = 5 cm and l = 13 cm, we get CSA = (22/7) × 5 × 13 = 1430/7 = 204.28 cm² (approx).
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.