Class 9 Maths - Surface Areas and Volumes NCERT Solutions

Chapter 13: Surface Areas and Volumes (NCERT Solutions)

Exercise 13.1 — Cuboid & Cube

Q1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Find the cost of sheet required for making the box if 1 m² of sheet costs &rupee; 20.

l = 1.5 m, b = 1.25 m, h = 65 cm = 0.65 m.
The box is open at the top, so the total sheet area = LSA + base area.
LSA = 2h(l + b) = 2 × 0.65 × (1.5 + 1.25) = 1.3 × 2.75 = 3.575 m².
Base area = l × b = 1.5 × 1.25 = 1.875 m².
Total area = 3.575 + 1.875 = 5.45 m².
Cost = 5.45 × 20 = &rupee; 109.

Q2. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of whitewashing the four walls and ceiling of the room if the cost is &rupee; 7.50 per m².

l = 5m, b = 4m, h = 3m.
Area of 4 walls = 2h(l + b) = 2 × 3 × (5 + 4) = 6 × 9 = 54 m².
Area of ceiling = l × b = 5 × 4 = 20 m².
Total area = 54 + 20 = 74 m².
Cost = 74 × 7.50 = &rupee; 555.

Q3. The floor of a rectangular hall has a perimeter of 250 m. If the cost of painting the four walls at the rate of &rupee; 10 per m² is &rupee; 15000, find the height of the hall.

Perimeter of floor = 2(l + b) = 250 m.
Area of 4 walls = Perimeter × h = 250h.
Cost = 250h × 10 = 15000.
2500h = 15000 ⇒ h = 6 m.

Q4. The paint in a certain container is sufficient to paint an area of 9.375 m². How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?

TSA of one brick = 2(lb + bh + hl) = 2(22.5×10 + 10×7.5 + 7.5×22.5)
= 2(225 + 75 + 168.75) = 2 × 468.75 = 937.5 cm² = 0.09375 m².
Number of bricks = 9.375 / 0.09375 = 100 bricks.

Q5. A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high. Which box has the greater lateral surface area and by how much?

LSA of cube = 4a² = 4 × 100 = 400 cm².
LSA of cuboid = 2h(l + b) = 2 × 8 × (12.5 + 10) = 16 × 22.5 = 360 cm².
The cube has greater LSA by 400 - 360 = 40 cm².

Q6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high. What is the area of the glass? How much tape is needed for all the 12 edges?

TSA = 2(lb + bh + hl) = 2(30×25 + 25×25 + 25×30) = 2(750 + 625 + 750) = 2 × 2125 = 4250 cm².
Total length of 12 edges = 4(l + b + h) = 4(30 + 25 + 25) = 4 × 80 = 320 cm.

Q8. Two cubes each of 10 cm edge are joined end to end. Find the surface area of the resulting cuboid.

When two cubes (side 10 cm each) are joined, the resulting cuboid has:
l = 20 cm, b = 10 cm, h = 10 cm.
TSA = 2(lb + bh + hl) = 2(200 + 100 + 200) = 2 × 500 = 1000 cm².

Exercise 13.2 — Cylinder

Q1. 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.

CSA = 2πrh = 88.
2 × (22/7) × r × 14 = 88.
88r = 88 ⇒ r = 1 cm.
Diameter = 2r = 2 cm.

Q2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required?

r = 70 cm = 0.7 m, h = 1 m.
TSA = 2πr(r + h) = 2 × (22/7) × 0.7 × (0.7 + 1) = 2 × (22/7) × 0.7 × 1.7.
= 2 × 22 × 0.1 × 1.7 = 2 × 3.74 = 7.48 m².

Q5. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².

r = 42 cm, h = 120 cm.
CSA (area per revolution) = 2πrh = 2 × (22/7) × 42 × 120 = 31680 cm².
Area of playground = 500 × 31680 = 15840000 cm² = 1584 m².

Exercise 13.3 — Cone

Q1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

r = 10.5/2 = 5.25 cm, l = 10 cm.
CSA = πrl = (22/7) × 5.25 × 10 = (22/7) × 52.5 = 165 cm².

Q4. A conical tent is 10 m high and the radius of its base is 24 m. Find: (i) slant height of the tent (ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is &rupee; 70.

r = 24 m, h = 10 m.
(i) l = √(r² + h²) = √(576 + 100) = √676 = 26 m.
(ii) CSA = πrl = (22/7) × 24 × 26 = 13728/7 ≈ 1961.14 m².
Cost = 1961.14 × 70 = &rupee; 1,37,280.

Q7. A joker's cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

r = 7 cm, h = 24 cm.
l = √(49 + 576) = √625 = 25 cm.
CSA of 1 cap = πrl = (22/7) × 7 × 25 = 550 cm².
Area for 10 caps = 10 × 550 = 5500 cm².

Exercise 13.4 — Sphere & Hemisphere

Q1. Find the surface area of a sphere of (i) radius 10.5 cm (ii) diameter 14 cm (iii) radius 1 m.

SA = 4πr².

(i) r = 10.5 cm: SA = 4 × (22/7) × 10.5² = 4 × (22/7) × 110.25 = 1386 cm².

(ii) diameter = 14 cm ⇒ r = 7 cm: SA = 4 × (22/7) × 49 = 616 cm².

(iii) r = 1 m: SA = 4 × (22/7) × 1 = 12.57 m².

Q5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of &rupee; 16 per 100 cm².

r = 10.5/2 = 5.25 cm.
Inner CSA of hemisphere = 2πr² = 2 × (22/7) × 5.25² = 2 × (22/7) × 27.5625 = 173.25 cm².
Cost = (173.25 / 100) × 16 = &rupee; 27.72.

Exercise 13.5 to 13.8 — Volumes (Key Questions)

Volume of Cuboid: A wooden box 3 m long, 2 m wide, 0.75 m tall. Find its volume.

Volume = l × b × h = 3 × 2 × 0.75 = 4.5 m³.

Volume of Cylinder (Ex 13.6 Q1): Circumference of base of a cylindrical vessel is 132 cm and height is 25 cm. Find volume.

2πr = 132 ⇒ r = 132 × 7 / (2 × 22) = 21 cm.
Volume = πr²h = (22/7) × 441 × 25 = 34650 cm³.

Volume of Cone (Ex 13.7 Q1): Find the volume of a right circular cone with radius 6 cm, height 7 cm.

Volume = (1/3)πr²h = (1/3) × (22/7) × 36 × 7 = 264 cm³.

Volume of Sphere (Ex 13.8 Q1): Find the volume of a sphere of radius 11.2 cm.

Volume = (4/3)πr³ = (4/3) × (22/7) × (11.2)³
= (4/3) × (22/7) × 1404.928 = 5887.32 cm³.

Ex 13.8 Q5: A hemispherical tank of radius 1.75 m is full of water. It is emptied into a cuboid tank 7 m long, 11 m wide. Find height of water in tank.

Volume of hemisphere = (2/3)πr³ = (2/3) × (22/7) × (1.75)³
= (2/3) × (22/7) × 5.359375 = 11.2292 m³.
Volume of water in cuboid = 7 × 11 × h = 11.2292.
h = 11.2292 / 77 = 0.1458 m ≈ 14.58 cm.

Class 9 Maths - Surface Areas and Volumes Practice

Chapter 13: Surface Areas and Volumes (Practice Questions)

RD Sharma / Extra Practice

Q1. The total surface area of a cube is 1176 cm². Find its volume.

TSA = 6a² = 1176 ⇒ a² = 196 ⇒ a = 14 cm.
Volume = a³ = 14³ = 2744 cm³.

Q2. A cylinder has CSA of 264 cm² and volume of 924 cm³. Find its radius and height.

CSA = 2πrh = 264 ...(1)
Volume = πr²h = 924 ...(2)
Dividing (2) by (1): r/2 = 924/264 = 3.5 ⇒ r = 7 cm.
From (1): 2 × (22/7) × 7 × h = 264 ⇒ 44h = 264 ⇒ h = 6 cm.

Q3. A sphere and a cube have equal volumes. Find the ratio of sphere's radius to cube's side.

(4/3)πr³ = a³
r³/a³ = 3/(4π)
r/a = (3/(4π))^1/3 = (3/4π)^1/3.

Q4. How many spherical balls of radius 1 cm can be made from a solid sphere of radius 8 cm?

Volume of large sphere = (4/3)π(8)³ = (4/3)π × 512.
Volume of small sphere = (4/3)π(1)³ = (4/3)π.
Number = 512 / 1 = 512 balls.

Q5. Find the volume of iron required to make a hollow cylindrical pipe of inner radius 4 cm, outer radius 5 cm, and length 1.4 m.

h = 1.4 m = 140 cm, R = 5 cm, r = 4 cm.
Volume = πh(R² - r²) = (22/7) × 140 × (25 - 16) = (22/7) × 140 × 9 = 3960 cm³.

Q6. The slant height of a cone is 13 cm and radius of base is 5 cm. Find its height and volume.

h = √(l² - r²) = √(169 - 25) = √144 = 12 cm.
Volume = (1/3)πr²h = (1/3) × (22/7) × 25 × 12 = 314.28 cm³.

Q7. If diameter of a sphere is decreased by 25%, by what percent does its surface area decrease?

Original radius = r, so original SA = 4πr².
New diameter = 0.75d, so new radius = 0.75r.
New SA = 4π(0.75r)² = 4π × 0.5625r².
Decrease = 4πr² - 4π × 0.5625r² = 4πr² × 0.4375.
Percentage decrease = 43.75% ≈ 43.75%.

Q8. A tank is 4 m long and 3 m wide. Water is poured into it to a depth of 1.5 m. What is the volume of water in litres?

Volume = 4 × 3 × 1.5 = 18 m³.
1 m³ = 1000 litres.
Volume = 18 × 1000 = 18,000 litres.

Q9. A cone has radius 3 cm and height 4 cm. Find its TSA.

l = √(3² + 4²) = √25 = 5 cm.
TSA = πr(r + l) = (22/7) × 3 × (3 + 5) = (22/7) × 24 = 75.43 cm².

Q10. Find the ratio of the total surface areas of a cube and a sphere if they have equal volumes.

Let a³ = (4/3)πr³ ⇒ a = r(4π/3)^1/3.
TSA of cube = 6a². TSA of sphere = 4πr².
Ratio = 6a² / 4πr² = (6/4π) × (4π/3)^2/3 = π^1/3/(6^1/3) (simplified form).

Q11. The radius and height of a cone are in the ratio 3:4. If its volume is 301.44 cm³, find the radius. (use π = 3.14)

Let r = 3k, h = 4k.
Volume = (1/3) × 3.14 × (3k)² × 4k = (1/3) × 3.14 × 36k³ = 37.68k³ = 301.44.
k³ = 8 ⇒ k = 2.
Radius = 3k = 6 cm.

Q12. A cylinder and a cone have the same base and height. What is the ratio of their volumes?

Volume of cylinder = πr²h.
Volume of cone = (1/3)πr²h.
Ratio = πr²h : (1/3)πr²h = 3 : 1.

Class 9 Maths - Surface Areas and Volumes Summary

Chapter 13: Surface Areas and Volumes (Formulas)

1. Cuboid

LSA = 2h(l + b)
TSA = 2(lb + bh + hl)
Volume = l × b × h
Diagonal = √(l² + b² + h²)

2. Cube (side = a)

LSA = 4a²
TSA = 6a²
Volume = a³
Diagonal = a√3

3. Cylinder (r = radius, h = height)

CSA = 2πrh
TSA = 2πr(r + h)
Volume = πr²h

4. Cone (r = radius, h = height, l = slant height)

Slant height: l = √(r² + h²)
CSA = πrl
TSA = πr(r + l)
Volume = (1/3)πr²h

5. Sphere & Hemisphere

Sphere (r = radius):

Surface Area = 4πr²
Volume = (4/3)πr³

Hemisphere:

CSA = 2πr²
TSA = 3πr²
Volume = (2/3)πr³

6. Key Relations

1 m³ = 1,000 litres = 1,000,000 cm³
1 cm³ = 1 mL
Volume(cylinder) / Volume(cone) = 3 : 1 (same base and height)

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