RowQ
The Vault
RowQ
The Vault
CBSE Class 9 Maths · 11 questions · 26 marks
Every solid object you meet has two numbers worth knowing: how much wrapping it needs and how much it holds. Surface area answers the first, volume answers the second, and this chapter builds a small toolkit of formulas for cuboids, cylinders, cones and spheres. The real skill is spotting which of the two a word problem is actually asking for.
The total surface area of a cube is 216 cm². Its volume is:
Answer
The volume is 216 cm³. For a cube of edge a, total surface area = 6a². 6a² = 216, so a² = 36 and a = 6 cm. Volume = a³ = 6³ = 216 cm³. (The numerical match with the surface area is a coincidence that happens only at a = 6.)
The curved surface area of a cylinder of radius 7 cm and height 10 cm, taking π = 22/7, is:
Answer
The curved surface area is 440 cm². Curved surface area of a cylinder = 2πrh. = 2 × (22/7) × 7 × 10 The 7 in the radius cancels the 7 in the denominator: = 2 × 22 × 10 = 440 cm².
A cone has base radius 5 cm and slant height 13 cm. Its volume is:
Answer
The volume is 100π cm³. First find the vertical height from l² = r² + h². 13² = 5² + h², so 169 = 25 + h² and h² = 144, giving h = 12 cm. Volume = (1/3)πr²h = (1/3)π(5²)(12) = (1/3)π(25)(12). (1/3)(12) = 4, so volume = 25 × 4 × π = 100π cm³.
If the radius of a sphere is doubled, its surface area becomes:
Answer
The surface area becomes 4 times the original. Surface area of a sphere = 4πr². Replacing r by 2r gives 4π(2r)² = 4π(4r²) = 4 × (4πr²). So the new surface area is 4 times the old one. (Note the volume would become 8 times, since volume depends on r³.)
Assertion (A): A solid hemisphere of radius 7 cm has a total surface area of 462 cm², taking π = 22/7. Reason (R): The total surface area of a solid hemisphere is 3πr², because the curved part contributes 2πr² and the flat circular face contributes πr².
Answer
Both A and R are true and R is the correct explanation of A. Reason R correctly splits the hemisphere's surface into the curved dome, 2πr², and the flat circular base, πr², giving 3πr². Applying it: 3 × (22/7) × 7² = 3 × (22/7) × 49 = 3 × 22 × 7 = 462 cm². So Assertion A is true and it follows directly from the formula in R.
A rectangular water tank measures 2 m by 1.5 m by 1 m. Find its capacity in litres.
Answer
Volume = length × breadth × height = 2 × 1.5 × 1 = 3 m³. Since 1 m³ = 1000 litres, capacity = 3 × 1000 = 3000 litres.
A conical heap of sand has base radius 7 m and vertical height 24 m. Taking π = 22/7, find its slant height, curved surface area and total surface area.
Answer
Slant height l = √(r² + h²) = √(7² + 24²) = √(49 + 576) = √625 = 25 m. Curved surface area = πrl = (22/7)(7)(25) = 22 × 25 = 550 m². Base area = πr² = (22/7)(49) = 154 m². Total surface area = 550 + 154 = 704 m².
The volume of a sphere is 36π cm³. Find its radius and hence its surface area.
Answer
Volume of a sphere = (4/3)πr³. (4/3)πr³ = 36π Dividing both sides by π gives (4/3)r³ = 36, so r³ = 36 × 3/4 = 27 and r = 3 cm. Surface area = 4πr² = 4π(3²) = 36π cm².
A closed cylindrical water tank has internal diameter 2.8 m and height 3.5 m. Taking π = 22/7, find (a) its capacity in litres, (b) the area of metal sheet needed for its curved surface, and (c) the cost of painting only the curved surface at ₹25 per m².
Answer
Step 1: The radius is half the diameter, so r = 2.8/2 = 1.4 m, and h = 3.5 m. (a) Volume = πr²h = (22/7)(1.4)(1.4)(3.5). (22/7)(1.4) = 22 × 0.2 = 4.4. So volume = 4.4 × 1.4 × 3.5 = 4.4 × 4.9 = 21.56 m³. Capacity in litres = 21.56 × 1000 = 21560 litres. (b) Curved surface area = 2πrh = 2 × (22/7) × 1.4 × 3.5. (22/7)(1.4) = 4.4, so the expression is 2 × 4.4 × 3.5 = 30.8 m². (c) Cost = 30.8 × 25 = ₹770.
(a) A solid metal sphere of radius 6 cm is melted and recast into small solid cones, each of base radius 2 cm and vertical height 3 cm. Find how many complete cones are formed. (b) Find the total surface area of one such cone, leaving your answer in terms of π and √13.
Answer
(a) Volume of the sphere = (4/3)πr³ = (4/3)π(6³) = (4/3)π(216). (4/3)(216) = 288, so the sphere's volume is 288π cm³. Volume of one cone = (1/3)πr²h = (1/3)π(2²)(3) = (1/3)π(4)(3) = 4π cm³. Since recasting does not change the total volume, the number of cones = 288π/4π = 72. So 72 complete cones are formed. (b) For one cone, slant height l = √(r² + h²) = √(2² + 3²) = √(4 + 9) = √13 cm. Total surface area = πr(l + r) = π(2)(√13 + 2) = (2√13 + 4)π cm². Numerically, √13 ≈ 3.606, so the area ≈ (7.211 + 4)π ≈ 11.211 × 3.1416 ≈ 35.2 cm².
Read the following and answer the questions that follow: A travelling theatre group pitches a conical tent for its evening shows. The tent has a base diameter of 14 m and a vertical height of 24 m. Canvas for the curved surface costs ₹80 per square metre; the ground is left bare. Take π = 22/7. (a) Find the slant height of the tent. (b) Find the area of canvas required for the curved surface. (c) Find the total cost of the canvas. (d) Find the volume of air enclosed by the tent.
Answer
(a) Radius r = 14/2 = 7 m and height h = 24 m. Slant height l = √(r² + h²) = √(49 + 576) = √625 = 25 m. (b) Canvas area = curved surface area = πrl = (22/7)(7)(25) = 22 × 25 = 550 m². (c) Cost = 550 × 80 = ₹44000. (d) Volume = (1/3)πr²h = (1/3)(22/7)(49)(24). (22/7)(49) = 154, and (1/3)(24) = 8. Volume = 154 × 8 = 1232 m³.
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