Surface Areas and Volumes CBSE Class 9 Mathematics Worksheets Set 05

Exercise I

Question 1. The dimensions of the drawing room of Ahmad are 8 m × 6 m × 4.5 m. It has one door measuring 3 m × 2 m and two windows each measuring 2 m × 1 m. He wants the interior walls painted at Rs. 8.50 per sq m. How much will Ahmad have to pay?
Answer:
Area of four walls = 2(8 + 6) × 4.5 = 126 m².
Door and windows = 6 + 2 × 2 = 10 m², so the area to paint = 116 m².
Cost = 116 × 8.50 = Rs. 986.
In simple words: Subtract the door and windows from the wall area, then multiply by the rate.

Teacher's Note:
1. The ceiling and floor are not included.

Question 2. A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of two of them are 6 cm and 8 cm, find the edge of the third.
Answer:
Volume of the big cube = 1728 cm³. Two small cubes: 216 + 512 = 728 cm³.
Third cube = 1728 − 728 = 1000 cm³, so its edge is 10 cm.
In simple words: Subtract the two known volumes and take the cube root.

Teacher's Note:
1. Volume is conserved when melting.

Question 3. Find the volume of a rectangular box whose adjacent faces have areas 50 m², 30 m² and 20 m².
Answer:
V² = 50 × 30 × 20 = 30000.
V = √30000 ≈ 173.2 m³.
In simple words: Multiply the three face areas and take the square root.

Teacher's Note:
1. The edges are lb, bh, hl, so their product is V².

Question 4. How many bricks, each 18 cm by 12 cm by 10 cm, are needed to build a wall 15 m long, 6 dm thick and 6.5 m high when 1/10 of its volume is occupied by mortar? Also find the cost of the bricks to the nearest rupee at Rs. 1100 per 1000 bricks.
Answer:
Wall volume = 15 × 0.6 × 6.5 = 58.5 m³. Brick part = 9/10 of it = 52.65 m³ = 52,650,000 cm³.
Volume of one brick = 18 × 12 × 10 = 2160 cm³.
Bricks = 52,650,000 ÷ 2160 = 24375.
Cost = 24375 × 1.10 = 26812.5, which is Rs. 26813 to the nearest rupee.
In simple words: Take 9/10 of the wall volume and divide by one brick's volume.

Teacher's Note:
1. 1 m³ = 1,000,000 cm³.

Question 5. A plate 81 cm × 9 cm × 1 cm is melted and recast into a cube. Find the difference in the surface areas of the two solids.
Answer:
Volume = 729 cm³, so the cube has edge 9 cm and surface area 6 × 81 = 486 cm².
Plate: 2(81 × 9 + 9 × 1 + 1 × 81) = 2(729 + 9 + 81) = 1638 cm².
Difference = 1638 − 486 = 1152 cm².
In simple words: Find the cube's edge from the volume, then compare the two surface areas.

Teacher's Note:
1. The plate has the larger surface area.

Question 6. The cost of painting the four walls of a swimming pool at Rs. 0.70 per m² is Rs. 157.50. The length and breadth of the pool are in the ratio 4 : 1 and its height is 5 m. Find the length and breadth.
Answer:
Area painted = 157.50 ÷ 0.70 = 225 m².
2(l + b) × 5 = 225, so l + b = 22.5 m.
With l : b = 4 : 1, b = 22.5 ÷ 5 = 4.5 m and l = 18 m.
In simple words: Get the wall area from the cost, then use the ratio.

Teacher's Note:
1. Walls only; no floor.

Question 7. The curved surface area of a cylinder is 264 m² and its volume is 924 m³. Find its diameter and height.
Answer:
Divide volume by curved area: πr²h ÷ 2πrh = r/2 = 924 ÷ 264 = 3.5, so r = 7 m.
Then 2 × (22/7) × 7 × h = 264 gives h = 6 m.
Diameter = 14 m.
In simple words: The ratio V : CSA equals r/2.

Teacher's Note:
1. This shortcut avoids solving two equations.

Question 8. The trunk of a tree is cylindrical with circumference 176 cm and length 3 m. Find the volume of timber obtainable.
Answer:
2πr = 176, so r = 28 cm. Length = 300 cm.
V = (22/7) × 28 × 28 × 300 = 739200 cm³ = 0.74 m³ (0.7392).
In simple words: Radius from circumference, then πr²h.

Teacher's Note:
1. Convert cm³ to m³ by dividing by 10⁶.

Question 9. A rectangular sheet 30 cm × 18 cm can be rolled into the curved surface of a cylinder in two ways: along its length or along its breadth. Find the ratio of the volumes of the two cylinders.
Answer:
Rolled so that 30 cm is the circumference: r = 30/(2π), h = 18, V1 = πr²h = 30² × 18/(4π).
Rolled so that 18 cm is the circumference: V2 = 18² × 30/(4π).
V1 : V2 = (30 × 30 × 18) : (18 × 18 × 30) = 30 : 18 = 5 : 3.
In simple words: Volume = (circumference)² × height ÷ 4π.

Teacher's Note:
1. The longer side as circumference gives the bigger volume.

Question 10. The external diameter of a cylindrical iron pipe 25 dm long is 14 cm and the thickness of the iron is 1.75 cm. Find the weight of the pipe if 1 cm³ of iron weighs 10 g.
Answer:
R = 7 cm, r = 7 − 1.75 = 5.25 cm, length = 250 cm.
Volume = (22/7) × 250 × (49 − 27.5625) = (22/7) × 250 × 21.4375 = 16843.75 cm³.
Weight = 16843.75 × 10 g = 168437.5 g = 168.4375 kg.
In simple words: Hollow-cylinder volume times density.

Teacher's Note:
1. Wording in the source is garbled; the data used are external diameter 14 cm, length 25 dm.

Question 11. Two cylindrical vessels of radii 15 cm and 10 cm and heights 35 cm and 15 cm are filled with water. The water is poured into a new cylindrical vessel 25 cm high. Find the radius of the new vessel.
Answer:
Total volume = π(225 × 35 + 100 × 15) = 9375π cm³.
πr² × 25 = 9375π, so r² = 375 and r = √375 ≈ 19.36 cm.
In simple words: Add the volumes and equate with the new cylinder.

Teacher's Note:
1. √375 = 5√15.

Exercise II

Question 12. Two cones have heights in the ratio 1 : 3 and base radii in the ratio 3 : 1. Find the ratio of their volumes.
Answer:
V is proportional to r²h, so V1 : V2 = (9 × 1) : (1 × 3) = 3 : 1.
In simple words: Volume goes with radius squared times height.

Teacher's Note:
1. Square the radius ratio.

Question 13. The area of the base of a cone is 346.5 cm² and its curved surface area is 577.5 cm². Find its volume.
Answer:
πr² = 346.5 gives r² = 110.25 and r = 10.5 cm.
πrl = 577.5 gives l = 577.5 ÷ (22/7 × 10.5) = 17.5 cm.
h = √(306.25 − 110.25) = 14 cm.
V = ⅓ × 346.5 × 14 = 1617 cm³.
In simple words: Find r, then l, then h.

Teacher's Note:
1. CSA / base area = l / r = 5/3.

Question 14. The height of a cone is 5 cm. Find the height of another cone whose volume is sixteen times its volume and whose radius is equal to the diameter of the first cone's base.
Answer:
Let the first radius be r; the second has radius 2r and height h.
⅓π(4r²)h = 16 × ⅓πr² × 5, so 4h = 80 and h = 20 cm.
In simple words: Radius doubles, so its square is 4 times.

Teacher's Note:
1. The source key prints metres; the unit is cm.

Question 15. A military tent is a right circular cone 21 dm high with base diameter 4 m. If 10 men sleep in it, find the average air space per man in dm³.
Answer:
r = 2 m = 20 dm, h = 21 dm.
V = ⅓ × (22/7) × 400 × 21 = 8800 dm³.
Per man = 8800 ÷ 10 = 880 dm³.
In simple words: Convert to dm, find the cone volume and divide by 10.

Teacher's Note:
1. 1 m = 10 dm.

Question 16. A tent is a right circular cylinder surmounted by a cone. The base diameter is 24 m and the height of the cylinder is 11 m. If the vertex of the cone is 16 m above the ground, find the area of canvas needed.
Answer:
r = 12 m, cone height = 16 − 11 = 5 m, slant height = √(144 + 25) = 13 m.
Canvas = curved area of cylinder + cone = 2π × 12 × 11 + π × 12 × 13 = 12π × 35.
= 1320 m².
In simple words: Add the curved surfaces; there is no base or joint.

Teacher's Note:
1. 12, 5, 13 is a Pythagorean triple.

Question 17. The volumes of two spheres are in the ratio 64 : 27. Find their radii if the sum of the radii is 21 cm.
Answer:
Radii ratio = ∛64 : ∛27 = 4 : 3.
Sum 21 cm gives radii 12 cm and 9 cm.
In simple words: Cube roots give the ratio of radii.

Teacher's Note:
1. Volume goes with r³.

Question 18. The internal diameter of a hemispherical tank is 14 m. It already holds 50 m³ of water. How much more water is needed to fill it?
Answer:
r = 7 m. Capacity = ⅔ × (22/7) × 343 = 718.67 m³.
Required = 718.67 − 50 = 668.67 m³.
In simple words: Hemisphere volume minus water already present.

Teacher's Note:
1. Capacity is (2/3)πr³.

Question 19. A sphere of diameter 6 cm is dropped into a cylindrical vessel of diameter 12 cm partly filled with water. If the sphere is completely submerged, by how much does the water level rise?
Answer:
Sphere volume = (4/3)π × 27 = 36π cm³.
Rise: π × 36 × h = 36π, so h = 1 cm.
In simple words: Displaced water equals the sphere's volume.

Teacher's Note:
1. Cylinder radius is 6 cm.

Question 20. Two metallic cones of heights 4.1 cm and 4.3 cm and base radius 2.1 cm each are melted and recast into a sphere. Find the diameter of the sphere.
Answer:
Total volume = ⅓π × 4.41 × (4.1 + 4.3) = ⅓π × 4.41 × 8.4 = 12.348π cm³.
(4/3)πr³ = 12.348π, so r³ = 9.261 and r = 2.1 cm.
Diameter = 4.2 cm.
In simple words: Add the cone volumes and equate with the sphere.

Teacher's Note:
1. 2.1³ = 9.261.

Question 21. A hemisphere of lead of radius 7 cm is melted and recast into a right circular cone of height 49 cm. Find the radius of the cone.
Answer:
⅔π × 343 = ⅓πr² × 49, so r² = 2 × 343 ÷ 49 = 14.
r = √14 ≈ 3.74 cm.
In simple words: Equate the two volumes.

Teacher's Note:
1. 343 = 7 × 49.

Question 22. The largest possible sphere is carved out of a solid cube of side 40 m. Find the surface area and volume of the sphere. (Use π = 3.14.)
Answer:
Diameter = 40 m, so r = 20 m.
Surface area = 4 × 3.14 × 400 = 5024 m².
Volume = (4/3) × 3.14 × 8000 = 33493.33 m³.
In simple words: The sphere's diameter equals the cube's edge.

Teacher's Note:
1. r = 20 m.