Chapter-5 (Area And Volume) Exercise Solutions

Area and Volume: Exercise Solutions

Mathematics • Grade 10 • Exercises 5.1 – 5.4 and Mixed Exercise

Notes: π = 22/7 unless stated. Where a figure is not visible in the book text, its dimensions were worked out so that the solution agrees with the textbook answer key. Misprints in the book are pointed out in small print.

Exercise 5.1: Pyramid

1. For the square based pyramid GWXYZ (G apex, O centre of base, M midpoint of XY): (a) What do GO and GM indicate? (b) If XY = a cm, what is OM? (c) Write the relation of vertical height h, slant height l and side of base a. (d) What do GZ, GY, GX, GW represent? Are they equal?

(a) GO = vertical height, GM = slant height.
(b) OM = half of XY = a/2 cm.
(c) In right-angled triangle GOM: l² = h² + (a/2)².
(d) They are the lateral edges of the pyramid. Yes, all four are equal in a right pyramid with a square base.

2. (a) Total surface area of a square based pyramid with base side m cm and slant height n cm. (b) Lateral surface area with base side p cm and slant height q cm. (c) Volume in terms of r and h if base side is r cm and height h cm. (d) Area of the triangular surfaces if base area is z and total surface area is y square units.

(a) TSA = a² + 2al = (m² + 2mn) cm²
(b) LSA = 2al = 2pq cm²
(c) V = ⅓a²h = ⅓r²h cm³
(d) Triangular surfaces = TSA − base = (y − z) sq. units

3. (a) Volume of a square based pyramid with base area 64 cm² and height 15 cm. (b) TSA = 285 cm², area of triangular surfaces = 192 cm²; find base area. (c) Find (i) slant height and edge if PO = 8 cm, OM = 12 cm; (ii) vertical height and edge if SA = 13 cm, NO = 10 cm.

(a) V = ⅓ × 64 × 15 = 320 cm³
(b) Base area = TSA − LSA = 285 − 192 = 93 cm²
(c)(i) h = PO = 8, OM = a/2 = 12 ⇒ a = 24.
l = √(8² + 12²) = √208 = 4√13 cm. Half diagonal = 12√2, so e = √(8² + 288) = √352 = 4√22 cm.
(ii) Slant height SA = 13, base side NO = 10 ⇒ a/2 = 5.
h = √(13² − 5²) = √144 = 12 cm. e = √(l² + (a/2)²) = √(169 + 25) = √194 cm.

4. Find the total surface area and volume of each square based pyramid.

(a) a = 12, l = 10: h = √(100 − 36) = 8. TSA = 144 + 2 × 12 × 10 = 384 cm²; V = ⅓ × 144 × 8 = 384 cm³
(b) a = 10, h = 12: l = √(144 + 25) = 13. TSA = 100 + 260 = 360 cm²; V = ⅓ × 100 × 12 = 400 cm³
(c) a = 10, l = 13: h = √(169 − 25) = 12. TSA = 360 cm²; V = 400 cm³
(d) a = 12, h = 8: l = 10. TSA = 384 cm²; V = 384 cm³
(e) a = 16, h = 6: l = √(36 + 64) = 10. TSA = 256 + 320 = 576 cm²; V = ⅓ × 256 × 6 = 512 cm³
(f) a = 16, l = 17: h = √(289 − 64) = 15. TSA = 256 + 544 = 800 cm²; V = ⅓ × 256 × 15 = 1280 cm³

5. Sophiya has a pot shaped like a square based pyramid, height 25 cm and base length 30 cm. How much water can it contain?

V = ⅓ × 30² × 25 = 7500 cm³ = 7500 ÷ 1000 = 7.5 litres

6. TSA of a square based pyramid is 800 cm², side of base 16 cm. Find the area of triangular surfaces and the volume.

Triangular surfaces = 800 − 256 = 544 cm². 2 × 16 × l = 544 ⇒ l = 17. h = √(289 − 64) = 15. V = ⅓ × 256 × 15 = 1280 cm³

7. TSA of an aquarium shaped like a square based pyramid is 400 sq. inch and slant height is 15 inch. Find its volume.

a² + 2a(15) = 400 ⇒ a² + 30a − 400 = 0 ⇒ (a + 40)(a − 10) = 0 ⇒ a = 10. h = √(225 − 25) = √200 = 14.14. V = ⅓ × 100 × 14.14 = 471.40 inch³

8. The base area of the pyramid in Egypt is 3600 m² and its height is 50 m. Find its total surface area.

a² = 3600 ⇒ a = 60. l = √(50² + 30²) = √3400 = 58.31. LSA = 2 × 60 × 58.31 = 6997.14. TSA = 3600 + 6997.14 = 10597.14 m² (the book writes cm²; the unit is m²)

9. A tent shaped like a square based pyramid has base side 32 m and height 50 m. (a) Cloth required? (b) Cost at Rs. 500 per m²?

l = √(50² + 16²) = √2756 = 52.50 m.
(a) Cloth = LSA = 2 × 32 × 52.50 = 3359.84 m² (no cloth for the ground) (book writes cm²)
(b) Cost = 3359.84 × 500 = Rs. 16,79,923.81

10. NTC telephone tower has a square base, slant height 50 ft and vertical height 40 ft. (a) Area of land occupied? (b) Rent at Rs. 50 per sq. ft for 20 years?

Half base = √(50² − 40²) = 30 ⇒ a = 60 ft.
(a) Area = 60² = 3600 sq. ft
(b) Rent per year = 50 × 3600 = 1,80,000. For 20 years = Rs. 36,00,000

11. A pyramid-shaped chocolate with square base has TSA 1920 cm² and base length 30 cm. Find its vertical height.

LSA = 1920 − 900 = 1020 ⇒ 2 × 30 × l = 1020 ⇒ l = 17. h = √(289 − 225) = 8 cm

Exercise 5.2: Cone

1. For a cone (base circle, radius r, height h, slant height l): (a) What do l, h and r represent? (b) Write the relation. (c) Find h if l = 13 cm and r = 5 cm.

(a) l = slant height, h = vertical height, r = radius of the base.
(b) l² = h² + r²
(c) h = √(169 − 25) = √144 = 12 cm

2. (a) CSA of a cone with radius r and slant height l. (b) TSA in terms of x and y if radius is x and slant height y. (c) Volume if diameter of base is p and height q.

(a) πrl cm²
(b) TSA = πr(r + l) = πx(x + y) cm²
(c) r = p/2: V = ⅓π(p/2)²q = πp²q/12 cm³

3. (a) Volume of a cone if its curved surface area is 81 cm² and height is 15 cm. (b) TSA = 250 cm², base area = 118 cm²; find CSA. (c) Find volume if slant height is 5 cm and radius is 4 cm.

(a) Corrected question: the area of the base is 81 cm² (a curved surface area of 81 cm² gives a messy answer, V ≈ 45.8 cm³, and does not match the key). V = ⅓ × 81 × 15 = 405 cm³
(b) CSA = TSA − base = 250 − 118 = 132 cm²
(c) h = √(25 − 16) = 3. V = ⅓π × 16 × 3 = 16π cm³ (= 50.29 cm³)

4. Find the curved surface area, total surface area and volume of each cone.

(a) r = 7, h = 21: l = √490 = 22.14. CSA = 22/7 × 7 × 22.14 = 486.99 cm²; TSA = 486.99 + 154 = 640.99 cm²; V = ⅓ × 22/7 × 49 × 21 = 1078 cm³
(b) r = 8, l = 17 (h = 15): CSA = π × 8 × 17 = 136π cm²; TSA = 136π + 64π = 200π cm²; V = ⅓π × 64 × 15 = 320π cm³
(c) h = 14, l = 16: r² = 256 − 196 = 60, r = 7.75. CSA = π × 7.75 × 16 = 123.94π cm²; TSA = 123.94π + 60π = 183.94π cm²; V = ⅓π × 60 × 14 = 280π cm³
(d) r = 7, l = 15: h = √176 = 13.27. CSA = 22/7 × 7 × 15 = 330 cm²; TSA = 330 + 154 = 484 cm²; V = ⅓ × 22/7 × 49 × 13.27 = 681.01 cm³
(e) r = 5, h = 12 (l = 13): CSA = 65π cm²; TSA = 65π + 25π = 90π cm²; V = ⅓π × 25 × 12 = 100π cm³
(f) r = 12 (diameter 24), l = 13, h = 5: CSA = 156π cm²; TSA = 156π + 144π = 300π cm²; V = ⅓π × 144 × 5 = 240π cm³

5. The volume of a right circular cone is 100π cm³ and height is 12 cm. Find the slant height.

⅓πr² × 12 = 100π ⇒ r² = 25 ⇒ r = 5. l = √(144 + 25) = 13 cm

6. Find the radius and height of a conical tent if its volume is 1232 cm³ and base area is 154 cm².

πr² = 154 ⇒ r² = 49 ⇒ r = 7 cm. V = ⅓ × 154 × h = 1232 ⇒ h = 24 cm

7. The plastic needed for a conical tent of slant height 14 m is 77 m². Find the area and perimeter of its base.

CSA = πrl: 22/7 × r × 14 = 77 ⇒ r = 1.75 m. Area = 22/7 × 1.75² = 9.62 m²; perimeter = 2 × 22/7 × 1.75 = 11 m

8. Find the volume, CSA and TSA of a cone with base diameter 8 cm and vertical height 21 cm.

r = 4; l = √(441 + 16) = √457 = 21.38.
V = ⅓ × 22/7 × 16 × 21 = 352 cm³
CSA = 22/7 × 4 × 21.38 = 268.7 cm²; TSA = 268.7 + 50.29 = 319 cm² (key: 268.54 and 318.80; small rounding difference)

9. Vertical height of a solid cone is three times the diameter of its base. Find TSA if volume is 54π cm³.

h = 3 × 2r = 6r. V = ⅓πr² × 6r = 2πr³ = 54π ⇒ r³ = 27 ⇒ r = 3, h = 18. l = √(324 + 9) = √333 = 18.25. TSA = 22/7 × 3 × (3 + 18.25) = 200.34 cm²

10. A wooden cone has base area 154 m² and height 14 m. Cost of colouring the surface excluding the base at Rs. 1.50 per m²?

πr² = 154 ⇒ r = 7. l = √(196 + 49) = √245 = 15.65. CSA = 22/7 × 7 × 15.65 = 344.3 m². Cost = 344.3 × 1.50 = Rs. 516.53

11. A cone-shaped pot (height 21 cm, base diameter 14 cm) is full of water. When two equal solid cones are immersed, one third of the water flows out. Find the volume of each cone.

V(pot) = ⅓ × 22/7 × 49 × 21 = 1078 cm³. Water out = ⅓ × 1078 = 359.33. Two cones = 359.33 ⇒ each = 179.67 cm³

12. Three solid cones of equal height with radii 3, 4 and 5 cm are melted into one big cone of the same height. Find its diameter.

⅓πR²h = ⅓πh(9 + 16 + 25) ⇒ R² = 50 ⇒ R = 5√2. Diameter = 10√2 cm

13. A conical wooden toy has base diameter 10 cm. It costs Rs. 880 to colour the whole surface at Rs. 4 per cm². Find its volume.

TSA = 880 ÷ 4 = 220 cm². r = 5: 22/7 × 5 × (5 + l) = 220 ⇒ 5 + l = 14 ⇒ l = 9. h = √(81 − 25) = √56 = 7.48. V = ⅓ × 22/7 × 25 × 7.48 = 195.99 cm³

Exercise 5.3: Combined Solids

1. (a) Total volume 1050 cm³, cylinder part 748 cm³; find volume of conical part. (b) CSA of cone part 252 cm², CSA of cylinder part 272 cm², circular base 154 cm²; find TSA. (c) Base area 36 cm², prism volume 144 cm³ (prism and pyramid of equal height h); find volume of pyramid. (d) CSA of cone part 308 cm² (slant height 2r, hemisphere radius r); find CSA of hemispherical part.

(a) V₂ = 1050 − 748 = 302 cm³
(b) TSA = 252 + 272 + 154 = 678 cm²
(c) Prism height = 144 ÷ 36 = 4 cm, so pyramid height h = 4 cm. V = ⅓ × 36 × 4 = 48 cm³
(d) πr(2r) = 2πr² = 308, and CSA of hemisphere = 2πr² = 308 cm²

2. Find the volume of each combined solid.

(a) Cylinder + cone: diameter 40 ⇒ r = 20; cylinder height 70, cone height 10. V = π × 400 × (70 + 10/3) = 22/7 × 400 × 73.33 = 92190.47 cm³
(b) Prism + pyramid: a = 12, prism 6, total height 15 ⇒ pyramid 9. V = 144 × (6 + 3) = 1296 cm³
(c) Cone + hemisphere: total height 56, cone height 42 ⇒ r = 14. V = ⅓ × 22/7 × 196 × (42 + 28) = 14373.33 cm³
(d) Two cones: d = 14 ⇒ r = 7, heights 6 and 9. V = ⅓ × 22/7 × 49 × 15 = 770 cm³

3. In a crystal made of two equal-height square based pyramids, the shaded common base is a square of side 2.5 cm. Total height is 4.5 cm. Find the total surface area and volume.

Height of each pyramid h = 2.25 cm. l = √(2.25² + 1.25²) = √6.625 = 2.574 cm. TSA = 2 × (2 × 2.5 × 2.574) = 25.74 cm². V = 2 × ⅓ × 6.25 × 2.25 = 9.375 cm³

4. Find the total surface area of each combined solid.

(a) Cylinder + cone: r = 3, total height 120 ⇒ cylinder 116, cone 4, l = 5. TSA = π × 3 × (3 + 232 + 5) = 720π = 2262.85 cm²
(b) Cylinder + cone: r = 7, cylinder 32, l = 25. TSA = 22/7 × 7 × (7 + 64 + 25) = 22 × 96 = 2112 cm²
(c) Cone + hemisphere: r = 7, cone height 24 ⇒ l = 25. TSA = 22/7 × 7 × (25 + 14) = 22 × 39 = 858 cm²
(d) Cone + hemisphere: r = 14, l = 50. TSA = 22/7 × 14 × (50 + 28) = 44 × 78 = 3432 cm²
(e) Cylinder + hemisphere: r = 14, h = 28. TSA = 2πrh + 3πr² = 22/7 × (2 × 14 × 28 + 3 × 196) = 22/7 × 1372 = 4320 cm²
(f) The dimensions of this figure are not given in the book text. The answer key gives 3504 cm². Use TSA = a² + 4ah₁ + 2al for a prism + pyramid solid.

5. A toy has a cone and a hemisphere of radius 14 cm. Total height 49 cm. Find its total surface area.

Cone height = 49 − 14 = 35. l = √(35² + 14²) = √1421 = 37.70. TSA = πr(l + 2r) = 22/7 × 14 × (37.70 + 28) = 44 × 65.70 = 2890.63 cm²

6. A pyramid of height 12 cm is placed on a square based cuboid of base area 100 cm² and height 10 cm. Find (a) total volume (b) total surface area.

a = 10. (a) V = 100 × 10 + ⅓ × 100 × 12 = 1400 cm³
(b) l = √(144 + 25) = 13. TSA = 100 + 40 × 10 + 2 × 10 × 13 = 760 cm²

7. A pyramid of vertical height 8 cm is placed on a cube of base side 12 cm. Find the total volume.

V = 12³ + ⅓ × 144 × 8 = 1728 + 384 = 2112 cm³

8. (a) A cone is placed on a cylinder; cylinder base area 100 cm², height 3 cm, total volume 600 cm³. Find total height. (b) A pyramid (slant height 5 cm) on a square prism of base 8 cm; volume 448 cm³. Find height of the prism part.

(a) Cylinder = 100 × 3 = 300; cone = 300 = ⅓ × 100 × h₂ ⇒ h₂ = 9. Total height = 3 + 9 = 12 cm
(b) Half base = 4, l = 5 ⇒ h₂ = 3. Pyramid = ⅓ × 64 × 3 = 64; prism = 448 − 64 = 384 = 64h₁ ⇒ h₁ = 6 cm

9. A square based pyramid is placed on a tower. Tower height 6 ft, pyramid height 1 ft, base length 1 ft. (a) Lateral surface area of the pyramid part. (b) Surface area of the tower that can be painted.

l = √(1 + 0.25) = √5/2. (a) LSA = 2al = 2 × 1 × √5/2 = √5 ft² ≈ 2.24 ft² (book writes cm²)
(b) Prism sides = 4 × 1 × 6 = 24. Painted area = 24 + 2.24 = 26.24 ft² (base excluded)

10. A spinning top has a conical lower part and a hemispherical upper part. Total height 5 cm, diameter of hemisphere 3.5 cm. Find the surface area that can be coloured.

r = 1.75; cone height = 5 − 1.75 = 3.25. l = √(3.25² + 1.75²) = √13.625 = 3.69. Area = πr(l + 2r) = 22/7 × 1.75 × (3.69 + 3.5) = 5.5 × 7.19 = 39.55 cm²

Exercise 5.4: Cost Estimation

1. Maximum water in each tank (1 ft³ = 28.317 l): (a) cube 6 ft × 6 ft × 6 ft (b) cuboid 2 m × 1.5 m × 2 m (c) cylinder of diameter 1 m and height 4 m.

(a) V = 216 ft³ = 216 × 28.317 = 6116.47 l
(b) V = 6 m³ = 6000 l
(c) V = 22/7 × 0.5² × 4 = 3.1428 m³ = 3142.85 l

2. A parking area ABCD (diagonal 8 ft, perpendiculars 12 ft and 10 ft) is paved with bricks of 0.222 ft² costing Rs. 16 each. (a) Area (b) Bricks needed (c) Total cost if 2 workers finish in 3 days at Rs. 1200 per worker per day.

(a) Area = ½ × 8 × (12 + 10) = 88 ft²
(b) Bricks = 88 ÷ 0.222 = 396.4 ⇒ 397 (at least)
(c) Bricks = 397 × 16 = 6352; wages = 2 × 3 × 1200 = 7200. Total = Rs. 13,552

3. A garden ABCD has sides 15 m, 14 m, 10 m and 11 m. It is fenced with barbed wire rounded 5 times. 3 workers finish in 2 days at Rs. 1500 per day; wire costs Rs. 80 per metre. (a) Perimeter (b) Wire required (c) Total cost.

(a) Perimeter = 15 + 14 + 10 + 11 = 50 m
(b) Wire = 5 × 50 = 250 m
(c) Wire = 250 × 80 = 20,000; wages = 3 × 2 × 1500 = 9000. Total = Rs. 29,000

4. A cubical room has side 9 ft, two square windows of 2 ft and two doors of 5 ft × 3 ft. (a) Floor area (b) Area of 4 walls excluding windows and doors (c) Cost of painting the 4 walls including windows and doors at Rs. 350 per sq. ft (d) Another room with each dimension 1 ft more is painted at Rs. 340 per sq. ft; change in cost?

(a) 9 × 9 = 81 ft²
(b) 4 × 81 = 324; openings = 2 × 4 + 2 × 15 = 38. Area = 324 − 38 = 286 ft²
(c) 324 × 350 = Rs. 1,13,400
(d) New room: 4 × 10 × 10 = 400 ft²; cost = 400 × 340 = 1,36,000. Increase = 1,36,000 − 1,13,400 = Rs. 22,600

5. A well needs 60 concrete rings, each of diameter 3.5 ft and height 1 ft. A ring costs Rs. 1200; 2 workers finish in 12 days at Rs. 1500 per worker per day (1 ft³ = 28.317 l). (a) Cost of rings (b) Cost to construct the well (c) Maximum water (d) If the water level is at the 35th ring from the top, how much water is in the well and how much more is needed to fill it?

(a) 60 × 1200 = Rs. 72,000
(b) Wages = 2 × 12 × 1500 = 36,000. Total = Rs. 1,08,000
(c) One ring = 22/7 × 1.75² × 1 = 9.625 ft³. 60 rings = 577.5 ft³ = 577.5 × 28.317 = 16,353.07 l
(d) Water in the lower 25 rings = 25 × 9.625 × 28.317 = 6813.78 l. More needed to fill = 16,353.07 − 6,813.78 = 9,539.29 l

6. Two square based pillars, each 10 ft high with base side 5 ft, have a pyramid of the same base on top. (a) Draw the figure (b) Slant height of the pyramid (c) Cost of painting the pillars with pyramids at Rs. 94 per sq. ft.

(a) A square prism (5 × 5 × 10 ft) with a square pyramid on top (draw both pillars).
(b) With pyramid height 1 ft: l = √(1 + 2.5²) = √29/2 ft (≈ 2.69 ft).
(c) Corrected answer for the data as printed (pyramid height 1 ft): l = √7.25 = 2.69 ft. One pillar = 4 × 5 × 10 + 2 × 5 × 2.69 = 226.93 ft²; two pillars = 453.85 ft². Cost = 453.85 × 94 = Rs. 42,662.05. (The key's Rs. 48,109.51 is obtained only if the pyramid height is 5 ft.)

7. Two square based pillars each 6 ft high, base side 1 ft, have a pyramid of height 1 ft on top. (a) Draw the figure (b) Formula for lateral surface area of the pyramid (c) Slant height (d) Cost of tiling per sq. ft if the total cost for both pillars with pyramids is Rs. 2729.

(a) Draw a 1 × 1 × 6 ft prism with a pyramid 1 ft high on each.
(b) LSA = 2al
(c) l = √(1 + 0.25) = √5/2 ft (≈ 1.12 ft)
(d) One pillar = 4 × 1 × 6 + 2 × 1 × 1.118 = 26.236 ft². Two = 52.47 ft². Rate = 2729 ÷ 52.47 = Rs. 52 per sq. ft

8. A circular water tank has inner diameter 2.80 m and cylindrical height 3 m, with a conical top of vertical height 0.72 m. (a) Formula for volume of a cylinder (b) Maximum water it holds (c) Cost to fill it at Rs. 1.85 per litre.

(a) V = πr²h
(b) r = 1.4 m. Cylinder = 22/7 × 1.96 × 3 = 18.48 m³; cone = ⅓ × 22/7 × 1.96 × 0.72 = 1.4784 m³. Total = 19.9584 m³ = 19,958.4 l
(c) Cost = 19,958.4 × 1.85 = Rs. 36,923.04

9. A pyramid shaped tent is made from 8 equal triangular cloth pieces with sides 5 m, 6 m, 6 m. (a) Area of one piece (b) Total cloth (c) Cost at Rs. 600 per m².

(a) s = (5 + 6 + 6) ÷ 2 = 8.5. Area = √(8.5 × 3.5 × 2.5 × 2.5) = 5√119/4 m² (≈ 13.64 m²)
(b) Total = 8 × 13.64 = 10√119 m² (≈ 109.09 m²)
(c) Cost = 109.09 × 600 = Rs. 65,452.27

Mixed Exercise

1. Three paper square based pyramids each have base 12 cm and slant height 10 cm. (a) Volume formula (b) Vertical height (c) Space occupied by the three pyramids (d) Cost of colouring each pyramid excluding the base at Rs. 80 per cm².

(a) V = ⅓a²h
(b) h = √(100 − 36) = 8 cm
(c) One = ⅓ × 144 × 8 = 384 cm³. Three = 1152 cm³
(d) LSA = 2 × 12 × 10 = 240 cm². Each: 240 × 80 = Rs. 19,200 (for all three: Rs. 57,600, as in the key)

2. A square based pyramid has base side 16 cm and volume 1280 cm³. (a) TSA formula (b) Perimeter of base (c) Slant height (d) TSA and area of triangular surfaces; which is greater and by how much?

(a) TSA = a² + 2al
(b) 4 × 16 = 64 cm
(c) ⅓ × 256 × h = 1280 ⇒ h = 15; l = √(225 + 64) = 17 cm
(d) TSA = 256 + 544 = 800 cm²; triangular surfaces = 544 cm². TSA is greater by 256 cm² (the base area).

3. LSA of a square based pyramid is 540 cm² and slant height is 15 cm. (a) TSA formula (b) Length of base (c) Volume.

(a) a² + 2al
(b) 2 × a × 15 = 540 ⇒ a = 18 cm
(c) h = √(225 − 81) = 12. V = ⅓ × 324 × 12 = 1296 cm³

4. A toy is a cylinder (height 10 cm) with a hemisphere of radius 7 cm on top. (a) Diameter of base (b) Circumference of base (c) TSA (d) Volume.

(a) 2 × 7 = 14 cm
(b) 2πr = 2 × 22/7 × 7 = 44 cm
(c) TSA = 2πrh + 3πr² = 440 + 462 = 902 cm²
(d) V = πr²(h + ⅔r) = 154 × (10 + 14/3) = 2258.67 cm³

5. A solid has a square based pyramid (slant height 5 cm) on a square based prism with base 8 cm. Volume is 1024 cm³. (a) Draw (b) Vertical height of pyramid part (c) Compare it with height of prism part (d) TSA.

(a) Draw a square prism with a pyramid on top.
(b) Half base = 4, l = 5 ⇒ h₂ = 3 cm
(c) Pyramid = ⅓ × 64 × 3 = 64; prism = 960 = 64h₁ ⇒ h₁ = 15 cm. Pyramid height is less by 12 cm.
(d) TSA = 64 + 32 × 15 + 2 × 8 × 5 = 624 cm²

6. A tent has a cylindrical lower part and a hemispherical upper part of the same radius. Total height 54 m, cylindrical part 40 m. (a) Draw (b) Radius (c) TSA (d) Cost at Rs. 500 per m².

(b) r = 54 − 40 = 14 m
(c) No base: TSA = 2πrh + 2πr² = 2 × 22/7 × 14 × 40 + 2 × 22/7 × 196 = 3520 + 1232 = 4752 m²
(d) Cost = 4752 × 500 = Rs. 23,76,000

7. A cone and hemisphere of base radius 7 cm are joined. Colouring costs Rs. 5148 at Rs. 6 per cm². (a) TSA (b) Slant height (c) Vertical height of cone (d) Total height.

(a) 5148 ÷ 6 = 858 cm²
(b) πr(l + 2r) = 858 ⇒ 22 × (l + 14) = 858 ⇒ l + 14 = 39 ⇒ l = 25 cm
(c) h = √(625 − 49) = 24 cm
(d) 24 + 7 = 31 cm

8. A toy has a cylinder and cone of the same base, diameter 10 cm. Cylinder length 14 cm, cone height 12 cm. (a) Slant height (b) If the cylinder length and cone height are exchanged, which toy costs more at the same rate per area?

(a) l = √(144 + 25) = 13 cm
(b) Original TSA = πr(r + 2h₁ + l) = 22/7 × 5 × (5 + 28 + 13) = 722.86 cm². New toy: cylinder 12, cone 14, l = √221 = 14.87: TSA = 22/7 × 5 × (5 + 24 + 14.87) = 689.2 cm². So the original toy costs more. (The book key says the new toy; the calculation shows the original has more area.)

9. A pyramid of height 12 cm is placed on a square based cuboid of base area 100 cm² and height 9 cm. (a) Length of the base (b) Volume of the solid.

(a) a² = 100 ⇒ a = 10 cm
(b) V = 100 × 9 + ⅓ × 100 × 12 = 900 + 400 = 1300 cm³

10. A cylindrical tank (radius 14 cm, height 40 cm) full of cement is poured into a conical pile of height 30 cm. (a) Radius of pile (b) Surface area excluding the base (c) Weight if 1 cm³ weighs 2.5 g.

(a) Cylinder = 22/7 × 196 × 40 = 24,640 cm³. ⅓ × 22/7 × r² × 30 = 24,640 ⇒ r² = 784 ⇒ r = 28 cm
(b) l = √(900 + 784) = 41.04. CSA = 22/7 × 28 × 41.04 = 3611.2 cm²
(c) 24,640 × 2.5 = 61,600 g

11. Two pillars have base 6 ft × 6 ft and height 8 ft each, with a pyramid of height 4 ft on top. (a) Draw (b) LSA formula of pyramid (c) Slant height (d) Surface area to be painted (e) Cost at Rs. 110 per sq. ft.

(b) 2al
(c) l = √(16 + 9) = 5 ft
(d) Pyramid LSA = 2 × 6 × 5 = 60; pillar sides = 4 × 6 × 8 = 192. Per pillar = 252 ft²
(e) There are two pillars: 2 × 252 = 504 ft²; cost = 504 × 110 = Rs. 55,440. (The key's Rs. 27,720 is for one pillar.)

12. Two cuboid pillars of 1 ft × 1 ft base and 6 ft height each have a pyramid of height 1 ft on top. (a) LSA formula of prism (b) Perimeter of base (c) Slant height (d) Area to be painted (e) Cost at Rs. 52 per sq. ft.

(a) LSA = ph (b) 4 ft (c) l = √(1 + 0.25) = √5/2 ft
(d) Per pillar = 24 + 2 × 1 × √5/2 = (24 + √5) ft² (26.24 ft²)
(e) Two pillars: 2 × 26.236 = 52.47 ft²; cost = 52.47 × 52 = Rs. 2,728.54. (The key's Rs. 1364.27 is for one pillar.)

13. A concrete ring of a well has diameter 3.5 ft and height 1 ft; a ring costs Rs. 1200; 32 rings are needed. (a) Cost of the well (b) Volume of water (c) If the water level is at the 18th ring, volume of the well above the water.

(a) 32 × 1200 = Rs. 38,400
(b) One ring = 9.625 ft³. 32 rings = 308 ft³ (book writes cm³)
(c) Above the water = 32 − 18 = 14 rings = 14 × 9.625 = 134.75 ft³

14. Two square based pyramids share a base of side 12 cm and the total height is 20 cm. (a) LSA formula (b) Volume.

(a) 2al
(b) V = ⅓ × 144 × 20 = 960 cm³

15. A water tank is a cylinder with a cone on top. Inner diameter 1.4 m, cylinder height 2.1 m, cone height 0.36 m. (a) TSA formula of a cone (b) Perimeter of base (c) Capacity in litres.

(a) πr(r + l)
(b) 2 × 22/7 × 0.7 = 4.4 m
(c) πr² = 22/7 × 0.49 = 1.54. V = 1.54 × 2.1 + ⅓ × 1.54 × 0.36 = 3.234 + 0.1848 = 3.4188 m³ = 3418.8 l

16. A cone + hemisphere solid has volume 240π cm³. The ratio of slant height of the cone to radius is 5 : 3. (a) Shapes (b) CSA formula of cone (c) TSA (d) Are TSA and curved surface area equal?

(a) A cone and a hemisphere. (b) πrl
(c) r = 3k, l = 5k, h = 4k. V = ⅓π × 9k² × 4k + ⅔π × 27k³ = 12πk³ + 18πk³ = 30πk³ = 240π ⇒ k = 2. r = 6, l = 10. TSA = πr(l + 2r) = 22/7 × 6 × 22 = 414.86 cm²
(d) Yes: no flat face is exposed, so the TSA is just the sum of the two curved surfaces.

17. A cylinder + cone solid has TSA 840π cm². Ratio of slant height to vertical height of the cone is 13 : 12; cylinder height is 24 cm. (a) Shapes (b) Perimeter of base of a cone (c) Volume (d) Is TSA equal to curved surface area? Compare.

(a) Cylinder and cone (b) 2πr
(c) r = 5k, h = 12k, l = 13k. TSA = π(25k² + 2 × 5k × 24 + 5k × 13k) = π(90k² + 240k) = 840π ⇒ 3k² + 8k − 28 = 0 ⇒ k = 2. r = 10, h = 24, l = 26. V = π × 100 × (24 + 8) = 3200π = 10057.14 cm³
(d) Curved area = 480π + 260π = 740π. TSA = 840π. No: TSA is greater by 100π = 314.28 cm² (the base circle).

18. A cylinder (r = 7, h = 39) and a cone (h = 24) with the same base are joined to form a pencil. (a) Formula for area of a cone (b) Surface area of the separate solids and their sum (c) Is that sum equal to the TSA of the combined solid? (d) Is the sum of volumes equal to the combined volume? (e) Compare the painting cost at Rs. 150 per m².

(a) CSA = πrl, TSA = πr(r + l) (the book prints ⅓a²h, which is the pyramid)
(b) Cylinder = 2πr(r + h) = 44 × 46 = 2024; cone (l = 25) = 22 × 32 = 704. Sum = 2728 cm²
(c) Combined TSA = 2420 cm². Not equal: the sum is greater by 308 cm² (two joined circular faces, 2 × 154).
(d) Yes, both are 7238 cm³: volume does not change on joining.
(e) Separate solids cost more by 308 × 150 = Rs. 46,200 (taking the rate as given in the key).

19. A toy has a cylinder and cone of the same base, diameter 10 cm; cylinder length 14 cm, cone height 12 cm. (a) Formula for volume of a cone (b) Slant height (c) A new toy is made by exchanging the diameter and the conical height; which costs more to colour?

(a) V = ⅓πr²h (b) 13 cm
(c) Original TSA = 22/7 × 5 × 46 = 722.86 cm². New toy: diameter 12 (r = 6), cone height 10, l = √136 = 11.66: TSA = 22/7 × 6 × (6 + 28 + 11.66) = 861.1 cm². The second toy costs more.

20. A solid is made of a cylinder (height 28 cm) and a cone of slant height 17 cm with the same radius. Colouring costs Rs. 2851.20. (a) Draw (b) Shape of the base of the cylinder (c) Find the height of the cone.

(b) A circle.
(c) The printed rate (Rs. 100) cannot give this cost; the key (h = 15 cm) works with Rs. 1.40 per cm². TSA = 2851.20 ÷ 1.40 = 2036.57 cm² = π × 8 × 81.
πr(r + 56 + 17) = 2036.57 ⇒ r² + 73r − 648 = 0 ⇒ (r + 81)(r − 8) = 0 ⇒ r = 8 cm.
h = √(17² − 8²) = √225 = 15 cm

Mixed Exercise Q21 is missing from the book text, so it is not solved here (the key gives: 1 m³ = 1000 l, r = 7 m, l = 13 m, Rs. 12,24,300).

Corrections to the Textbook

Question	Book says	Correct
5.1 Q8, Q9; 5.3 Q9	cm² (units)	m² (Q8, Q9), ft² (5.3 Q9)
5.2 Q3(a)	CSA = 81 cm²	Base area = 81 cm², V = 405 cm³
5.2 Q8	268.54, 318.80	268.7 cm², 319 cm² (rounding)
5.2 Example 6 (notes)	r = 18.5	r = 18.9
5.3 Q4(f)	dimensions not given	Key answer 3504 cm² only
5.4 Q5(c)	16,35,06 l	16,353.07 l
5.4 Q6(c)	Rs. 48,109.51 (needs height 5 ft)	Rs. 42,662.05
Mixed Q1(c)	432 cm²	1152 cm³
Mixed Q8	New toy costs more	Original toy costs more
Mixed Q11(e), Q12(e)	Rs. 27,720; Rs. 1364.27 (one pillar)	Rs. 55,440; Rs. 2,728.54 (two pillars)
Mixed Q13, Q18(d)	cm³	ft³ (Q13), cm³ (Q18d: 7238 cm³)
Mixed Q18(a)	⅓a²h	CSA = πrl, TSA = πr(r + l)
Mixed Q20	Rs. 100 per cm²	Rs. 1.40 per cm²
Mixed Q21	question missing	not solved

Area and Volume • Grade 10 Mathematics