Unit 5: Area and Volume
Square-Based Pyramid and Cone — WordPress-ready exercise solutions reconstructed from the supplied 37-page handwritten notes.
Original Scanned PDF – View Notes
Square-Based Pyramid – Important Formulae
Exercise 5.1 – Square-Based Pyramid
Basic Pyramid Problems
1(a). Area of base = 64 cm² and height = 15 cm. Find volume.
V = (1/3) × area of base × height
= (1/3) × 64 × 15
∴ V = 320 cm³
1(b). TSA = 285 cm² and LSA = 192 cm². Find area of base.
TSA = LSA + area of base
285 = 192 + area of base
Area of base = 285 − 192
∴ Area of base = 93 cm²
2. Relations shown in the square-pyramid diagram
Pages 2–3 apply Pythagoras’ theorem to the vertical height, slant height and half of the base side. The same relation is used throughout the exercise:
Total Surface Area and Volume from Given Measurements
3(a). Slant height = 10 cm, side of base = 12 cm
10² = h² + (12/2)²
100 = h² + 36
h = 8 cm
TSA = 12² + 2 × 12 × 10 = 384 cm²
V = (1/3) × 12² × 8 = 384 cm³
3(b). Height = 12 cm, side of base = 10 cm
l² = 12² + 5² = 169
l = 13 cm
TSA = 10² + 2 × 10 × 13 = 360 cm²
V = (1/3) × 10² × 12 = 400 cm³
3(c). Slant height = 13 cm, side of base = 10 cm
13² = h² + 5²
169 = h² + 25
h = 12 cm
TSA = 360 cm²
V = 400 cm³
3(d). Height = 8 cm, side of base = 12 cm
l² = 8² + 6² = 100
l = 10 cm
TSA = 12² + 2 × 12 × 10 = 384 cm²
V = (1/3) × 12² × 8 = 384 cm³
3(e). Height = 6 cm, side of base = 16 cm
l² = 6² + 8² = 100
l = 10 cm
TSA = 16² + 2 × 16 × 10 = 576 cm²
V = (1/3) × 16² × 6 = 512 cm³
3(f). Height = 15 cm, slant height = 17 cm, side = 16 cm
TSA = 16² + 2 × 16 × 17 = 800 cm²
V = (1/3) × 16² × 15 = 1,280 cm³
Finding Missing Dimensions
4. Side of base = 30 cm, perpendicular height = 25 cm
V = (1/3) × 30² × 25
∴ V = 7,500 cm³
5. TSA = 800 cm² and side of base = 16 cm
800 = 16² + 2 × 16 × l
800 − 256 = 32l
l = 17 cm
17² = h² + 8²
h = 15 cm
Volume = (1/3) × 16² × 15 = 1,280 cm³
6. TSA = 400 square units and slant height = 15 units
400 = a² + 2a(15)
a² + 30a − 400 = 0
Positive root: a = 10
15² = h² + 5²
h = 10√2
V = (1/3) × 100 × 10√2 = (1000√2)/3
7. Base area = 3,600 m², perpendicular height = 50 m
a = √3600 = 60 m
l² = 50² + 30² = 3400
l = 10√34 m
TSA = 60² + 2 × 60 × 10√34
= 3600 + 1200√34 m²
11. TSA = 1,920 cm² and side of base = 30 cm
1,920 = 30² + 2 × 30 × l
1,020 = 60l
l = 17 cm
17² = h² + 15²
h² = 64
∴ h = 8 cm
Cost and Rent Problems
8. Square pyramid: side = 32 m, height = 50 m; cost of covering lateral faces = Rs 500 per m²
l² = 50² + 16² = 2756
l = 2√689 m
LSA = 2 × 32 × 2√689 = 128√689 m²
Required cost = LSA × 500
The handwritten source gives a total close to Rs 16.8 lakh.
9. Slant height = 50 ft, vertical height = 40 ft
50² = 40² + (a/2)²
(a/2)² = 900
a = 60 ft
Area of base = 60² = 3,600 ft²
At Rs 50 per ft², annual rent = Rs 1,80,000
For 20 years = Rs 36,00,000
Cone – Important Formulae
Exercise 5.2 – Cone
Basic Cone Problems
1(a). Area of base = 81 cm² and height = 15 cm
V = (1/3) × area of base × height
= (1/3) × 81 × 15
∴ V = 405 cm³
1(b). TSA = 250 cm² and base area = 118 cm²
TSA = base area + CSA
250 = 118 + CSA
∴ CSA = 132 cm²
1(c). Slant-height/radius problem
Page 19 uses the relation l² = h² + r² first and then applies V = (1/3)πr²h.
Volume, Curved Surface Area and Total Surface Area
2(a). Height = 21 cm, diameter = 14 cm
r = 7 cm
V = (1/3) × (22/7) × 7² × 21 = 1,078 cm³
l = √(21² + 7²) = 7√10 cm
CSA = πrl
TSA = πr(l + r)
2(b). Slant height = 17 cm, radius = 8 cm
17² = h² + 8²
h = 15 cm
V = (1/3)π(8²)(15)
CSA = π(8)(17)
TSA = π(8)(17 + 8)
2(c). Height = 14 cm, slant height = 16 cm
r² = 16² − 14² = 60
r = 2√15 cm
V = (1/3)π(60)(14) = 280π cm³
CSA = πr(16)
TSA = πr(16 + r)
2(d). Radius = 7 cm, slant height = 15 cm
h = √(15² − 7²) = 4√11 cm
CSA = π × 7 × 15 = 105π cm²
TSA = π × 7 × (15 + 7) = 154π cm²
2(e). Radius = 5 cm, height = 12 cm
l = √(12² + 5²) = 13 cm
CSA = π × 5 × 13 = 65π cm²
TSA = π × 5 × (13 + 5) = 90π cm²
V = (1/3)π × 25 × 12 = 100π cm³
2(f). Height = 8 cm, diameter = 24 cm
r = 12 cm
l = √(8² + 12²) = 4√13 cm
CSA = π × 12 × 4√13
TSA = π × 12(12 + 4√13)
V = (1/3)π × 12² × 8 = 384π cm³
Finding Radius, Height and Slant Height
5. Volume = 100π cm³ and height = 12 cm
100π = (1/3)πr² × 12
r² = 25
r = 5 cm
l = √(12² + 5²) = 13 cm
6. Base area = 154 cm² and volume = 1,232 cm³
πr² = 154
Using π = 22/7, r² = 49
∴ r = 7 cm
1,232 = (1/3) × 154 × h
∴ h = 24 cm
9. Vertical height is three times the diameter; volume = 54π cm³
h = 3d = 6r
54π = (1/3)πr²(6r)
54 = 2r³
r³ = 27
∴ r = 3 cm
h = 18 cm
l = √(18² + 3²) = 3√37 cm
TSA = πr(r + l) = 9π(1 + √37) cm²
Cost, Sheet and Shade Problems
7. Conical shade: curved surface area = 77 m², slant height = 14 m
CSA = πrl
77 = (22/7) × r × 14
r = 1.75 m
Perimeter of base = 2πr = 11 m
Area of base = πr² ≈ 9.625 m²
8. Diameter = 8 cm and height = 21 cm
r = 4 cm
V = (1/3)π × 4² × 21 = 112π cm³ = 352 cm³
l = √(21² + 4²) = √457 cm
CSA = π × 4 × √457
TSA = π × 4(√457 + 4)
10. Cardboard cone: base area = 154 m², vertical height = 14 m, rate = Rs 7.50 per m²
πr² = 154 ⇒ r = 7 m
l = √(14² + 7²) = 7√5 m
Area to be covered excluding base = CSA = πrl
= 154√5 ≈ 344.35 m²
Cost ≈ 344.35 × 7.50
∴ Cost ≈ Rs 2,582.63
13. Conical toy: base diameter = 10 cm; coloring costs Rs 880 at Rs 4 per cm²
r = 5 cm
Total area colored = 880/4 = 220 cm²
TSA = πr(r + l)
220 = π × 5(5 + l)
The source obtains l = 9 cm.
h = √(9² − 5²) = √56 = 2√14 cm
V = (1/3)π × 25 × 2√14
∴ V ≈ 195.9 cm³
Combined Cone Problems
11. Cone-shaped pot and overflowing water
Pages 34–35 calculate the volume of a conical pot and then take a fraction of the contained water to find the amount in smaller cones.
The source records:
Volume of water considered ≈ 359.33 cm³
Volume of one smaller cone ≈ 179.665 cm³
12. Three cones of radii 3 cm, 4 cm and 5 cm are melted into one cone of the same height
(1/3)πh(3² + 4² + 5²) = (1/3)πhR²
R² = 9 + 16 + 25 = 50
R = 5√2 cm
Diameter = 2R
∴ Diameter = 10√2 cm
Discussion
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