1 Right Circular Cone
What is it? A right circular cone is a solid shape generated by rotating a right-angled triangle around one of its perpendicular sides. It looks like an ice-cream cone or a birthday cap.
- Height (h): The straight distance from the vertex (top tip) to the center of the circular base.
- Radius (r): The radius of the circular base.
- Slant Height (l): The distance from the vertex to any point on the edge of the circular base.
Derivation of Surface Area: If you cut a paper cone straight along its side and open it flat, it looks like a sector of a circle (like a piece of round cake).
- If you cut this paper into hundreds of tiny pieces, each piece looks like a small triangle.
- The height of each small triangle is equal to the slant height (l) of the cone.
- The base of all these triangles combined forms the circumference of the cone’s base (2πr).
- Adding the areas of all these small triangles gives the curved surface area.
Formulas for Cone:
- Slant Height (l): Using Pythagoras theorem on the height and radius:
- l = √(r² + h²)
- Curved Surface Area (CSA): This is the area of just the curved part (excluding the bottom circle).
- CSA = πrl
- Total Surface Area (TSA): This includes the curved area plus the area of the circular base.
- TSA = Curved Area + Base Area
- TSA = πrl + πr² = πr(l + r)
- Volume: To compare a cone with a cylinder that has the same base radius and same height:
- Experiment: If you fill the cone with sand and pour it into the cylinder, it takes exactly 3 full cones to fill the cylinder completely.
- Therefore, the volume of a cone is one-third of the volume of a cylinder.
- Volume = (1/3)πr²h
2 Sphere
What is it? A sphere is a 3-dimensional solid figure which is like a ball. Every point on its surface is at the same distance (radius) from the center.
Derivation of Surface Area: This formula is found through a practical activity involving string.
- Take a ball (sphere) and wind a string tightly around it until it is completely covered.
- Measure the radius of the ball.
- Draw 4 flat circles on a paper with that same radius.
- If you unwind the string from the ball, you will see that it exactly fills those 4 circles.
- This proves the surface area of a sphere is 4 times the area of a circle.
Formulas for Sphere:
- Surface Area: Since it equals 4 circles:
- Surface Area = 4 × (πr²) = 4πr²
- Volume: This is found by measuring water displacement. If you dip spheres of different sizes into a full container of water, the amount of water that overflows represents the volume of the sphere.
- Volume = (4/3)πr³
3 Hemisphere
What is it? If you slice a solid sphere exactly through the middle, you get two equal halves. Each half is called a hemisphere.
Faces of a Hemisphere: Unlike a sphere which has only one curved face, a solid hemisphere has two faces:
- A curved face (the bowl part).
- A flat face (the circular top or base).
Formulas for Hemisphere:
- Curved Surface Area (CSA): This is exactly half the surface area of a full sphere.
- CSA = (1/2) × 4πr² = 2πr²
- Total Surface Area (TSA): This is the curved area plus the area of the flat circular top.
- TSA = Curved Area + Area of Circle
- TSA = 2πr² + πr² = 3πr²
- Volume: Since a hemisphere is half of a sphere, its volume is half of the sphere’s volume.
- Volume = (1/2) × (4/3)πr³ = (2/3)πr³
Activities
Page 1 Activity: Rotating a Right-Angled Triangle
Question: What happens? Do you recognize the shape that the triangle is forming as it rotates around the string?
Answer: When the right-angled triangle is rotated about one of its perpendicular sides, it creates a solid 3-dimensional shape. This shape is called a Right Circular Cone.
Question: Does it remind you of the time you had eaten an ice-cream heaped into a container of that shape?
Answer: Yes, the shape formed resembles an ice-cream cone.
Page 6 Activity: String on a Sphere
Observation: This activity asks what you achieved by winding string around a ball (sphere) and then filling four circles with the same string.
Answer: This shows that the surface area of a sphere is equal to 4 times the area of a circle with the same radius.
Formula: Surface Area of Sphere = 4πr²
Page 9 Activity: Filling Cylinder with Cone
Observation: Comparing the volume of a cone and a cylinder with the same base radius and height.
Answer: It takes exactly 3 full cones of sand to fill the cylinder completely.
This means the volume of a cone is one-third (1/3) the volume of the cylinder.
Examples (In-Text)
Example 1: Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.
Given:
Radius (r) = 7 cm
Slant height (l) = 10 cm
Formula: Curved Surface Area = πrl
Solution:
CSA = (22/7) × 7 × 10
CSA = 22 × 10
CSA = 220 cm²
Example 2: The height of a cone is 16 cm and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone (Use π = 3.14).
Given:
Height (h) = 16 cm
Radius (r) = 12 cm
Step 1: Find slant height (l) using Pythagoras theorem:
l = √(h² + r²)
l = √(16² + 12²)
l = √(256 + 144)
l = √400
l = 20 cm
Step 2: Curved Surface Area
CSA = πrl
CSA = 3.14 × 12 × 20
CSA = 37.68 × 20
CSA = 753.6 cm²
Step 3: Total Surface Area
TSA = πr(l + r)
TSA = 3.14 × 12 × (20 + 12)
TSA = 37.68 × 32
TSA = 1205.76 cm²
Example 3: A corn cob, shaped somewhat like a cone, has the radius of its broadest end as 2.1 cm and length (height) as 20 cm. If each 1 cm² of the surface of the cob carries an average of four grains, find how many grains you would find on the entire cob.
Given:
Radius (r) = 2.1 cm
Height (h) = 20 cm
Grains per cm² = 4
Step 1: Find slant height (l):
l = √(r² + h²)
l = √((2.1)² + 20²)
l = √(4.41 + 400)
l = √404.41
l ≈ 20.11 cm
Step 2: Curved Surface Area:
CSA = πrl
CSA = (22/7) × 2.1 × 20.11
CSA = 22 × 0.3 × 20.11
CSA = 132.726 cm²
Step 3: Number of grains:
Number of grains = Area × grains per cm²
Number of grains = 132.726 × 4
Number of grains = 530.904
Answer: Approximately 531 grains
Example 4: Find the surface area of a sphere of radius 7 cm.
Given:
Radius (r) = 7 cm
Formula: Surface Area = 4πr²
Solution:
Surface Area = 4 × (22/7) × 7 × 7
Surface Area = 4 × 22 × 7
Surface Area = 616 cm²
Example 5: Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm.
Given:
Radius (r) = 21 cm
(i) Curved Surface Area:
CSA = 2πr²
CSA = 2 × (22/7) × 21 × 21
CSA = 2 × 22 × 3 × 21
CSA = 2772 cm²
(ii) Total Surface Area:
TSA = 3πr²
TSA = 3 × (22/7) × 21 × 21
TSA = 3 × 22 × 3 × 21
TSA = 4158 cm²
Example 6: The hollow sphere, in which the circus motorcyclist performs his stunts, has a diameter of 7 m. Find the area available to the motorcyclist for riding.
Given:
Diameter = 7 m
Radius (r) = 3.5 m
Formula: Area available = Surface Area of sphere = 4πr²
Solution:
Area = 4 × (22/7) × 3.5 × 3.5
Area = 4 × 22 × 0.5 × 3.5
Area = 88 × 0.5 × 3.5
Area = 154 m²
Example 7: A hemispherical dome of a building needs to be painted. If the circumference of the base of the dome is 17.6 m, find the cost of painting it, given the cost of painting is ₹ 5 per 100 cm².
Given:
Circumference of base = 17.6 m
Cost of painting = ₹ 5 per 100 cm²
Step 1: Find radius:
2πr = 17.6
2 × (22/7) × r = 17.6
(44/7) × r = 17.6
r = (17.6 × 7) / 44
r = 2.8 m
Step 2: Curved Surface Area (to be painted):
CSA = 2πr²
CSA = 2 × (22/7) × 2.8 × 2.8
CSA = 44 × 0.4 × 2.8
CSA = 49.28 m²
Step 3: Convert to cm²:
49.28 m² = 49.28 × 10000 = 492800 cm²
Step 4: Calculate Cost:
Cost = (Area / 100) × Rate
Cost = (492800 / 100) × 5
Cost = 4928 × 5
Cost = ₹ 24,640
Example 8: The height and the slant height of a cone are 21 cm and 28 cm respectively. Find the volume of the cone.
Given:
Height (h) = 21 cm
Slant height (l) = 28 cm
Step 1: Find radius (r):
r = √(l² – h²)
r = √(28² – 21²)
r = √(784 – 441)
r = √343
r = 7√7 cm
Step 2: Calculate Volume:
Volume = (1/3)πr²h
Volume = (1/3) × (22/7) × (7√7)² × 21
Volume = (1/3) × (22/7) × 343 × 21
Volume = 22 × 49 × 7
Volume = 7546 cm³
Example 9: Monica has a piece of canvas whose area is 551 m². She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and the wastage incurred while cutting, amounts to approximately 1 m², find the volume of the tent that can be made with it.
Given:
Total canvas area = 551 m²
Wastage = 1 m²
Radius (r) = 7 m
Step 1: Effective area for tent (CSA):
CSA = 551 – 1 = 550 m²
Step 2: Find slant height (l):
CSA = πrl
550 = (22/7) × 7 × l
550 = 22 × l
l = 550 / 22
l = 25 m
Step 3: Find height (h):
h = √(l² – r²)
h = √(25² – 7²)
h = √(625 – 49)
h = √576
h = 24 m
Step 4: Calculate Volume:
Volume = (1/3)πr²h
Volume = (1/3) × (22/7) × 7 × 7 × 24
Volume = 22 × 7 × 8
Volume = 1232 m³
Example 10: Find the volume of a sphere of radius 11.2 cm.
Given:
Radius (r) = 11.2 cm
Formula: Volume = (4/3)πr³
Solution:
Volume = (4/3) × (22/7) × 11.2 × 11.2 × 11.2
Volume = 5887.32 cm³ (approx)
Example 11: A shot-putt is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm³, find the mass of the shot-putt.
Given:
Radius (r) = 4.9 cm
Density = 7.8 g/cm³
Step 1: Calculate Volume:
Volume = (4/3)πr³
Volume = (4/3) × (22/7) × 4.9 × 4.9 × 4.9
Volume ≈ 493 cm³
Step 2: Calculate Mass:
Mass = Volume × Density
Mass = 493 × 7.8
Mass = 3845.4 g
Mass = 3.85 kg (approx)
Example 12: A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?
Given:
Radius (r) = 3.5 cm
Formula: Volume = (2/3)πr³
Solution:
Volume = (2/3) × (22/7) × 3.5 × 3.5 × 3.5
Volume = 89.8 cm³ (approx)
Exercise 11.1
1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Given:
Diameter = 10.5 cm
Radius (r) = 5.25 cm
Slant height (l) = 10 cm
Formula: Curved Surface Area = πrl
Solution:
CSA = (22/7) × 5.25 × 10
CSA = 22 × 0.75 × 10
CSA = 165 cm²
2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Given:
Diameter = 24 m
Radius (r) = 12 m
Slant height (l) = 21 m
Formula: Total Surface Area = πr(l + r)
Solution:
TSA = (22/7) × 12 × (21 + 12)
TSA = (22/7) × 12 × 33
TSA = 8712 / 7
TSA = 1244.57 m²
3. Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) radius of the base and (ii) total surface area of the cone.
Given:
CSA = 308 cm²
Slant height (l) = 14 cm
(i) Find radius:
CSA = πrl
308 = (22/7) × r × 14
308 = 44 × r
r = 308 / 44
r = 7 cm
(ii) Total Surface Area:
TSA = πr(l + r)
TSA = (22/7) × 7 × (14 + 7)
TSA = 22 × 21
TSA = 462 cm²
4. 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 ₹ 70.
Given:
Height (h) = 10 m
Radius (r) = 24 m
(i) Find slant height:
l = √(r² + h²)
l = √(24² + 10²)
l = √(576 + 100)
l = √676
l = 26 m
(ii) Cost of canvas:
Canvas required = Curved Surface Area = πrl
CSA = (22/7) × 24 × 26
CSA = 13728 / 7 m²
Cost = Area × Rate
Cost = (13728 / 7) × 70
Cost = 13728 × 10
Cost = ₹ 1,37,280
5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14).
Given:
Height (h) = 8 m
Radius (r) = 6 m
Width of tarpaulin = 3 m
Extra for wastage = 20 cm = 0.2 m
Step 1: Find slant height:
l = √(h² + r²)
l = √(8² + 6²)
l = √(64 + 36)
l = √100
l = 10 m
Step 2: Area of tarpaulin = CSA of tent:
CSA = πrl
CSA = 3.14 × 6 × 10
CSA = 188.4 m²
Step 3: Length of tarpaulin:
Length = Area / Width
Length = 188.4 / 3
Length = 62.8 m
Step 4: Add extra length for wastage:
Total length = 62.8 + 0.2
Total length = 63 m
6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹ 210 per 100 m².
Given:
Slant height (l) = 25 m
Diameter = 14 m
Radius (r) = 7 m
Rate = ₹ 210 per 100 m²
Step 1: Curved Surface Area:
CSA = πrl
CSA = (22/7) × 7 × 25
CSA = 22 × 25
CSA = 550 m²
Step 2: Calculate Cost:
Cost = (Area / 100) × Rate
Cost = (550 / 100) × 210
Cost = 5.5 × 210
Cost = ₹ 1155
7. 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.
Given:
Radius (r) = 7 cm
Height (h) = 24 cm
Number of caps = 10
Step 1: Find slant height:
l = √(r² + h²)
l = √(7² + 24²)
l = √(49 + 576)
l = √625
l = 25 cm
Step 2: Area of 1 cap (CSA):
CSA = πrl
CSA = (22/7) × 7 × 25
CSA = 550 cm²
Step 3: Area for 10 caps:
Total Area = 10 × 550
Total Area = 5500 cm²
8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹ 12 per m², what will be the cost of painting all these cones? (Use π = 3.14 and take √1.04 = 1.02)
Given:
Diameter = 40 cm, so Radius (r) = 20 cm = 0.2 m
Height (h) = 1 m
Number of cones = 50
Cost of painting = ₹ 12 per m²
Step 1: Find slant height:
l = √(r² + h²)
l = √(0.2² + 1²)
l = √(0.04 + 1)
l = √1.04
l = 1.02 m (given)
Step 2: CSA of 1 cone:
CSA = πrl
CSA = 3.14 × 0.2 × 1.02
CSA = 0.64056 m²
Step 3: CSA of 50 cones:
Total CSA = 50 × 0.64056
Total CSA = 32.028 m²
Step 4: Calculate Cost:
Cost = Area × Rate
Cost = 32.028 × 12
Cost = ₹ 384.34 (approx)
Exercise 11.2
1. Find the surface area of a sphere of radius: (i) 10.5 cm (ii) 5.6 cm (iii) 14 cm
Formula: Surface Area = 4πr²
(i) r = 10.5 cm:
Area = 4 × (22/7) × 10.5 × 10.5
Area = 1386 cm²
(ii) r = 5.6 cm:
Area = 4 × (22/7) × 5.6 × 5.6
Area = 394.24 cm²
(iii) r = 14 cm:
Area = 4 × (22/7) × 14 × 14
Area = 2464 cm²
2. Find the surface area of a sphere of diameter: (i) 14 cm (ii) 21 cm (iii) 3.5 m
Formula: Surface Area = 4πr²
(i) Diameter = 14 cm → Radius = 7 cm:
Area = 4 × (22/7) × 7 × 7
Area = 616 cm²
(ii) Diameter = 21 cm → Radius = 10.5 cm:
Area = 4 × (22/7) × 10.5 × 10.5
Area = 1386 cm²
(iii) Diameter = 3.5 m → Radius = 1.75 m:
Area = 4 × (22/7) × 1.75 × 1.75
Area = 38.5 m²
3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Given:
Radius (r) = 10 cm
Formula: TSA = 3πr²
Solution:
TSA = 3 × 3.14 × 10 × 10
TSA = 3 × 314
TSA = 942 cm²
4. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Given:
Initial radius (r₁) = 7 cm
Final radius (r₂) = 14 cm
Solution:
Surface Area Ratio = (4πr₁²) / (4πr₂²)
Ratio = (r₁/r₂)²
Ratio = (7/14)²
Ratio = (1/2)²
Ratio = 1/4
Answer: Ratio is 1:4
5. 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 ₹ 16 per 100 cm².
Given:
Diameter = 10.5 cm
Radius (r) = 5.25 cm
Rate = ₹ 16 per 100 cm²
Step 1: Inner CSA:
CSA = 2πr²
CSA = 2 × (22/7) × 5.25 × 5.25
CSA = 173.25 cm²
Step 2: Calculate Cost:
Cost = (173.25 / 100) × 16
Cost = ₹ 27.72
6. Find the radius of a sphere whose surface area is 154 cm².
Given:
Surface Area = 154 cm²
Solution:
4πr² = 154
4 × (22/7) × r² = 154
(88/7) × r² = 154
r² = (154 × 7) / 88
r² = 49/4
r = √(49/4)
r = 7/2
r = 3.5 cm
7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Given:
Let diameter of earth = d
Diameter of moon = d/4
Radius of earth = R
Radius of moon = R/4
Solution:
Ratio = (Area of Moon) / (Area of Earth)
Ratio = (4π(R/4)²) / (4πR²)
Ratio = (R²/16) / R²
Ratio = 1/16
Answer: Ratio is 1:16
8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Given:
Inner radius = 5 cm
Thickness = 0.25 cm
Step 1: Find outer radius:
Outer radius (r) = 5 + 0.25 = 5.25 cm
Step 2: Outer CSA:
CSA = 2πr²
CSA = 2 × (22/7) × 5.25 × 5.25
CSA = 173.25 cm²
9. A right circular cylinder just encloses a sphere of radius r. Find (i) surface area of the sphere, (ii) curved surface area of the cylinder, (iii) ratio of the areas obtained in (i) and (ii).
Given:
Radius of sphere = r
For cylinder enclosing sphere: Radius = r, Height = 2r
(i) Surface Area of sphere:
Surface Area = 4πr²
(ii) CSA of cylinder:
CSA = 2πrh
CSA = 2πr(2r)
CSA = 4πr²
(iii) Ratio:
Ratio = (4πr²) / (4πr²)
Ratio = 1:1
Exercise 11.3
1. Find the volume of the right circular cone with (i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm.
Formula: V = (1/3)πr²h
(i) r = 6 cm, h = 7 cm:
V = (1/3) × (22/7) × 6 × 6 × 7
V = 22 × 2 × 6
V = 264 cm³
(ii) r = 3.5 cm, h = 12 cm:
V = (1/3) × (22/7) × 3.5 × 3.5 × 12
V = 22 × 0.5 × 3.5 × 4
V = 154 cm³
2. Find the capacity in litres of a conical vessel with (i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm.
(i) r = 7 cm, l = 25 cm:
Step 1: Find height:
h = √(l² – r²)
h = √(25² – 7²)
h = √(625 – 49)
h = √576 = 24 cm
Step 2: Volume:
V = (1/3)πr²h
V = (1/3) × (22/7) × 7 × 7 × 24
V = 1232 cm³
Step 3: Convert to litres:
Capacity = 1232 / 1000 = 1.232 litres
(ii) h = 12 cm, l = 13 cm:
Step 1: Find radius:
r = √(l² – h²)
r = √(13² – 12²)
r = √(169 – 144)
r = √25 = 5 cm
Step 2: Volume:
V = (1/3)πr²h
V = (1/3) × (22/7) × 5 × 5 × 12
V = 2200/7 cm³ ≈ 314.28 cm³
Step 3: Convert to litres:
Capacity = (2200/7) / 1000 = 11/35
Capacity = 0.314 litres (approx)
3. The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)
Given:
Volume = 1570 cm³
Height (h) = 15 cm
Solution:
(1/3) × 3.14 × r² × 15 = 1570
3.14 × 5 × r² = 1570
15.7 × r² = 1570
r² = 1570 / 15.7
r² = 100
r = √100
r = 10 cm
4. If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.
Given:
Volume = 48π cm³
Height (h) = 9 cm
Solution:
(1/3)πr²h = 48π
(1/3) × r² × 9 = 48
3r² = 48
r² = 16
r = 4 cm
Diameter = 2 × 4 = 8 cm
5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Given:
Diameter = 3.5 m
Radius (r) = 1.75 m
Depth/Height (h) = 12 m
Solution:
Volume = (1/3)πr²h
Volume = (1/3) × (22/7) × 1.75 × 1.75 × 12
Volume = 22 × 0.25 × 1.75 × 4
Volume = 38.5 m³
Since 1 m³ = 1 kilolitre:
Capacity = 38.5 kilolitres
6. The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find (i) height of the cone (ii) slant height of the cone (iii) curved surface area of the cone.
Given:
Volume = 9856 cm³
Diameter = 28 cm
Radius (r) = 14 cm
(i) Find height:
(1/3) × (22/7) × 14 × 14 × h = 9856
(1/3) × 22 × 2 × 14 × h = 9856
(616/3) × h = 9856
h = (9856 × 3) / 616
h = 48 cm
(ii) Find slant height:
l = √(r² + h²)
l = √(14² + 48²)
l = √(196 + 2304)
l = √2500
l = 50 cm
(iii) Curved Surface Area:
CSA = πrl
CSA = (22/7) × 14 × 50
CSA = 22 × 2 × 50
CSA = 2200 cm²
7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Given:
Sides of triangle: 5 cm, 12 cm, 13 cm
Revolving about 12 cm side
Analysis:
Height (h) = 12 cm (axis of rotation)
Radius (r) = 5 cm (perpendicular side)
Solution:
Volume = (1/3)πr²h
Volume = (1/3) × π × 5 × 5 × 12
Volume = 100π cm³ (or approx 314.28 cm³)
8. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
Given:
Revolving about 5 cm side
Height (h) = 5 cm (axis of rotation)
Radius (r) = 12 cm (perpendicular side)
Step 1: Calculate Volume:
Volume = (1/3)πr²h
Volume = (1/3) × π × 12 × 12 × 5
Volume = 240π cm³
Step 2: Find Ratio:
Ratio (Q7 Vol : Q8 Vol) = 100π : 240π
Ratio = 100 : 240
Ratio = 10 : 24
Ratio = 5:12
9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Given:
Diameter = 10.5 m
Radius (r) = 5.25 m
Height (h) = 3 m
Step 1: Volume:
Volume = (1/3)πr²h
Volume = (1/3) × (22/7) × 5.25 × 5.25 × 3
Volume = 22 × 0.75 × 5.25
Volume = 86.625 m³
Step 2: Find slant height for canvas area:
l = √(r² + h²)
l = √(5.25² + 3²)
l = √(27.5625 + 9)
l = √36.5625
l ≈ 6.05 m
Step 3: Canvas Area (CSA):
Area = πrl
Area = (22/7) × 5.25 × 6.05
Area = 99.825 m² (approx)
Exercise 11.4
1. Find the volume of a sphere whose radius is (i) 7 cm (ii) 0.63 m.
Formula: Volume = (4/3)πr³
(i) r = 7 cm:
V = (4/3) × (22/7) × 7 × 7 × 7
V = 1437.33 cm³ (approx)
(ii) r = 0.63 m:
V = (4/3) × (22/7) × 0.63 × 0.63 × 0.63
V = 1.05 m³ (approx)
2. Find the amount of water displaced by a solid spherical ball of diameter (i) 28 cm (ii) 0.21 m.
Note: Water displaced = Volume of sphere
(i) Diameter = 28 cm, Radius = 14 cm:
V = (4/3) × (22/7) × 14 × 14 × 14
V = 11498.67 cm³
(ii) Diameter = 0.21 m, Radius = 0.105 m:
V = (4/3) × (22/7) × 0.105 × 0.105 × 0.105
V = 0.004851 m³
3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³?
Given:
Diameter = 4.2 cm
Radius (r) = 2.1 cm
Density = 8.9 g/cm³
Step 1: Calculate Volume:
Volume = (4/3) × (22/7) × 2.1 × 2.1 × 2.1
Volume = 38.808 cm³
Step 2: Calculate Mass:
Mass = Volume × Density
Mass = 38.808 × 8.9
Mass = 345.39 g (approx)
4. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Given:
Let Earth radius = R
Moon radius = R/4
Solution:
Volume of Earth = (4/3)πR³
Volume of Moon = (4/3)π(R/4)³ = (4/3)π(R³/64)
Fraction = Volume of Moon / Volume of Earth
Fraction = [(4/3)π(R³/64)] / [(4/3)πR³]
Fraction = 1/64
Answer: 1/64
5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Given:
Diameter = 10.5 cm
Radius (r) = 5.25 cm
Step 1: Calculate Volume:
Volume = (2/3)πr³
Volume = (2/3) × (22/7) × 5.25 × 5.25 × 5.25
Volume = 303.1875 cm³
Step 2: Convert to litres:
Capacity = 303.1875 / 1000
Capacity = 0.303 litres (approx)
6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
Given:
Inner radius (r) = 1 m
Thickness = 1 cm = 0.01 m
Outer radius (R) = 1 + 0.01 = 1.01 m
Solution:
Volume of iron = Outer Volume – Inner Volume
Volume = (2/3)πR³ – (2/3)πr³
Volume = (2/3) × (22/7) × [(1.01)³ – 1³]
Volume = (44/21) × [1.030301 – 1]
Volume = (44/21) × 0.030301
Volume = 0.06348 m³ (approx)
7. Find the volume of a sphere whose surface area is 154 cm².
Given:
Surface Area = 154 cm²
Step 1: Find radius:
4πr² = 154
4 × (22/7) × r² = 154
r² = (154 × 7) / 88
r² = 12.25
r = 3.5 cm
Step 2: Calculate Volume:
Volume = (4/3)πr³
Volume = (4/3) × (22/7) × 3.5 × 3.5 × 3.5
Volume = 179.67 cm³
8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹ 4989.60. If the cost of white-washing is ₹ 20 per square metre, find the (i) inside surface area of the dome, (ii) volume of the air inside the dome.
Given:
Total Cost = ₹ 4989.60
Rate = ₹ 20 per m²
(i) Inside Surface Area:
Area = Total Cost / Rate
Area = 4989.60 / 20
Area = 249.48 m²
(ii) Volume of air inside:
First, find radius:
2πr² = 249.48
2 × (22/7) × r² = 249.48
r² = (249.48 × 7) / 44
r² = 39.69
r = 6.3 m
Volume = (2/3)πr³
Volume = (2/3) × (22/7) × 6.3 × 6.3 × 6.3
Volume = 523.9 m³ (approx)
9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’. Find the (i) radius r’ of the new sphere, (ii) ratio of S and S’.
Given:
Number of small spheres = 27
Radius of each small sphere = r
Surface area of small sphere = S
(i) Find radius r’ of new sphere:
Volume of 27 small spheres = Volume of 1 big sphere
27 × (4/3)πr³ = (4/3)π(r’)³
27r³ = (r’)³
r’ = ∛(27r³)
r’ = 3r
(ii) Ratio S : S’:
S = 4πr²
S’ = 4π(r’)² = 4π(3r)² = 4π(9r²) = 36πr²
Ratio = S / S’
Ratio = 4πr² / 36πr²
Ratio = 1/9
Answer: Ratio is 1:9
10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule?
Given:
Diameter = 3.5 mm
Radius (r) = 1.75 mm
Solution:
Volume = (4/3)πr³
Volume = (4/3) × (22/7) × 1.75 × 1.75 × 1.75
Volume = 22.46 mm³ (approx)