CDC Mathematics (Book 9) Solution
Cylinder and Sphere
Surface area of the cylinder
Exercise 7.1
1. If radius and height of cylinder is x cm and y cm, find:
(a) Circumference of base
(b) Area of the base
(c) Curved surface area
(d) Total surface area
(e) Volume
Solution:
Here,
Radius of the cylinder (r) = x cm
Height of the cylinder (h) = y cm
(a) Circumference of base = ?
(b) Area of the base = ?
(c) Curved surface area = ?
(d) Total surface area = ?
(e) Volume of cylinder = ?
By formula,
(a) Circumference of base = 2Ï€r
= 2Ï€x cm
(b) Area of the base = Ï€r²
= Ï€x² cm²
(c) Curved surface area = 2Ï€rh
= 2Ï€xy cm²
(d) Total surface area = 2Ï€r(r + h)
= 2Ï€x(x + y) cm²
(e) Volume of cylinder = Ï€r²h
= Ï€x²y cm³
2. If radius of cylinder is x cm and height is y cm, then find curved surface area and volume of the prism:
(a) Curved surface area
(b) Volume
Solution:
Here,
Radius of the cylinder (r) = x cm
Height of the cylinder (h) = y cm
(a) Curved surface area = ?
(b) Volume of cylinder = ?
By formula,
(a) Curved surface area = 2Ï€rh
= 2Ï€xy cm²
(b) Volume of cylinder = Ï€r²h
= Ï€x²y cm³
5. If the radius of the cylinder is r unit and height h unit, find:
(a) Volume
(b) Curved surface area
(c) Plane surface area
(d) Total surface area
Solution:
Here,
Radius of the cylinder (r) = r unit
Height of the cylinder (h) = h unit
(a) Volume of cylinder = ?
(b) Curved surface area = ?
(c) Plane surface area = ?
(d) Total surface area = ?
By formula,
(a) Volume of cylinder = Ï€r²h
= Ï€r²h cubic unit
(b) Curved surface area = 2Ï€rh
= 2Ï€rh sq. unit
(c)Plane surface area (both bases) = 2Ï€r²
= 2Ï€r² sq. unit
(d) Total surface area = 2Ï€r(r + h)
= 2Ï€r(r + h) sq. unit
6. The circumference of the base of a cylinder is C unit and the sum of radius and height is S unit, find the total surface area of the cylinder:
(a) Total surface area
Solution:
Here,
Circumference of the base = C unit
Sum of radius and height (r + h) = S unit
(a) Total surface area = ?
Circumference of the base = 2Ï€r
or, C = 2Ï€r
or, r = C / (2Ï€) unit
By formula,
(a) Total surface area = 2Ï€r(r + h)
= 2Ï€ (C / (2Ï€)) (S)
= C S sq. unit
7(a). Find the plane surface area, curved surface area, and volume of the following prism: radius (r) = 7 cm and height (h) = 5 cm:
(a) Plane surface area
(b) Curved surface area
(c) Volume
Solution:
Here,
Radius of the cylinder (r) = 7 cm
Height of the cylinder (h) = 5 cm
(a) Plane surface area = ?
(b) Curved surface area = ?
(c) Volume of cylinder = ?
By formula,
(a) Plane surface area (both bases) = 2Ï€r²
= 2 x 22/7 × (7)²
= 2 x 22/7 × 49
= 308 cm²
(b) Curved surface area = 2Ï€rh
= 2 × 22/7 × 7 × 5
= 44 × 5
= 220 cm²
(c) Volume of cylinder = Ï€r²h
= 22/7 × (7)² × 5
= 22/7 × 49 × 5
= 154 × 5
= 770 cm³
7(b). Find the plane surface area, curved surface area, and volume of the following prism: radius (r) = 3.5 m and height (h) = 6 m:
(a) Plane surface area
(b) Curved surface area
(c) Volume
Solution:
Here,
Radius of the cylinder (r) = 3.5 m
Height of the cylinder (h) = 6 m
(a) Plane surface area = ?
(b) Curved surface area = ?
(c) Volume of cylinder = ?
By formula,
(a) Plane surface area (both bases) = 2Ï€r²
= 2 x 22/7 × (3.5)²
= 2 x 22/7 × 12.25
= 77 m²
(b) Curved surface area = 2Ï€rh
= 22/7 × 3.5 × 6
= 22 × 0.5 × 6
= 44 × 3
= 132 m²
(c) Volume of cylinder = Ï€r²h
= 22/7 × (3.5)² × 6
= 22/7 × 12.25 × 6
= 22 × 1.75 × 6
= 38.5 × 6
= 231 m³
7(c). Find the plane surface area, curved surface area, and volume of the following prism: radius (r) = 2 ft and height (h) = 7 ft:
(a) Plane surface area
(b) Curved surface area
(c) Volume
Solution:
Here,
Radius of the cylinder (r) = 2 ft
Height of the cylinder (h) = 7 ft
(a) Plane surface area = ?
(b) Curved surface area = ?
(c) Volume of cylinder = ?
By formula,
(a) Plane surface area (both bases) =2Ï€r²
= 2 × 22/7 × (2)²
= 2 × 22/7 × 4
= 25.14 ft²
(b) Curved surface area = 2Ï€rh
= 2 × 22/7 × 2 × 7
= 44 × 2
= 88 ft²
(c) Volume of cylinder = Ï€r²h
= 22/7 × (2)² × 7
= 22/7 × 4 × 7
= 22 × 4
= 88 ft³
8(a). Find the area of the base, plane surface area, curved surface area, total surface area, and volume of the following prism:
(a) Area of the base
(b) Plane surface area
(c) Curved surface area
(d) Total surface area
(e) Volume
Solution:
Here,
Diameter of the cylinder = 14 cm
Radius of the cylinder (r) = 14 / 2 = 7 cm
Height of the cylinder (h) = 12 cm
(a) Area of the base = ?
(b) Plane surface area = ?
(c) Curved surface area = ?
(d) Total surface area = ?
(e) Volume of cylinder = ?
By formula,
(a) Area of the base = Ï€r²
= 22/7 × (7)²
= 22/7 × 49
= 154 cm²
(b) Plane surface area (both bases) = 2Ï€r²
= 2 x 22/7 × (7)²
= 2 x 22/7 × 49
= 308 cm²
(c) Curved surface area = 2Ï€rh
= 2 × 22/7 × 7 × 12
= 44 × 12
= 528 cm²
(d) Total surface area = 2Ï€r(r + h)
= 2 × 22/7 × 7 × (7 + 12)
= 44 × 19
= 836 cm²
(e) Volume of cylinder = Ï€r²h
= 22/7 × (7)² × 12
= 22/7 × 49 × 12
= 154 × 12
= 1848 cm³
8(b). Find the area of the base, plane surface area, curved surface area, total surface area, and volume of the following prism: diameter = 14 cm and height = 21 cm:
(a) Area of the base
(b) Plane surface area
(c) Curved surface area
(d) Total surface area
(e) Volume
Solution:
Here,
Diameter of the cylinder = 14 cm
Radius of the cylinder (r) = 14 / 2 = 7 cm
Height of the cylinder (h) = 21 cm
(a) Area of the base = ?
(b) Plane surface area = ?
(c) Curved surface area = ?
(d) Total surface area = ?
(e) Volume of cylinder = ?
By formula,
(a) Area of the base = Ï€r²
= 22/7 × (7)²
= 22/7 × 49
= 154 cm²
(b) Plane surface area (both bases) = 2Ï€r²
= 2 × 22/7 × (7)²
= 2 × 22/7 × 49
= 2 × 154
= 308 cm²
(c) Curved surface area = 2Ï€rh
= 2 × 22/7 × 7 × 21
= 44 × 21
= 924 cm²
(d) Total surface area = 2Ï€r(r + h)
= 2 × 22/7 × 7 × (7 + 21)
= 44 × 28
= 1232 cm²
(e) Volume of cylinder = Ï€r²h
= 22/7 × (7)² × 21
= 22/7 × 49 × 21
= 154 × 21
= 3234 cm³
8(c). Find the area of the base, plane surface area, curved surface area, total surface area, and volume of the following prism: radius = 7 cm and height = 14 cm:
(a) Area of the base
(b) Plane surface area
(c) Curved surface area
(d) Total surface area
(e) Volume
Solution:
Here,
Radius of the cylinder (r) = 7 cm
Height of the cylinder (h) = 14 cm
(a) Area of the base = ?
(b) Plane surface area = ?
(c) Curved surface area = ?
(d) Total surface area = ?
(e) Volume of cylinder = ?
By formula,
(a) Area of the base = Ï€r²
= 22/7 × (7)²
= 22/7 × 49
= 154 cm²
(b) Plane surface area (both bases) = 2Ï€r²
= 2 × 22/7 × (7)²
= 2 × 22/7 × 49
= 2 × 154
= 308 cm²
(c) Curved surface area = 2Ï€rh
= 2 × 22/7 × 7 × 14
= 44 × 14
= 616 cm²
(d) Total surface area = 2Ï€r(r + h)
= 2 × 22/7 × 7 × (7 + 14)
= 44 × 21
= 924 cm²
(e) Volume of cylinder = Ï€r²h
= 22/7 × (7)² × 14
= 22/7 × 49 × 14
= 154 × 14
= 2156 cm³
8(d). Find the area of the base, plane surface area, curved surface area, total surface area, and volume of the following prism: radius = 2.8 cm and height = 21 cm:
(a) Area of the base
(b) Plane surface area
(c) Curved surface area
(d) Total surface area
(e) Volume
Solution:
Here,
Radius of the cylinder (r) = 2.8 cm
Height of the cylinder (h) = 21 cm
(a) Area of the base = ?
(b) Plane surface area = ?
(c) Curved surface area = ?
(d) Total surface area = ?
(e) Volume of cylinder = ?
By formula,
(a) Area of the base = Ï€r²
= 22/7 × (2.8)²
= 22/7 × 7.84
= 22 × 1.12
= 24.64 cm²
(b) Plane surface area (both bases) = 2Ï€r²
= 2 × 22/7 × (2.8)²
= 2 × 22/7 × 7.84
= 2 × 22 × 1.12
= 2 × 24.64
= 49.28 cm²
(c) Curved surface area = 2Ï€rh
= 2 × 22/7 × 2.8 × 21
= 44 × 0.4 × 21
= 44 × 8.4
= 369.6 cm²
(d) Total surface area = 2Ï€r(r + h)
= 2 × 22/7 × 2.8 × (2.8 + 21)
= 44 × 0.4 × 23.8
= 44 × 9.52
= 418.88 cm²
(e) Volume of cylinder = Ï€r²h
= 22/7 × (2.8)² × 21
= 22/7 × 7.84 × 21
= 22 × 1.12 × 21
= 24.64 × 21
= 517.44 cm³
9. If the circumference of the base of a cylinder is 176 cm, and the height is 30 cm, find the curved surface area and total surface area of the cylinder:
(a) Curved surface area
(b) Total surface area
Solution:
Here,
Circumference of the base = 176 cm
Height of the cylinder (h) = 30 cm
(a) Curved surface area = ?
(b) Total surface area = ?
Circumference of the base = 2Ï€r
or, 176 = 2 × 22/7 × r
or, r = 176 × 7 / (2 × 22)
or, r = 1232 / 44
or, r = 28 cm
By formula,
(a) Curved surface area = 2Ï€rh
= 2 × 22/7 × 28 × 30
= 44 × 4 × 30
= 5280 cm²
(b) Total surface area = 2Ï€r(r + h)
= 2 × 22/7 × 28 × (28 + 30)
= 44 × 4 × 58
= 10208 cm²
10. If the circumference of the base of a cylinder is 88 cm, and the sum of radius and height is 24 cm, then find:
(a) Area of the base
(b) Curved surface area
(c) Total surface area
(d) Volume
Solution:
Here,
Circumference of the base = 88 cm
Sum of radius and height (r + h) = 24 cm
(a) Area of the base = ?
(b) Curved surface area = ?
(c) Total surface area = ?
(d) Volume of cylinder = ?
Circumference of the base = 2Ï€r
or, 88 = 2 × 22/7 × r
or, r = 88 × 7 / (2 × 22)
or, r = 616 / 44
or, r = 14 cm
Since r + h = 24 cm,
or, 14 + h = 24
or, h = 24 - 14
or, h = 10 cm
By formula,
(a) Area of the base = Ï€r²
= 22/7 × (14)²
= 22/7 × 196
= 22 × 28
= 616 cm²
(b) Curved surface area = 2Ï€rh
= 2 × 22/7 × 14 × 10
= 44 × 2 × 10
= 880 cm²
(c) Total surface area = 2Ï€r(r + h)
= 2 × 22/7 × 14 × (14 + 10)
= 44 × 2 × 24
= 2112 cm²
(d) Volume of cylinder = Ï€r²h
= 22/7 × (14)² × 10
= 22/7 × 196 × 10
= 22 × 28 × 10
= 6160 cm³
11. If the sum of the radius and height of the cylinder is 34 cm, and the total surface area is 2992 cm², then find the volume of the prism:
(a) Volume
Solution:
Here,
Sum of radius and height (r + h) = 34 cm
Total surface area = 2992 cm²
(a) Volume of cylinder = ?
Total surface area = 2Ï€r(r + h)
or, 2992 = 2 × 22/7 × r × 34
or, 2992 = 44/7 × r × 34
or, 2992 = 44 × 34/7 × r
or, 2992 = 1496/7 × r
or, r = 2992 × 7 / 1496
or, r = 20944 / 1496
or, r = 14 cm
Since r + h = 34 cm,
or, 14 + h = 34
or, h = 34 - 14
or, h = 20 cm
By formula,
(a) Volume of cylinder = Ï€r²h
= 22/7 × (14)² × 20
= 22/7 × 196 × 20
= 22 × 28 × 20
= 12320 cm³
12. The sum of the diameter and height of the cylinder is 28 cm, and the curved surface area is 462 cm², find the total surface area.
Solution:
Given: 2r + h = 28 cm,
or, h = 28 - 2r
CSA = 462 cm²
We know that:
CSA = 2×(22/7)×r×h
or, 462 = 2×(22/7)×r×h
or, 462 x 7/44 = r (28 - 2r)
or, 147/2 = 28r - 2r²
or, 2(2r² - 28r) + 147 = 0
or, 4r² - 56r + 147 = 0
or, 4r² - 42r - 14r + 147 = 0
2r(2r - 21) - 7 (r - 21) = 0
(2r - 21) (2r - 7) = 0
Either (2r - 21)= 0 ...... (i) or (2r - 7) = 0 .....(ii)
From the equation (i),
2r - 21= 0
or, r = 21/2 = 10.5 cm
From the equation (ii),
2r - 7 = 0
or, r = 3.5 cm
Since r = 3.5, h= (28 - 2r)= 28 - 21= 7 cm
For Case 1, r = 10.5 cm, h = 7 cm
TSA = 2×(22/7) × 10.5× 7 = 1155 cm²
For case 2, r = 3.5 cm → h = 28 - 7 = 21 cm
TSA = 2×(22/7)×3.5 × 21 = 539 cm²
Hence, TSA = 1155 cm² or 539 cm²
13. The ratio of the radius of the base and height of a cylinder is 1:3, and their sum is 56 cm, find the curved surface area and total surface area of the cylinder:
(a) Curved surface area
(b) Total surface area
Solution:
Here,
Ratio of radius to height (r:h) = 1:3
r = x (say), then h = 3x
Sum of radius and height (r + h) = 56 cm
x + 3x = 56
or, 4x = 56
or, x = 56 / 4
or, x = 14 cm
So, r = 14 cm, h = 3 × 14 = 42 cm
By formula,
(a) Curved surface area = 2Ï€rh
= 2 × 22/7 × 14 × 42
= 44 × 2 × 42
= 3696 cm²
(b) Total surface area = 2Ï€r(r + h)
= 2 × 22/7 × 14 × (14 + 42)
= 44 × 2 × 56
= 4928 cm²
14. The ratio of the radius of the base and height of a cylinder is 1:3 and its curved surface area is 924 cm², find the area of the base and volume of the cylinder:
(a) Area of the base
(b) Volume
Solution:
Here,
Ratio of radius to height (r:h) = 1:3
r = x (say), then h = 3x
Curved surface area = 924 cm²
Curved surface area = 2Ï€rh
or, 924 = 2 × 22/7 × x × 3x
or, 924 = 44/7 × 3x²
or, 924 = 132/7 × x²
or, x² = 924 × 7 / 132
or, x² = 6474 / 132
or, x² = 49
or, x = √49
or, x = 7 cm
So, r = 7 cm, h = 3 × 7 = 21 cm
By formula,
(a) Area of the base = Ï€r²
= 22/7 × (7)²
= 22/7 × 49
= 154 cm²
(b) Volume of cylinder = Ï€r²h
= 22/7 × (7)² × 21
= 22/7 × 49 × 21
= 154 × 21
= 3234 cm³
15. The curved surface area of a cylindrical log is three times more than the area of the base. If the sum of the radius of the base and height of the log is 25 cm, find the volume of the cylindrical log.
Solution:
Here,
Curved surface area = 3 × Area of the base
Sum of radius and height (r + h) = 25 cm
(a) Volume of cylinder = ?
We know that,
Curved surface area = 2Ï€rh = 3Ï€r²
2h = 3r
or, h = 3r/2
Substituting h in r + h = 25,
r + (3r/2) = 25
or, 5r/2 = 25
or, r = 10 cm
So, h = (3×10)/2 = 15 cm
By formula,
Volume of cylinder = Ï€r²h
= 22/7 × (10)² × 15
= 22/7 × 100 × 15
= 33000/7 cm³
= 4714.29 cm³
16. The inner circumference of the following cylindrical pots are 21 cm and 14 cm and the heights are 14 cm and 21 cm respectively. Find how much water can be filled in these pots.
Solution:
Here,
Pot 1: Inner circumference = 21 cm, Height (h₁) = 14 cm
Pot 2: Inner circumference = 14 cm, Height (h₂) = 21 cm
(a) Total volume of water = ?
For Pot 1:
Inner circumference = 2Ï€r₁
21 = 2 × 22/7 × r₁
or, r₁ = 21 × 7 / 44 = 3.3409 cm ≈ 3.34 cm
For Pot 2:
Inner circumference = 2Ï€r₂
14 = 2 × 22/7 × r₂
or, r₂ = 14 × 7 / 44 = 2.2273 cm ≈ 2.23 cm
By formula,
Volume of Pot 1 = Ï€r₁²h₁
= 22/7 × (3.3409)² × 14 ≈ 491.07 cm³ = 0.491 l
Volume of Pot 2 = Ï€r₂²h₂
= 22/7 × (2.2273)² × 21 ≈ 327.38 cm³ = 0.327 l
Hence, total volume of water = Volume of Pot 1 + Volume of Pot 2
= 0.491 l + 0.327 l
= 0.818 l
(Note: Since the translation is incorrect, the answer we obtain doesn't match the textbook solution.)
The question should be written as follows:
16. The inner circumference of the following cylindrical pots are 21 cm and 14 cm and the heights are 14 cm and 21 cm respectively. Find which pot can hold more water.?
Difference = Pot 1 - Pot 2
The difference in volume is 0.164 l 0.164 l (or 163.69 cm³)Hence, Pot 1 holds more water than Pot 2.
17. The students in a school are requested to participate in the competition of making cylindrical pencil cases. The radius and height of each pencil case must be 3 cm and 10.5 cm respectively. If 35 students participate in the competition, find how much cardboard paper was used in that competition.
Solution:
Here,
Radius of each pencil case (r) = 3 cm
Height (h) = 10.5 cm
Number of competitors = 35
Total cardboard required = ?
Since the pencil case is a cylinder with a base (open at the top), the cardboard required for one penholder is the curved surface area plus the area of the base:
Curved surface area = 2Ï€rh
= 2 × 22/7 × 3 × 10.5
= 198 cm²
Area of the base = Ï€r²
= 22/7 × 3 × 3
= 198/7 cm²
Curved surface area of the penholder + area of the base is:
= 198 + 198/7
= 1584/7 cm²
There are total 35 competitors, so the total cardboard required by them for the competition is:
= 1584/7 × 35
= 7920 cm²
Hence, total cardboard required = 7920 cm²
18. A hospital provides mushroom soup on a cylindrical glass with a 7 cm diameter for his patient each day. If the mushroom soup filled the 6 cm height of the glass, find how much soup is prepared by the hospital daily for their 250 patients:
Solution:
Here,
Radius of the cylindrical glass (r) = 7 cm = 3.5 cm
Height of soup in the glass (h) = 6 cm
Number of patients = 250
(a) Total soup prepared daily = ?
Volume of soup in one glass = Ï€r²h
= 22/7 × (3.5)² × 6
= 231 cm³
Since 1 liter = 1000 cm³,231 cm³ = 0. 231 liters
Total soup prepared daily = 0.231 × 250 = 57.75 l
19. Mankumari planned to prepare a cylindrical bucket from the mat to store 1.4 cu. meter crops. If the breadth of the mat is 1.1 meter, then how long mat is needed to store 1.4 cu. meter crops:
Solution:
Here,
Volume of the cylindrical bucket = 1.4 m³
Breadth of the mat = 1.1 m (height of the cylinder, h)
Length of the mat = ?
Volume of the cylinder = Ï€r²h
or, 1.4 = 22/7 × r² × 1.1
or, r² = 1.4 × 7 / (22 × 1.1)
or, r² = 9.8 / 24.2
or, r² = 0.405
or, r = √0.405
or, r ≈ 0.636 m
Now, length of the mat = 2Ï€r
= 2 × 22/7 × 0.636
= 44/7 × 0.636
= 4 cm
Hence, length of the mat = 4 m
Surface area of sphere
Surface area of hemisphere
Exercise 7.2
1. If the radius of a sphere is x unit, then find:
Solution,
Here, radius of sphere (r) = x unit
a) Surface area of sphere = ?
b) Volume of sphere = ?
c) Circumference of the large circle = ?
d) Area of the large circle = ?
We know that,
a) Surface area of sphere (A) = 4Ï€r²
= 4Ï€x² sq. unit
b) Volume of sphere (V) = 4/3 Ï€r³
= 4/3 Ï€x³ cu. unit
c) Circumference of the large circle (C) = 2Ï€r
= 2Ï€x unit
d) Area of the large circle = Ï€r²
= Ï€x² sq. unit
2. If the area of the large circle of a sphere is a sq. unit, then find the surface area of the sphere:
Solution,
Here, area of the large circle = a sq. unit
Surface area of sphere = ?
We know that,
Area of the large circle = Ï€r²
or, a = Ï€r²
or, r² = a/Ï€
Surface area of sphere (A) = 4Ï€r²
= 4Ï€ × (a/Ï€)
= 4a sq. unit
3. If the surface area of a sphere is 4x sq. unit, find the total surface area of one hemisphere formed from the sphere:
Solution,
Here, surface area of sphere = 4x sq. unit
Total surface area of one hemisphere = ?
We know that,
Surface area of sphere (A) = 4Ï€r²
or, 4x = 4Ï€r²
or, r² = x/Ï€
Total surface area of hemisphere = 3Ï€r²
= 3Ï€ × (x/Ï€)
= 3x sq. unit
4. If the area of the large circle is y sq. unit, find the total surface area of the sphere:
Solution,
Here, area of the large circle = y sq. unit
Total surface area of sphere = ?
We know that,
Area of the large circle = Ï€r²
or, y = Ï€r²
or, r² = y/Ï€
Total surface area of sphere (A) = 4Ï€r²
= 4Ï€ × (y/Ï€)
= 4y sq. unit
5. If the diameter of a sphere is d cm, find:
Solution,
Here, diameter of sphere (d) = d cm
radius of sphere (r) = d/2 cm
a) Circumference of the large circle = ?
b) Area of the large circle = ?
c) Volume of sphere = ?
d) Surface area of sphere = ?
We know that,
a) Circumference of the large circle (C) = 2Ï€r
= 2Ï€ × (d/2)
= πd cm
b) Area of the large circle = Ï€r²
= Ï€ × (d/2)²
= Ï€d²/4 cm²
c) Volume of sphere (V) = 4/3 Ï€r³
= 4/3 Ï€ × (d/2)³
= Ï€d³/6 cm³
d) Surface area of sphere (A) = 4Ï€r²
= 4Ï€ × (d/2)²
= Ï€d² cm²
6. Find, how many materials (TPE lather - Thermoplastic Elastomer Lather) are needed to make a volleyball of the size given in the picture.
Solution,
Here, radius of the volleyball (r) = 10.5 cm
Surface area of volleyball = ?
We know that,
Surface area of sphere (A) = 4Ï€r²
= 4 × 22/7 × (10.5)² cm²
= 1386 cm²
So, surface area of volleyball (A) = 1386 cm²
7. Kazakhstan Pavilion and Science of Noor Alam, Kazakhstan is in a spherical shape. A science exhibition of 100 countries was organized in this hall in 2017. If the diameter of the spherical hall is 80 meters and the outer surface of the hall is covered by glass, find the surface area of the hall covered by the hall.
Solution,
Here, diameter of the spherical hall (d) = 80 m
radius of the spherical hall (r) = d/2 = 80/2 = 40 m
Surface area of the hall = ?
We know that,
Surface area of sphere (A) = 4Ï€r²
= 4 × 22/7 × (40)² m²
= 4 × 22/7 × 1600 m²
= 201600/7 m²
= 20114 m²
So, surface area of the hall (A) = 20114 m²
8. Find the surface area and volume of the given spheres and hemispheres:
Solution,
a) For sphere with radius 7 cm
Here, radius of sphere (r) = 7 cm
Surface area of sphere = ?
Volume of sphere = ?
Surface area of sphere (A) = 4Ï€r²
= 4 × 22/7 × (7)² cm²
= 4 × 22/7 × 49 cm²
= 616 cm²
Volume of sphere (V) = 4/3 Ï€r³
= 4/3 × 22/7 × (7)³ cm³
= 4/3 × 22/7 × 343 cm³
= 4312/3 cm³
= 1437.33 cm³
So, surface area = 616 cm² and volume = 1437.33 cm³
b) For sphere with diameter 14 cm
Here, diameter of sphere (d) = 14 cm
radius of sphere (r) = d/2 = 14/2 = 7 cm
Surface area of sphere = ?
Volume of sphere = ?
Surface area of sphere (A) = 4Ï€r²
= 4 × 22/7 × (7)² cm²
= 4 × 22/7 × 49 cm²
= 616 cm²
Volume of sphere (V) = 4/3 Ï€r³
= 4/3 × 22/7 × (7)³ cm³
= 4/3 × 22/7 × 343 cm³
= 4312/3 cm³
= 1437.33 cm³
So, surface area = 616 cm² and volume = 1437.33 cm³
c) For hemisphere with diameter 28 cm
Here, diameter of hemisphere (d) = 28 cm
radius of hemisphere (r) = d/2 = 28/2 = 14 cm
Total surface area of hemisphere = ?
Volume of hemisphere = ?
Total surface area of hemisphere (TSA) = 3Ï€r²
= 3 × 22/7 × (14)² cm²
= 3 × 22/7 × 196 cm²
= 1848 cm²
Volume of hemisphere (V) = 2/3 Ï€r³
= 2/3 × 22/7 × (14)³ cm³
= 2/3 × 22/7 × 2744 cm³
= 17248/3 cm³
= 5749.33 cm³
So, total surface area = 1848 cm² and volume = 5749.33 cm³
d) For hemisphere with circumference of the base C = 88 cm
Here, circumference of the base (C) = 88 cm
radius of hemisphere (r) = ?
Total surface area of hemisphere = ?
Volume of hemisphere = ?
We know that,
Circumference of the base (C) = 2Ï€r
88 = 2 × 22/7 × r
r = 88 × 7 / (2 × 22)
r = 14 cm
Total surface area of hemisphere (TSA) = 3Ï€r²
= 3 × 22/7 × (14)² cm²
= 3 × 22/7 × 196 cm²
= 1848 cm²
Volume of hemisphere (V) = 2/3 Ï€r³
= 2/3 × 22/7 × (14)³ cm³
= 2/3 × 22/7 × 2744 cm³
= 17248/3 cm³
= 5749.33 cm³
So, total surface area = 1848 cm² and volume = 5749.33 cm³
9. The diameter of a spherical ball is 35 cm, find the surface area and volume of the ball.
Solution,
Here, diameter of the spherical ball (d) = 35 cm
radius of the spherical ball (r) = d/2 = 35/2 cm
Surface area of the ball = ?
Volume of the ball = ?
Surface area of sphere (A) = 4Ï€r²
= 4 × 22/7 × (35/2)² cm²
= 4 × 22/7 × 1225/4 cm²
= 3850 cm²
Volume of sphere (V) = 4/3 Ï€r³
= 4/3 × 22/7 × (35/2)³ cm³
= 4/3 × 22/7 × 42875/8 cm³
= 134750/3 cm³
= 44916.67 cm³
So, surface area = 3850 cm² and volume = 44916.67 cm³ (
10. The circumference of the large circle of a hemisphere is 44 cm. Find the total surface area of the hemisphere.
Solution,
Here, circumference of the large circle (C) = 44 cm
radius of hemisphere (r) = ?
Total surface area of hemisphere = ?
We know that,
Circumference of the large circle (C) = 2Ï€r
44 = 2 × 22/7 × r
or, r = 44 × 7 / (2 × 22)
or, r = 7 cm
Total surface area of hemisphere (TSA) = 3Ï€r²
= 3 × 22/7 × (7)² cm²
= 3 × 22/7 × 49 cm²
= 462 cm²
So, total surface area = 462 cm²
11. The total surface area of the spherical solid object is 2464 cm². Find the diameter of the object.
Solution,
Here, total surface area of the spherical object = 2464 cm²
radius of sphere (r) = ?
diameter of sphere (d) = ?
We know that,
Surface area of sphere (A) = 4Ï€r²
or, 2464 = 4 × 22/7 × r²
r² = 2464 × 7 / (4 × 22)
r² = 196
r = 14 cm
Diameter (d) = 2r
= 2 × 14 cm
= 28 cm
So, diameter = 28 cm
12. If the volume of a sphere is 38808 m³, find its radius.
Solution,
Here, volume of the sphere (V) = 38808 m³
radius of sphere (r) = ?
We know that,
Volume of sphere (V) = 4/3 Ï€r³
or, 38808 = 4/3 × 22/7 × r³
or, r³ = 38808 × 3 × 7 / (4 × 22)
or, r³ = 9261
or, r = ∛9261
or, r = 21 m
So, radius = 21 m
13. (a) If the total surface area of the hemisphere is 243Ï€ cm², find its volume.
Solution,
Here, total surface area of the hemisphere = 243Ï€ cm²
radius of hemisphere (r) = ?
volume of hemisphere = ?
We know that,
Total surface area of hemisphere (TSA) = 3Ï€r²
or, 243Ï€ = 3Ï€r²
or, r² = 243 / 3
or, r² = 81
or, r = 9 cm
Volume of hemisphere (V) = 2/3 Ï€r³
= 2/3 × 22/7 × (9)³ cm³
= 1527.43 cm³
So, volume = 486Ï€ cm³
(b) If the volume of a sphere is 2304Ï€ cm³, find the surface area of the sphere.
Solution,
Here, volume of the sphere (V) = 2304Ï€ cm³
radius of sphere (r) = ?
surface area of sphere = ?
We know that,
Volume of sphere (V) = 4/3 Ï€r³
2304Ï€ = 4/3 Ï€r³
or, r³ = 2304 × 3 / 4
or, r³ = 1728
or, r = 12 cm
Surface area of sphere (A) = 4Ï€r²
= 4 × Ï€ × (12)² cm²
= 4 × Ï€ × 144 cm²
= 576Ï€ cm²
So, surface area = 576Ï€ cm²
14. The diameter of the moon is approximately one-fourth of the diameter of the sun. Find the ratio of their surface areas.
Solution,
Here, diameter of the moon = d₁
diameter of the sun = d₂
d₁/d₂ = 1/4
radius of the moon (r₁) = d₁/2
radius of the sun (r₂) = d₂/2
r₁/r₂ = (d₁/2) / (d₂/2) = d₁/d₂ = 1/4
Surface area of moon (A₁) = 4Ï€r₁²
Surface area of sun (A₂) = 4Ï€r₂²
Ratio of surface areas (A₁/A₂) = (4Ï€r₁²) / (4Ï€r₂²)
= (r₁/r₂)²
= (1/4)²
= 1/16
∴ A₁ : A₂ = 1 : 16
17. If the radius of a spherical balloon is increased from 7 cm to 14 cm, find the ratio of the surface areas.
Solution,
Original radius (r₁) = 7 cm
New radius (r₂) = 14 cm
Surface area of a sphere = 4 Ï€ r²
Original surface area (A₁) = 4 Ï€ (7)²
= 4 Ï€ × 49
= 196 Ï€ cm²
New surface area (A₂) = 4 Ï€ (14)²
= 4 Ï€ × 196
= 784 Ï€ cm²
Ratio of surface areas = A₁ : A₂
= 196 π : 784 π
= 196 : 784
= 1 : 4
So, the ratio of the surface areas is 1:4.
16. If the radius of the tennis ball is doubled on its original radius, find the total changes in their volume.
Solution,
Let the original radius of the tennis ball be r.
New radius after doubling = 2r
Original volume of the sphere = (4/3) Ï€ r³
New volume of the sphere = (4/3) Ï€ (2r)³
= (4/3) Ï€ × 8r³
= (32/3) Ï€ r³
Change in volume = New volume - Original volume
= (32/3) Ï€ r³ - (4/3) Ï€ r³
= (28/3) Ï€ r³
Ratio of new volume to original volume = ((32/3) Ï€ r³) / ((4/3) Ï€ r³)
= 32 / 4
= 8
So, the volume increases by a factor of 8, or the increase is 7 times the original volume.
18. Three spheres with radius of 2 cm, 12 cm, and 16 cm respectively are melted and formed into a single sphere. Find the diameter of the new sphere.
Solution,
Volume of the first sphere (r = 2 cm) = (4/3) Ï€ (2)³
= (4/3) Ï€ × 8 = (32/3) Ï€ cm³
Volume of the second sphere (r = 12 cm) = (4/3) Ï€ (12)³
= (4/3) Ï€ × 1728 = 2304 Ï€ cm³
Volume of the third sphere (r = 16 cm) = (4/3) Ï€ (16)³
= (4/3) Ï€ × 4096 = (16384/3) Ï€ cm³
Total volume = (32/3) π + 2304 π + (16384/3) π
= (32 π + 6912 π + 16384 π) / 3
= 23328 π / 3
= 7776 Ï€ cm³
The radius of the new sphere = R cm (say).
Volume of the new sphere = (4/3) Ï€ R³
or, (4/3) Ï€ R³ = 7776 Ï€
or, R³ = 7776 × (3/4)
or, R³ = 5832
or, R = 18 cm
Diameter of the new sphere = 2 × R
= 2 × 18 cm
= 36 cm
So, the diameter of the new sphere is 36 cm.
19. The total surface area of a hemispherical object is 243Ï€ cm², find the length of the perimeter and the volume of the object.
Solution,
Total surface area of the hemisphere = 243Ï€ cm²
Total surface area of a hemisphere = 3 Ï€ r²
3 Ï€ r² = 243 Ï€
or, r² = 243 / 3
or, r² = 81
or, r = 9 cm
Perimeter of the circular base = 2 π r
= 2 Ï€ × 9
= 18 π cm = 56.57 cm
Volume of the hemisphere = (2/3) Ï€ r³
= (2/3) Ï€ × (9)³
= (2/3) Ï€ × 729
= 486 Ï€ cm³ = 1527.43 cm³
So, the perimeter is 56.57 cm and the volume is 1527.43 cm³.
Vedanta Excel in Mathematics (Book 9) Solution
Unit 7: Mensuration (III): Cylinder and Sphere
7.1 Cylinder Looking back
Classwork-Exercise
Fill in the blanks with correct answers as quickly as possible.
a) The circumference of a circle with radius r units is 2Ï€r units.
b) The area of a circle with radius x units is Ï€x² sq. units.
c) The circumference of a circle with radius 7 cm is 44 cm.
d) The area of a circle with diameter 28 cm is 616 cm².
7.2 Cylinder
Example 1: The diameter of a cylindrical drum is 35 cm and its height is 30 cm. Find i) its curved surface area ii) total surface area iii) volume.
Solution,
Here, diameter of the cylindrical drum (d) = 35 cm
∴ The radius of the drum (r) = 35/2 = 17.5 cm
Height of the drum (h) = 30 cm
i) Now, the curved surface area (C.S.A.) = 2Ï€rh = 2 × 22/7 × 17.5 × 30 cm² = 3,300 cm²
ii) Also, the total surface area (T.S.A.) = 2Ï€r(r + h)
= 2 × 22/7 × 17.5 cm × (17.5 cm + 30 cm)
= 2 × 22/7 × 17.5 cm × 47.5 cm = 5,225 cm²
iii) And, volume (V) = Ï€r²h = 22/7 × 17.5 cm × 17.5 cm × 30 cm = 28,875 cm³
Example 2: The perimeter of the circular base of a cylinder having height 20 cm is 44 cm. Calculate its: i) curved surface area ii) total surface area iii) volume
Solution,
Here, the perimeter of circular base (C) = 44 cm
or,
2Ï€r = 44 cm
or,
2 × 22/7 × r = 44 cm
or,
r = 7 cm
Height of the cylinder (h) = 20 cm
i) Now, the curved surface area (C.S.A.) = 2Ï€rh = 2 × 22/7 × 7 cm × 20 cm = 880 cm²
ii) Also, the total surface area (T.S.A.) = 2Ï€r(r + h) = 2 × 22/7 × 7 cm × (7 cm + 20 cm)
= 2 × 22/7 × 7 cm × 27 cm = 1,188 cm²
iii) And, volume (V) = Ï€r²h = 22/7 × 7 cm × 7 cm × 20 cm = 3,080 cm³
EXERCISE 7.1
General section
1. a) If x be the radius and y be the height of the cylinder given alongside, write the formulae to find its:
Solution,
Here, radius of the cylinder (r) = x cm
Height of the cylinder (h) = y cm
i) Curved surface area (C.S.A.) = ?
ii) Total surface area (T.S.A.) = ?
iii) Volume = ?
We know that,
i) Curved surface area (C.S.A.) = 2Ï€rh = 2Ï€xy cm²
ii) Total surface area (T.S.A.) = 2Ï€r(r + h) = 2Ï€x(x + y) cm²
iii) Volume (V) = Ï€r²h = Ï€x²y cm³
1. b) A and a are the external and internal radii and B be the height of a hollow cylinder respectively. Write the formula to find its i) curved surface area (C.S.A.) ii) total surface area (T.S.A.) iii) volume of material contained by the cylinder.
Solution,
Here, external radius of the hollow cylinder (A) = A cm
Internal radius of the hollow cylinder (a) = a cm
Height of the hollow cylinder (b) = b cm
i) Curved surface area (C.S.A.) = ?
ii) Total surface area (T.S.A.) = ?
iii) Volume of material = ?
We know that,
i) Curved surface area (C.S.A.) = 2Ï€b(A + a) cm²
ii) Total surface area (T.S.A.) = 2Ï€(A + a)(b + A - a) cm²
iii) Volume of material (V) = Ï€b(A + a)(A - a) cm³
1. c) From the given half-cylinder, write the formulae to find i) curved surface area ii) lateral surface area iii) total surface area iv) volume.
Solution,
Here, radius of the half-cylinder (p) = p cm
Length of the half-cylinder (q) = q cm
i) Curved surface area = ?
ii) Lateral surface area = ?
iii) Total surface area = ?
iv) Volume = ?
We know that,
i) Curved surface area (C.S.A.) = Ï€pq cm²
ii) Lateral surface area = Ï€pq cm²
iii) Total surface area (T.S.A.) = Ï€p(q + 2) + Ï€p² cm²
iv) Volume (V) = 1/2 Ï€p²q cm³
2. a) The perimeter of the base of a cylinder is 44 cm and height is 10 cm, what is its curved surface area?
Solution,
Here, perimeter of the base (C) = 44 cm
or, 2Ï€r = 44 cm
or, 2 × 22/7 × r = 44 cm
or,r = 7 cm
Height of the cylinder (h) = 10 cm
Curved surface area (C.S.A.) = 2Ï€rh
= 2 × 22/7 × 7 cm × 10 cm
= 440 cm²
So, curved surface area = 440 cm²
2. b) What is the total surface area of a cylinder in which the circumference of the base is 22 inch and the sum of radius and height is 15 inch?
Solution,
Here, circumference of the base (C) = 22 inch
or, 2Ï€r = 22 inch
or, 2 × 22/7 × r = 22 inch
or, r = 7/2 inch
Given, r + h = 15 inch
or, 7/2 + h = 15 inch
or, h = 15 - 7/2 = 23/2 inch
Total surface area (T.S.A.) = 2Ï€r(r + h)
= 2 × 22/7 × 7/2 inch × (7/2 inch + 23/2 inch)
= 2 × 22/7 × 7/2 inch × 15 inch
= 330 inch²
So, total surface area = 330 inch²
2. c) What is the volume of a cylinder having area of base 38.5 cm² and height 20 cm?
Solution,
Here, area of the base = 38.5 cm²
Height of the cylinder (h) = 20 cm
Volume of the cylinder = ?
We know that,
Area of the base = Ï€r²
or, 38.5 = 22/7 × r²
or, r² = 38.5 × 7 / 22
or, r² = 12.25
or, r = 3.5 cm
Volume (V) = Ï€r²h
= 22/7 × 3.5 cm × 3.5 cm × 20 cm
= 770 cm³
So, volume = 770 cm³
3. a) Find the i) curved surface area, ii) total surface area, and iii) volume of the given cylinder with radius 7 cm and height 25 cm.
Solution,
Here, radius of the cylinder (r) = 7 cm
Height of the cylinder (h) = 25 cm
i) Curved surface area (C.S.A.) = 2Ï€rh
= 2 × 22/7 × 7 cm × 25 cm
= 1100 cm²
ii) Total surface area (T.S.A.) = 2Ï€r(r + h)
= 2 × 22/7 × 7 cm × (7 cm + 25 cm)
= 2 × 22/7 × 7 cm × 32 cm
= 1408 cm²
iii) Volume (V) = Ï€r²h
= 22/7 × 7 cm × 7 cm × 25 cm
= 3850 cm³
So, i) curved surface area = 1100 cm², ii) total surface area = 1408 cm², iii) volume = 3850 cm³
3. b) Find the i) curved surface area, ii) total surface area, and iii) volume of the given cylinder with radius 6 cm and height 28 cm.
Solution,
Here, radius of the cylinder (r) = 6 cm
Height of the cylinder (h) = 28 cm
i) Curved surface area (C.S.A.) = 2Ï€rh
= 2 × 22/7 × 6 cm × 28 cm
= 1056 cm²
ii) Total surface area (T.S.A.) = 2Ï€r(r + h)
= 2 × 22/7 × 6 cm × (6 cm + 28 cm)
= 2 × 22/7 × 6 cm × 34 cm
= 1280.57 cm²
iii) Volume (V) = Ï€r²h
= 22/7 × 6 cm × 6 cm × 28 cm
= 3168 cm³
So, i) curved surface area = 1056 cm², ii) total surface area = 1280.57 cm², iii) volume = 3168 cm³
3. c) Find the i) curved surface area, ii) total surface area, and iii) volume of the given cylinder with diameter 14 cm and height 20 cm.
Solution,
Here, diameter of the cylinder (d) = 14 cm
radius of the cylinder (r) = d/2 = 14/2 = 7 cm
Height of the cylinder (h) = 20 cm
i) Curved surface area (C.S.A.) = 2Ï€rh
= 2 × 22/7 × 7 cm × 20 cm
= 880 cm²
ii) Total surface area (T.S.A.) = 2Ï€r(r + h)
= 2 × 22/7 × 7 cm × (7 cm + 20 cm)
= 2 × 22/7 × 7 cm × 27 cm
= 1188 cm²
iii) Volume (V) = Ï€r²h
= 22/7 × 7 cm × 7 cm × 20 cm
= 3080 cm³
So, i) curved surface area = 880 cm², ii) total surface area = 1188 cm², iii) volume = 3080 cm³
3. d) Find the i) curved surface area, ii) total surface area, and iii) volume of the given cylinder with diameter 35 inch and height 50 inch.
Solution,
Here, diameter of the cylinder (d) = 35 inch
radius of the cylinder (r) = d/2 = 35/2 inch
Height of the cylinder (h) = 50 inch
i) Curved surface area (C.S.A.) = 2Ï€rh
= 2 × 22/7 × 35/2 inch × 50 inch
= 5500 inch²
ii) Total surface area (T.S.A.) = 2Ï€r(r + h)
= 2 × 22/7 × 35/2 inch × (35/2 inch + 50 inch)
= 2 × 22/7 × 35/2 inch × 135/2 inch
= 7425 inch²
iii) Volume (V) = Ï€r²h
= 22/7 × 35/2 inch × 35/2 inch × 50 inch
= 48125 inch³
So, i) curved surface area = 5500 inch², ii) total surface area = 7425 inch², iii) volume = 48125 inch³
Creative Section - A
4. a) There is a cylindrical wooden log in a meat-shop. If the radius of its circular base is 21 cm and height is 50 cm, i) find the curved surface area of the log, ii) find the total surface area of the log, iii) find the volume of the log.
Solution,
Here, radius of the cylindrical log (r) = 21 cm
Height of the cylindrical log (h) = 50 cm
i) Curved surface area (C.S.A.) = 2Ï€rh
= 2 × 22/7 × 21 cm × 50 cm
= 6600 cm²
ii) Total surface area (T.S.A.) = 2Ï€r(r + h)
= 2 × 22/7 × 21 cm × (21 cm + 50 cm)
= 2 × 22/7 × 21 cm × 71 cm
= 9372 cm²
iii) Volume (V) = Ï€r²h
= 22/7 × 21 cm × 21 cm × 50 cm
= 69300 cm³
So, i) curved surface area = 6600 cm², ii) total surface area = 9372 cm², iii) volume = 69300 cm³
4. b) From the can of vegetables given alongside, i) find the area of the label on the can, ii) find the area of the metal sheet used to make the can, iii) find the volume of the can.
Solution,
Here, radius of the can (r) = 1.75 inch
Height of the can (h) = 5 inch
i) Area of the label (C.S.A.) = 2Ï€rh
= 2 × 22/7 × 1.75 inch × 5 inch
= 55 inch²
ii) Area of the metal sheet (T.S.A.) = 2Ï€r(r + h)
= 2 × 22/7 × 1.75 inch × (1.75 inch + 5 inch)
= 2 × 22/7 × 1.75 inch × 6.75 inch
= 74.25 inch²
iii) Volume (V) = Ï€r²h
= 22/7 × 1.75 inch × 1.75 inch × 5 inch
= 48.125 inch³
So, i) area of the label = 55 inch², ii) area of the metal sheet = 74.25 inch², iii) volume = 48.125 inch³
5. a) A cylindrical tank has diameter 1.4 m and height 2 m. How many litres of water is required to fill the tank completely?
Solution,
Here, diameter of the tank (d) = 1.4 m
radius of the tank (r) = d/2 = 1.4/2 = 0.7 m
Height of the tank (h) = 2 m
Volume of the tank = Ï€r²h
= 22/7 × 0.7 m × 0.7 m × 2 m
= 22/7 × 7/10 m × 7/10 m × 2 m
= 22/7 × 49/100 m³ × 2
= 22/7 × 98/100 m³
= 3.08 m³
Since 1 m³ = 1000 litres,
Volume in litres = 3.08 × 1000 = 3080 litres
So, 3080 litres of water is required to fill the tank completely.
5. b) The diameter and height of a cylindrical vessel are equal. If its radius is 35 cm and it is completely filled with diesel, how many litres of diesel is filled in the vessel?
Solution,
Here, radius of the vessel (r) = 35 cm
Since diameter and height are equal, diameter = 2r = 2 × 35 = 70 cm
Height of the vessel (h) = 70 cm
Volume of the vessel = Ï€r²h
= 22/7 × 35 cm × 35 cm × 70 cm
= 22 × 5 cm × 35 cm × 70 cm
= 269500 cm³
Since 1 litre = 1000 cm³,
Volume in litres = 269500 / 1000 = 269.5 litres
So, 269.5 litres of diesel is filled in the vessel.
5. c) How many cubic meters of earth must be dug to make a well 20 m deep and 1.4 m diameter?
Solution,
Here, diameter of the well (d) = 1.4 m
radius of the well (r) = d/2 = 1.4/2 = 0.7 m
Depth of the well (h) = 20 m
Volume of earth to be dug = Ï€r²h
= 22/7 × 0.7 m × 0.7 m × 20 m
= 22/7 × 7/10 m × 7/10 m × 20 m
= 22/7 × 49/100 m³ × 20
= 22/7 × 980/100 m³
= 30.8 m³
So, 30.8 cubic meters of earth must be dug to make the well.
6. a) The circumference of the circular base of a closed cylindrical juice can of height 10 cm is 22 cm. How much metal sheet is required to make it? Find it.
Solution,
Here, circumference of the base (C) = 22 cm
or, 2Ï€r = 22 cm
or, 2 × 22/7 × r = 22 cm
or, r = 7/2 cm
Height of the juice can (h) = 10 cm
Metal sheet required = Total surface area (T.S.A.) = 2Ï€r(r + h)
= 2 × 22/7 × 7/2 cm × (7/2 cm + 10 cm)
= 2 × 22/7 × 7/2 cm × 27/2 cm
= 297 cm²
So, 297 cm² of metal sheet is required to make the juice can.
6. b) The height of a cylindrical drum is 20 cm and the area of its base is 346.5 sq. cm. If it is to be covered with leather, find the minimum quantity of leather sheet needed for it in sq. cm.
Solution,
Here, area of the base = 346.5 cm²
or, Ï€r² = 346.5 cm²
or, 22/7 × r² = 346.5
or, r² = 346.5 × 7/22
or, r² = 110.25
or, r = 10.5 cm
Height of the drum (h) = 20 cm
Minimum leather needed = Curved surface area (C.S.A.) = 2Ï€rh
= 2 × 22/7 × 10.5 cm × 20 cm
= 1320 cm²
So, the minimum quantity of leather sheet needed is 1320 cm².
7. a) The radius and height of a cylindrical container are in the ratio 5:7 and its curved surface area is 5500 sq.cm. Find its volume.Given:
Radius (r) = 5x
Height (h) = 7x
CSA = 5500 cm²
We know that,
CSA = 2Ï€rh
or, 5500 = 2Ï€rh
or, 5500 = 2Ï€(5x)(7x)
70Ï€x² = 5500
or, x² = 5500/70Ï€ = 550/7Ï€
or, x = 5.001 cm
V = Ï€r²h
= 22/7 x (5)²x 35
= 68750 cm³
7. b) The radius and height of a cylindrical jar are in the ratio 7:17 and its total surface area is 4224 sq.cm. Find its volume.
Solution,
Here, radius : height = 7 : 17
radius (r) = 7k cm and height (h) = 17k cm
Total surface area (T.S.A.) = 2Ï€r(r + h)
or, 2 × 22/7 × 7k × (7k + 17k) = 4224
or, 2 × 22/7 × 7k × 24k = 4224
or, 1056k² = 4224
or, k² = 4
or, k = 2 cm
So, radius (r) = 7k = 7 × 2 = 14 cm
Height (h) = 17k = 17 × 2 = 34 cm
Volume (V) = Ï€r²h
= 22/7 × 14 cm × 14 cm × 34 cm
= 20944 cm³
Hence, the volume of the jar is 20944 cm³.
7. c) A cylinder container water tank contains 4,62,000 litres of water and its radius is 3.5 m. Find its curved surface area.
Solution,
Here, volume of the tank = 462000 litres
Since 1 m³ = 1000 litres,
Volume in m³ = 462000 / 1000 = 462 m³
Radius of the tank (r) = 3.5 m
Volume = Ï€r²h
or, 462 = 22/7 × 3.5 × 3.5 × h
or, 462 = 22/7 × 7/2 × 7/2 × h
or, 462 = 77/2 × h
or, h = 462 × 2 / 77
or, h = 12 m
Curved surface area (C.S.A.) = 2Ï€rh
= 2 × 22/7 × 3.5 m × 12 m
= 264 m²
So, the curved surface area of the tank is 264 m².
7. d) The volume of a cylindrical can is 3.08 litre. If the diameter of its base is 14 cm, find the total surface area of the can.
Solution,
Here, volume of the can = 3.08 litres
Since 1 litre = 1000 cm³,
Volume in cm³ = 3.08 × 1000 = 3080 cm³
Diameter of the base (d) = 14 cm
radius (r) = d/2 = 14/2 = 7 cm
Volume = Ï€r²h
or, 3080 = 22/7 × 7 × 7 × h
or, 3080 = 154 × h
or, h = 3080 / 154
or, h = 20 cm
Total surface area (T.S.A.) = 2Ï€r(r + h)
= 2 × 22/7 × 7 cm × (7 cm + 20 cm)
= 2 × 22/7 × 7 cm × 27 cm
= 1188 cm²
So, the total surface area of the can is 1188 cm².
8. a) The area of curved surface of a cylinder is equal to 2/3 of the total surface area. If the total surface area is 924 cm², find the volume of the cylinder.
Solution,
Here, total surface area (T.S.A.) = 924 cm²
Curved surface area (C.S.A.) = 2/3 × T.S.A.
= 2/3 × 924 cm²
= 616 cm²
We know,
C.S.A. = 2Ï€rh
or, 2 × 22/7 × r × h = 616
or, 44/7 × r × h = 616
or, r × h = 616 × 7 / 44
or, r × h = 98
T.S.A. = 2Ï€r(r + h)
or, 924 = 2 × 22/7 × r × (r + h)
or, 924 × 7 / 44 = r × (r + h)
or, 147 = r × (r + h)
Since r × h = 98, h = 98/r
Substitute h in T.S.A.,
147 = r × (r + 98/r)
or, 147 = r² + 98
or, r² = 147 - 98
or, r² = 49
or, r = 7 cm
Then, h = 98/r = 98/7 = 14 cm
Volume (V) = Ï€r²h
= 22/7 × 7 cm × 7 cm × 14 cm
= 2156 cm³
So, the volume of the cylinder is 2156 cm³.
8. b) The number of volume of a cylinder is half its number of the total surface area. If the radius of its base is 7 cm, find the volume.
Solution,
Here, radius of the cylinder (r) = 7 cm
T.S.A. = A cm² (say)
Given, volume (V) = 1/2 × T.S.A.
or, V = 1/2 × A
So, A = 2V
We know that,
T.S.A. = 2Ï€r(r + h)
or, 2V = 2Ï€r(r + h)
or, V = πr(r + h)
Also, V = Ï€r²h
Equating the two expressions for V:
Ï€r²h = Ï€r(r + h)
or, r²h = r(r + h)
or, r²h = r² + rh
or, r²h - rh = r²
or, rh(r - 1) = r²
or, h(r - 1) = r
or, h = r / (r - 1)
Substituting r = 7 cm,
h = 7 / (7 - 1) = 7/6 cm
Volume (V) = Ï€r²h
= 22/7 × 7 cm × 7 cm × 7/6 cm
= 539/3 cm³
= 179.67 cm³
So, the volume of the cylinder is 179.67 cm³.
9. a) The sum of the radius of the base and the height of a solid cylinder is 37 cm. If the total surface area of the cylinder is 1,628 cm², calculate its volume.
Solution,
Here, radius (r) + height (h) = 37 cm
or, r + h = 37
Total surface area (T.S.A.) = 2Ï€r(r + h)
or, 2 × 22/7 × r × (r + h) = 1628
or, 2 × 22/7 × r × 37 = 1628
or, 44/7 × r × 37 = 1628
or, 44 × 37 / 7 × r = 1628
or, 44 × 37 × r = 1628 × 7
or, 1628 × r = 1628 × 7
or, r = 7 cm
Then, h = 37 - r = 37 - 7 = 30 cm
Volume (V) = Ï€r²h
= 22/7 × 7 cm × 7 cm × 30 cm
= 4620 cm³
So, the volume of the cylinder is 4620 cm³.
9. b) The circumference of the base of a cylindrical drum is 44 cm and the sum of its radius and height is 27 cm, find the volume of the cylinder.
Solution,
Here, circumference of the base = 44 cm
or, 2Ï€r = 44 cm
or, 2 × 22/7 × r = 44
or, r = 44 × 7 / 44
or, r = 7 cm
Given, r + h = 27 cm
or, 7 + h = 27
or, h = 27 - 7 = 20 cm
Volume (V) = Ï€r²h
= 22/7 × 7 cm × 7 cm × 20 cm
= 3080 cm³
So, the volume of the cylinder is 3080 cm³.
9. c) The curved surface area and volume of a cylinder are 880 cm² and 3,080 cm³ respectively, find its total surface area.
Solution,
Here, curved surface area (C.S.A.) = 880 cm²
Volume (V) = 3080 cm³
We know,
C.S.A. = 2Ï€rh
or, 2 × 22/7 × r × h = 880
or, 44/7 × r × h = 880
or, r × h = 880 × 7 / 44
or, r × h = 140
or, r = 140 / h
Now, Volume = Ï€r²h
or, 3080 = 22/7 × r² × h
3080 = 22/7 × r² × (140/r)
or, 3080 = 22/7 × r × 140
or, 3080 × 7 / (22 × 140) = r
or, r = 7 cm
Then, h = 140/r = 140/7 = 20 cm
Total surface area (T.S.A.) = 2Ï€r(r + h)
= 2 × 22/7 × 7 × (7 + 20)
= 2 × 22/7 × 7 × 27
= 1188 cm²
So, the total surface area of the cylinder is 1188 cm².
Question 10a: Calculate the lateral surface area, total surface area, and volume of the given half-cylinder with diameter 14 cm and height 20 cm.
Solution:
Here, radius (r) = 14 / 2 = 7 cm
height (h) = 20 cm
We know that,
LSA = rh (Ï€ + 2)
= 7 x 20 (22/7 + 2)
= 7 x 20 x (22/7 + 14/7)
= 7 x 20 x 36/7
= 720 cm²
Now, TSA = rh (Ï€ + 2) + Ï€r²
= 720 + 22/7 x (7)²
= 720 + 22/7 x 49
= 720 + 154
= 874 cm²
Next, V = 1/2 Ï€r²h
= 1/2 x 22/7 x (7)² x 20
= 1/2 x 22/7 x 49 x 20
= 22 x 7 x 10
= 1540 cm³
10b: The radius of the semi-circular base of a half-cylinder is 21 cm and its height is 50 cm. Find its lateral surface area, total surface area, and volume.
Solution:
Here, radius (r)= 21 cm
height (h) = 50 cm
We know that,
LSA = rh (Ï€ + 2)
= 21 x 50 (22/7 + 2)
= 21 x 50 x (22/7 + 14/7)
= 21 x 50 x 36/7
= 5400 cm²
Now, TSA = rh (Ï€ + 2) + Ï€r²
= 5400 + 22/7 x (21)²
= 5400 + 22/7 x 441
= 5400 + 1386
= 6786 cm²
Next, V = 1/2 Ï€r²h
= 1/2 x 22/7 x (21)² x 50
= 1/2 x (22/7) x 21 x 21 x 50
= 34650 cm³
10c: Mr. Binod made a vegetable tunnel as shown in the figure given aside. Find the volume of the tunnel with diameter 6 m and length 21 m.
Solution:
Here, radius (r) = 6 / 2 = 3 m
height (h) = 21 m
We know that,
V = 1/2 Ï€r²h
= 1/2 x 22/7 x (3)² x 21
= 1/2 x 22/7 x 9 x 21
= 22 x 27 / 2
= 297 m³
Question 11a: A hollow cylindrical metallic pipe is 28 cm long. If the external and internal diameters of the pipe are 12 cm and 8 cm respectively, find the volume of metal used in making the pipe.
Solution:
Here, external radius (R) = 12 / 2 = 6 cm
internal radius (r) = 8 / 2 = 4 cm
height (h)= 28 cm
We know that,
Volume of metal = Ï€ (R² - r²) h
= 22/7 x (6² - 4²) x 28
= 22/7 x (36 - 16) x 28
= 22/7 x 20 x 28
= 22 x 20 x 4
= 1760 cm³
Question 11b: The external and the internal radii of a cylindrical pipe of length 50 cm are 3 cm and 2.6 cm respectively. Find the volume of material contained by the pipe.
Solution:
Here, external radius (R) = 3 cm
internal radius (r) = 2.6 cm
height (h) = 50 cm
We know that,
Volume of material = Ï€ (R² - r²) h
= 22/7 x (3² - 2.6²) x 50
= 22/7 x (9 - 6.76) x 50
= 22/7 x 2.24 x 50
= 352 cm³
Question 11c: A metal pipe is 1.4 m long. Its internal diameter is 25 mm and thickness 1 mm. If 1 mm³ of the metal weighs 0.02 gm, find the weight of the pipe.
Solution:
Here, internal radius (r) 25 / 2 = 12.5 mm
thickness (t)= 1 mm
external radius (R)= 12.5 + 1 = 13.5 mm
height (h) 1.4 m = 1400 mm
We know that,
Volume of metal = Ï€ (R² - r²) h
= 22/7 x (13.5² - 12.5²) x 1400
= 22/7 x (182.25 - 156.25) x 1400
= 22/7 x 26 x 1400
= 114400 mm³
Weight of metal = volume x weight per mm³
= 114400 x 0.02
= 2288 gm
= 2.288 kg
Question 11d: A metal pipe has inner diameter of 5 cm. The pipe is 5 mm thick all round and 2 m long. What is the weight of the pipe if 1 cm³ of the metal weighs 7.7 g?
Solution:
Here, internal radius (r) = 5 / 2 = 2.5 cm
thickness (t) = 5 mm = 0.5 cm
external radius (R) 2.5 + 0.5 = 3 cm
height (h)= 2 m = 200 cm
We know that,
Volume of metal = Ï€ (R² - r²) h
= 22/7 x (3² - 2.5²) x 200
= 22/7 x (9 - 6.25) x 200
= 22/7 x 2.75 x 200
= 1728.6 cm³
Weight of metal = volume x weight per cm³
= 1728.6 x 7.7
= 13310 g
= 13.31 kg
Question 12a: The internal radius of a cylindrical bucket of height 50 cm is 21 cm. It is filled with water completely. If the water is poured in a rectangular vessel with internal length 63 cm and breadth 44 cm and it is completely filled with water, find the height of the vessel.
Solution:
Here, radius of cylindrical bucket (r) = 21 cm
height of cylindrical bucket (h) = 50 cm
length of rectangular vessel (l) = 63 cm
breadth of rectangular vessel (b) = 44 cm
We know that,
Volume of water in cylindrical bucket = Ï€ r² h
= 22/7 x (21)² x 50
= 22/7 x 441 x 50
= 69300 cm³
Volume of rectangular vessel = length x breadth x height
Volume of water remains the same, so 69300 = 63 x 44 x height
height = 69300 / (63 x 44)
= 69300 / 2772
= 25 cm
Question 12b: Vegetable ghee is stored in a cylindrical vessel of internal radius 1.4 m and height 1.5 m. If it is transferred into the rectangular tin cans 33 cm x 10 cm x 5 cm, how many cans are required to empty the vessel?
Solution:
Here, radius of cylindrical vessel (r) 1.4 m = 140 cm
height of cylindrical vessel (h) = 1.5 m = 150 cm
length of rectangular tin can (l) = 33 cm
breadth of rectangular tin can (b) = 10 cm
height of rectangular tin can (h) = 5 cm
We know that,
Volume of cylindrical vessel = Ï€ r² h
= 22/7 x (140)² x 150
= 22/7 x 19600 x 150
= 9240000 cm³
Volume of one rectangular tin can = length x breadth x height
= 33 x 10 x 5
= 1650 cm³
Number of cans = volume of cylindrical vessel / volume of one tin can
= 9240000 / 1650
= 5600
Question 13a: Mrs. Thapa has two cylindrical buckets. The bucket-A has radius of base 14 cm and height 30 cm. Likewise, bucket-B has the radius of base 16 cm and height 28 cm. Which of these buckets can hold more water and by how many litres?
Solution:
Here, radius of bucket-A (r1)= 14 cm
height of bucket-A (h1) = 30 cm
radius of bucket-B (r2) 16 cm
height of bucket-B (h2) 28 cm
We know that,
Volume of bucket-A = Ï€ r1² h1
= 22/7 x (14)² x 30
= 22/7 x 196 x 30
= 18480 cm³
Volume of bucket-B = Ï€ r2² h2
= 22/7 x (16)² x 28
= 22/7 x 256 x 28
= 22528 cm³
Comparison: 22528 cm³ (bucket-B) is greater than 18480 cm³ (bucket-A)
Difference in volume = 22528 - 18480
= 4048 cm³
Convert to litres (1 litre = 1000 cm³) = 4048 / 1000
= 4.048 litres
Creative Section - B
Question 13a: Mrs. Thapa has two cylindrical buckets. The bucket-A has radius of base 14 cm and height 30 cm. Likewise, bucket-B has the radius of base 16 cm and height 28 cm. Which of these buckets can hold more water and by how many litres?
Solution:
Here, radius of bucket-A (r1)= 14 cm
height of bucket-A (h1) = 30 cm
radius of bucket-B (r2) 16 cm
height of bucket-B (h2) 28 cm
We know that,
Volume of bucket-A = Ï€ r1² h1
= 22/7 x (14)² x 30
= 22/7 x 196 x 30
= 18480 cm³
Volume of bucket-B = Ï€ r2² h2
= 22/7 x (16)² x 28
= 22/7 x 256 x 28
= 22528 cm³
Comparison: 22528 cm³ (bucket-B) is greater than 18480 cm³ (bucket-A)
Difference in volume = 22528 - 18480
= 4048 cm³
= 4.048 litres
Question 13b: Mr. Rai has fixed two overhead cylindrical water tanks on the roof of his house. The diameter of the base of tank-X is 2.1 m and height is 3 m. Similarly, the diameter of the base of tank-Y is 3 m and height is 2.1 m. Which of these tanks holds more water and by how many litres? Find.
Solution:
Here, diameter of tank-X= 2.1 m
radius of tank-X (r1)= 2.1 / 2 = 1.05 m
height of tank-X (h1) = 3 m
diameter of tank-Y = 3 m
radius of tank-Y (r2) = 3 / 2 = 1.5 m
height of tank-Y (h2) = 2.1 m
We know that,
Volume of tank-X = Ï€ r1² h1
= 22/7 x (1.05)² x 3
= 22/7 x 1.1025 x 3
= 10.395 m³
Volume of tank-Y = Ï€ r2² h2
= 22/7 x (1.5)² x 2.1
= 22/7 x 2.25 x 2.1
= 14.85 m³
Comparison: 14.85 m³ (tank-Y) is greater than 10.395 m³ (tank-X)
Difference in volume = 14.85 - 10.395
= 4.455 m³
= 4455 litres
Question 14a: Mr. Chaudhary drives a tanker in a milk supply company. He supplies 5 tankers of milk every day. If the milk tank is cylindrical in shape having radius 0.8 m and length 3.5 m, how many liters of milk does he supply in each day?
Solution:
Here, radius of milk tank (r1)= 0.8 m
height of milk tank (h1) = 3.5 m
number of tankers = 5
We know that,
Volume of one milk tank = Ï€ r1² h1
= 22/7 x (0.8)² x 3.5
= 22/7 x 0.64 x 3.5
= 7.04 m³
Total volume of 5 tankers = 5 x 7.04
= 35.2 m³
= 35200 litres
Question 14b: Chari Maya is a juice seller. She serves the juice completely filled with the cylindrical glasses; each has diameter 7 cm and height 10 cm. If she sells 50 glasses of juice in a day, how many litres of juice does she sell in a day?
Solution:
Here, radius of glass (r1)= 7 / 2 = 3.5 cm
height of glass (h1) = 10 cm
number of glasses = 50
We know that,
Volume of one glass = Ï€ r1² h1
= 22/7 x (3.5)² x 10
= 22/7 x 12.25 x 10
= 385 cm³
Total volume of 50 glasses = 50 x 385
= 19250 cm³
= 19.25 litres
Question 15a: A roller of diameter 112 cm and length 150 cm takes 550 complete revolutions to level a playground. Calculate the area of the ground.
Solution:
Here, radius of roller (r1)= 112 / 2 = 56 cm
height of roller (h1) = 150 cm
number of revolutions = 550
We know that,
Curved surface area of roller = 2 π r1 h1
= 2 x 22/7 x 56 x 150
= 52800 cm²
Total area of ground = curved surface area x number of revolutions
= 52800 x 550
= 29040000 cm²
= 2904 m²
Question 15b: A temple has 14 cylindrical pillars made of concrete. If the radius of the base of each pillar is 20 cm and height is 5 m, how much concrete is required to make the pillars? Find it.
Solution:
Here, radius of pillar (r1)= 20 cm
height of pillar (h1) = 5 m = 500 cm
number of pillars = 14
We know that,
Volume of one pillar = Ï€ r1² h1
= 22/7 x (20)² x 500
= 22/7 x 400 x 500
= 628571.43 cm³
Total volume of 14 pillars = 14 x 628571.43
= 8800000 cm³
= 8800000 / 1000000 m³
= 8.8 m³
Area and Volume of a Sphere
EXERCISE 7.2
General section
a: Write the formula to find (i) surface area and (ii) volume of the given sphere.
Solution:
r= x cm
(i) Total surface area of the sphere = 4 Ï€ x² cm²
(ii) Volume of the sphere = 4/3 Ï€ x³ cm³
Question 1b: The radius of a hemisphere is p cm. Write the formula to find its (i) curved surface area (ii) total surface area (iii) volume.
Solution:
r= p cm
(i) Curved surface area of the hemisphere = 2 Ï€ p²cm²
(ii) Total surface area of the hemisphere = 3 Ï€ p²cm²
(iii) Volume of the hemisphere = 2/3 Ï€ p³ cm³
Question 1c: If the radius of a sphere is 1 cm, what are its surface area and the volume?
Solution:
r= 1 cm
We know that,
Total surface area of the sphere = 4 Ï€ r²
= 4 x 22/7 x (1)²
= 4 x 22/7 x 1
= 88/7
= 12.57 cm²
Now, volume of the sphere = 4/3 Ï€ r³
= 4/3 x 22/7 x (1)³
= 4/3 x 22/7 x 1
= 88/21
= 4.19 cm³
No comments:
Post a Comment