Class 9 MAHARASHTRA STATE TEXTBOOK BUREAU Solutions Maths Chapter 9 - Surface Area and Volume

Surface area of vertical faces of the box

= 2(l + b) x h

= 2(20+ 12) x 10

= 2 x 32 x 10

= 640 sq.cm.

Total surface area of the box

= 2 (l x b + b x h + l x h)

= 2(20 x 12+ 12 x 10 + 20 x 10)

= 2(240 + 120 + 200)

= 2 x 560

= 1120 sq.cm.

∴ The surface area of vertical faces and total surface area of the box are 640 sq.cm, and 1120 sq.cm, respectively.

Given: For cuboid shape box,

breadth (b) = 6 unit, height (h) = 5 unit, Total surface area = 500 sq. unit.

To find: Length of the box (l)

Total surface area of the box = 2 (l x b + b x h + l x h)

∴ 500 = 2 (6l + 6 x 5 + 5l)

∴  = (11l + 30)

∴ 250= 11l + 30

∴ 250 - 30= 11l

∴ 220 = 11l

∴ 220 = l

∴  = l

∴ l = 20 units

∴ The length of the box is 20 units. 

Given: Side of cube (l) = 4.5 cm

To find: Surface area of all vertical faces and the total surface area of the cube

Area of vertical faces of cube = 4l²

= 4 (4.5)² = 4 x 20.25 = 81 sq.cm.

Total surface area of the cube = 6l²

= 6 (4.5)²

= 6 x 20.25

= 121.5 sq.cm.

∴ The surface area of all vertical faces and the total surface area of the cube are 81 sq.cm, and 121.5 sq.cm, respectively. 

Given: Total surface area of cube = 5400 sq.cm.

To find: Surface area of all vertical faces of the cube

Total surface area of cube = 6l²

∴ 5400 = 6l²

∴  = l²

∴ l² = 900

Area of vertical faces of cube = 4l²

= 4 x 900 = 3600 sq.cm.

∴ The surface area of all vertical faces of the cube is 3600 sq.cm. 

Given: Breadth (b) = 1.5 m, height (h) = 1.15 m

Volume of cuboid = 34.50 cubic meter

To find: Length of the cuboid (l)

Volume of cuboid = l x b x h

∴ 34.50 = l x b x h

∴ 34.50 = l x 1.5 x 1.15

 ∴ l =  

 =  

 =  

 = 20

∴ The length of the cuboid is 20 m. 

Given: Length of edge of cube (l) = 7.5 cm

To find: Volume of a cube

Solution:

Volume of a cube = l³

= (7.5)³

= 421.875 ≈ 421.88 cubic cm

∴The volume of the cube is 421.88 cubic cm. 

Given: Radius (r) = 20 cm, height (h) = 13 cm

To find: Curved surface area and the total surface area of the cylinder

Curved surface area of cylinder = 2πrh

= 2 x 3.14 x 20 x 13

= 1632.8 sq.cm

Total surface area of cylinder = 2πr(r + h)

= 2 x 3.14 x 20(20 + 13)

= 2 x 3.14 x 20 x 33 = 4144.8 sq.cm

∴ The curved surface area and the total surface area of the cylinder are 1632.8 sq.cm and 4144.8 sq.cm respectively.

Given: Curved surface area of cylinder = 1980 sq.cm., radius (r) = 15 cm

To find: Height of the cylinder (h)

Solution:

Curved surface area of cylinder = 2πrh

∴ 1980 = 2 x x 15 x h

∴h =  cm

∴ h = 21 cm

∴ The height of the cylinder is 21 cm. 

Given: Height (h) = 12 cm, length (l) = 13 cm

To find: Radius of the base of the cone (r)

l²= r² + h²

∴ 13² = r² + 12²

∴ 169 = r² + 144

∴169 - 144 = r²

∴ r² = 25

∴ r =  … [Taking square root on both sides]

= 5 cm

∴ The radius of base of the cone is 5 cm.

Given: Radius (r) = 28 cm,

Total surface area of cone = 7128 sq.cm

To find: Volume of the cone

Total surface area of cone = πr (l + r)

∴ 7128= y x 28 x (l + 28)

∴ 7128 = 22 x 4 x (l +28)

∴ l + 28 =  

∴ l + 28 = 81

∴ l = 81 - 28

∴ l = 53cm 

Now, l² = r² + h²

∴ 53²= 28² + h²

∴ 2809 = 784 + h²

∴ 2809 - 784 = h²

∴ h² = 2025

∴ h =  …… [Taking square root on both sides]

= 45 cm

∴ Volume of Cone =  

 =  x  x 28² x 45

 =  x  x 28 x 28 x 45

 = 22 x 4 x 28 x 15

 = 36960 cubic.cm

∴ The volume of the cone is 36960 cubic.cm. 

Given: Radius (r) = 8 cm, curved surface area

of cone = 251.2 cm²

To find: Slant height (l) and the perpendicular height (h) of the cone

Curved surface area of cone = π r l

∴ 251.2 = 3.14 x 8 x l

 ∴ l =  

 =  

 =  

∴ l= 10 cm

Now, l2 = r2 + h²

∴ 10² = 82 + h²

∴ 100 = 64 + h²

∴ 100 - 64 = h²

∴ h² = 36

∴ h =  … [Taking square root on both sides]

= 6 cm

∴ The slant height and the perpendicular height of the cone are 10 cm and 6 cm respectively.

Given: Radius (r) = 6 m, length (l) = 8 m

To find: Total cost of making the cone

To find the total cost of making the cone of tin sheet, first we need to find the total surface area of the cone.

Total surface area of the cone = π r (l + r)

= x 6 x (8 + 6)

= x 6 x 14

= 22 x 6 x 2 = 264 sq.m

Rate of making the cone = Rs. 10 per sq.m

∴ Total cost = Total surface area x Rate of making the cone

= 264 x 10

= Rs. 2640

∴ The total cost of making the cone of tin sheet is Rs. 2640.

Given: Radius (r) = 20 cm,

Volume of cone = 6280 cubic cm

To find: Perpendicular height (h) of the cone

 Volume of cone =  π² h

∴ 6280 =  x 3.14 x 20² x h

∴ h =  

 = 

 =  = 15 cm

∴ The perpendicular height of the cone is 15 cm. 

Given: Length (l) =10 cm, curved surface area of the cone = 188.4 sq.cm

To find: Perpendicular height (h) of the cone

Curved surface area of the cone = π r l

∴ 188.4 = 3.14 x r x 10

∴ r =  

 =  

 =  

 = 6 cm

Now, l² = r² + h²

∴ 10² = 6² + h²

∴ 100 = 36 + h²

∴ 100 - 36 = h²

∴ h² = 64

∴ h = … [Taking square root on both sides]

 = 8 cm

∴ The perpendicular height of the cone is 8 cm. 

Given: Height (h) = 24 cm,

Volume of cone = 1232 cm³

To find: Surface area of the cone

Solution:

Volume of Cone =  π² h

∴ 1232 =  x  x r² x h

∴ r² =  

 =  

∴ r² = 49

∴ r = … [Taking square root on both sides]

= 7 cm

ii. Now, l² = r² + h²

∴ l² = 7² + 24²

= 49 + 576 = 625

∴ l = … [Taking square root on both sides]

= 25

Curved surface area of cone = π r l

= x 7 x 25

= 22 x 25

= 550 sq.cm

∴The surface area of the cone is 550 sq.cm. 

Given: Length (l) = 50 cm, curved surface area of cone = 2200 sq.cm

To find: Total surface area of the cone

Curved surface area of cone = π r l

∴ 2200 =  x r x 50

 r =  

 r =  = 14cm

Total surface area of cone = πr (l + r)

= 227 x 14 x (50 + 14)

= x 14 x 64

= 22 x 2 x 64

= 2816 sq.cm

∴ The total surface area of the cone is 2816 sq.cm. 

Given: For the tent,

height (h) = 18m,

number of people in the tent = 25,

area required for each person = 4 sq.m

To find: Volume of the tent

Every person needs an area of 4 sq.m, of the ground inside the tent. Surface area of the base of the tent = number of people in the tent × area required for each person

= 25 × 4

= 100 sq.m

Surface area of the base of the tent = πr²

∴ 100 = πr²

∴ πr²= 100

Volume of the tent=13πr²h

= x 100 x 18 …….[∵ πr²= 100]

= 100 x 6

= 600 cubic metre

∴ The volume of the tent is 600 cubic metre.

Given: Height of the heap (h) = 2.1 m.

diameter of the base (d) = 7.2 m

∴ Radius of the base (r) == 7.22= 3.6 m

To find: Volume of the heap of the fodder and polythene sheet required

Volume of the heap of fodder = 13πr²h

= 13 x x (3.6)² x 2.1

= 13 x x 3.6 x 3.6 x 2.1

= 1 x 22 x 1.2 x 3.6 x 0.3

= 28.51 cubic metre

Now, l² = r² + h²

= (3.6)² + (2.1)²

= 12.96 + 4.41

∴ l² =17.37

∴ l² = .. .[Taking square root on both sides]

= 4.17 m

Area of the polythene sheet needed to cover the heap of the fodder = Curved surface area of the conical heap

= π r l

= x 3.6 x 4.17

= 47.18 sq.m

∴ The volume of the heap of the fodder is 28.51 cubic metre and a polythene sheet of 47.18 sq.m will be required to cover it. 

Given: Radius (r) = 4 cm

To find: Surface area and volume of sphere

Surface area of sphere = 4πr²

= 4 x 3.14 x 4²

∴ Surface area of sphere = 200.96 sq.cm

Volume of sphere =  π r³

= x 3.14 x 4²

∴ Volume of sphere = 267.95 cubic cm

Given: Radius (r) = 9 cm

To find: Surface area and volume of sphere

Surface area of sphere = 4 π r²

= 4 x 3.14 x 9²

∴ Surface area of sphere = 1017.36 sq.cm

Volume of sphere =  π r²

 =  x 3.14 x 9³

 =  x 3.14 x 9 x 9 x 9

 = 4 x 3.14 x 3 x 9 x 9

∴ Volume of sphere = 3052.08 cubic cm

Given: Radius (r) = 3.5 cm

To find: Surface area and volume of sphere

Solution:

Surface area of sphere = 4πr²

= 4 x 3.14 x (3.5)²

∴ Surface area of sphere = 153.86 sq.cm

Volume of sphere = πr³

= x 3.14 x (3.5)³

∴ Volume of sphere = 179.50 cubic cm 

Given: Radius (r) = 5 cm

To find: Curved surface area and total surface area of hemisphere

Curved surface area of hemisphere = 2πr²

= 2 x 3.14 x 5²

= 2 x 3.14 x 25

= 50 x 3.14

= 157 sq.cm.

Total surface area of hemisphere = 3πr²

= 3 x 3.14 x 5²

= 235.5 sq.cm.

∴ The curved surface area and total surface area of hemisphere are 157 sq.cm, and 235.5 sq.cm, respectively. 

Given: Surface area of sphere = 2826 sq.cm.

To find: Volume of sphere

Surface area of sphere = 4πr²

∴ 2826 = 4 x 3.14 x r²

2826 = 282600 = 900

∴r² =  =  =  

∴ r² = 225

∴ r = ……. [Taking square root on both sides]

= 15 cm

Volume of sphere =  π r³

= x 3.14 x 15³

= x 3.14 x 15 x 15 x 15

= 4 x 3.14 x 5 x 15 x 15

= 14130 cubic cm.

∴ The volume of the sphere is 14130 cubic cm.

Given: Volume of sphere = 38808 cubic cm.

To find: Surface area of sphere

Volume of sphere =  π r³

∴ 38808 =  x  x r³

∴ r³ = 

 =  

∴ r³ = 441 x 21 = 21 x 21 x 21

∴ r = 21 cm … [Taking cube root on both sides]

Surface area of sphere = 4πr²

= 4 x x 21

= 4 x x 21 x 21

= 4 x 22 x 3 x 21

= 5544 sq.cm.

∴ The surface area of sphere is 5544 sq.cm.

Given: Volume of hemisphere = 18000π cubic cm.

To find: Diameter of the hemisphere

Volume of hemisphere = π r³

∴ 18000 π =  π r³

∴ 18000 =  r³

 r³ =  

 = 9000 x 3

∴ r³ = 27000

∴ r = …… [Taking cube root on both sides]

= 30 cm

Diameter = 2r

= 2 x 30 = 60 cm

∴ The diameter of the hemisphere is 60 cm. 

Given: For road roller,

diameter(d) = 0.9 m, length (h) = 1.4 m

To find: Area of a field pressed in 500 rotations

Since, area of field pressed in 1 rotation of road roller = curved surface area of road roller

∴ Curved surface area of the road roller = 2 π r h

= π d h …...[∵ d = 2r]

= x 0.9 x 1.4 7

= 22 x 0.9 x 0.2

= 3.96 sq.m.

Area of land pressed in 1 rotation = 3.96 sq.m.

∴Area of land pressed in 500 rotations = 500 x 3.96

= 1980 sq.m.

∴ 1980 sq.m, land will be pressed in 500 rotations of the road roller. 

Given: Thickness of the glass = 2 mm,

Outer length of the tank = 60.4 cm,

Outer breadth of the tank = 40.4 cm,

Outer height of the tank = 40.2 cm To find: Volume of water fish tank contains

Thickness of the glass = 2 mm.

= cm

= 0.2 cm

Outer length of the tank= 60.4 cm

∴ Inner length oldie tank (l) = Outer length - thickness oldie glass on both sides

= 60.4 - 0.2 - 0.2

= 60cm

Outer breadth oldie tank = 40.4 cm

∴ Inner breadth of the tank (b) = 40.4 - 0.2 - 0.2

= 40 cm

Outer height of the tank = 40.2 cm∴ Inner height of the tank (h) = 40.2 - 0.2

= 40 cm

Given: Ratio of radius of base and height of a cone = 5 : 12,

Volume = 314 cubic metre

To find: Perpendicular height (h) and slant height (l)

The ratio of radius and height of cone is 5 : 12

Let the common multiple be x.

∴ Radius of base (r) = 5x

Perpendicular height (h) = 12x

 Volume of cone =  π r² h

∴ 314 =  x 3.14 x (5x)² x (12x)

∴ 314 =  x 3.14 x (25x²) x (12x)

∴ x³ =  

 =  

∴ x³ = 1

∴ x = 1 … [Taking cube root on both sides]

∴ r = 5x = 5(1) = 5m

h = 12x = 12(1) = 12 m

Now, l² = r² + h²

= 5² + 12²

= 25 + 144

∴l² = 169

∴ l =  …… [Taking square root on both sides]

= 13 m The perpendicular height and slant height of the cone are 12 m and 13 m respectively.

Given: Volume of sphere = 904.32 cubic cm.

To find: Radius of a sphere

Solution:

Volume of sphere = π r³

∴ 904.32 =  x 3.14 x r³

∴ r³ =  

 =  

 =   r³ = 216

∴ r = …… [Taking cube root on both sides]

= 6 cm

∴ The radius of the sphere is 6 cm.

Given: Total surface area of cube = 864 sq. cm

To find: Volume of cube

Total surface area of cube = 6 l²

∴ 864 = 6 l² 

∴ l² =  

∴ l² = 144

∴ l =  … [Taking square root on both sides] = 12 cm

Volume of cube = l³

= 12³

= 1728 cubic cm.

∴ The volume of cube is 1728 cubic cm. 

Given: Total surface area of a cone = 616 sq.cm., slant height of the cone is three times the radius of its base To find: Slant height (l)

Let the radius of base be r cm.

∴ Slant height (l) = 3r cm

Total surface area of cone = π r (l + r)

∴ 616 = π r (l + r)

∴ 616 =  x r x (3r + r)

∴ 616 = x 4r²

∴ r² =  

 =  

∴ r² = 49

∴ r = …… [Taking square root on both sides] 

= 7

Slant height (l) = 3r = 3 x 7 = 21 cm

∴ The slant height of the cone is 21 cm.

Given: Inner diameter (d) = 4.2 m,

To find: depth (h) = 10 m,

rate of plastering = Rs. 52 per sq.m.

Inner surface area and total cost of plastering Inner curved surface area of the well = 2 π r h

= π d h …[∵ d = 2r]

= x 4.2 x 10

= x 42

= 22 x 6

= 132 sq.m.

Rate of plastering = Rs.52 per sq.m.

∴ Total cost = Curved surface area x Rate of plastering

= 132 x 52

= Rs.6864

∴ The cost of plastering the well from inside is Rs.6864. 

Given: For road roller,

diameter (d) = 1.4 m,

length (h) = 2.1 m

number of rotations required for levelling the ground = 500,

rate of levelling = Rs. 7 per sq. m.

To find: Area of ground levelled by the road roller and cost of levelling Since, area of ground levelled in 1 rotation of road roller = curved surface area of road roller

∴Curved surface area of the road roller = 2 π r h

= π d h …….[∵ d = 2r]

= x 1.4 x 2.1

= 22 x 0.2 x 2.1

= 9.24 sq.m.

Area of ground levelled in 1 rotation = 9.24 sq.m.

∴Area of ground levelled in 500 rotations = 9.24 x 500 = 4620 sq.m.

Rate of levelling Rs. 7 per sq.m.

∴Total cost = Area of ground levelled x Rate of levelling

= 4620 x 7

= Rs.32340

∴ The road roller levels 4620 sq.m. land in 500 rotation, and the cost of levelling is Rs.32340.