Maharashtra Board Class 9 Math Part 2 Solution Chapter 9 – Surface Area and Volume
Balbharati Maharashtra Board Class 9 Math Part 2 Solution Chapter 9: Surface Area and Volume. Marathi or English Medium Students of Class 9 get here Surface Area and Volume full Exercise Solution.
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Std |
Maharashtra Class 9 |
| Subject |
Math Part 2 Solution |
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Chapter |
Surface Area and Volume |
Practice Set 9.1
(1) Length, breadth and height of a cuboid shape box of medicine is 20cm, 12 cm and 10 cm respectively. Find the surface area of vertical faces and total surface area of this box.
Solution:
Given, l = 20 cm, b = 12 cm, h = 10 cm
Surface area of vertical faces = 2 (l + b) × n
2 (20 + 12) × 10
= 64 × 10 = 640 cm
Total surface area = 2 (lb + bh + hl)
= 2 (240 + 120 + 200)
= 2 (560)
= 1120 cm2
(2) Total surface area of a box of cuboid shape is 500 sq. unit. Its breadth and height is 6 unit and 5 unit respectively. What is the length of that box?
Solution:
Given, total surface area of a cuboid = 500 sq unit, b = 6 unit, h = 5 unit
We know, total surface area of cuboid = 2 (lb + bh + hl)
=) 500 = 2 (6l + 30 + 5l)
=) 11l = 250 – 30
=) 220
=) l = 20 units
(3) Side of a cube is 4.5 cm. Find the surface area of all vertical faces and total surface area of the cube.
Solution:
Given, l = 4.5 cm
Surface area of vertical faces of cuboid cubs = 4l2
= 4 × (4.5)2
= 81 cm2
Total surface area of cubs = 6l2 = 6 × (4.5)2 = 121.5 cm2
(4) Total surface area of a cube is 5400 sq. cm. Find the surface area of all vertical faces of the cube.
Solution:
Given, total surface area of cube = 5400 cm2
=) 6l2 = 5400
=) l2 = 900 cm2
∴ Surface area of all vertical faces of cube = 4l2 = 4 × 900 = 3600 cm2
(5) Volume of a cuboid is 34.50 cubic metre. Breadth and height of the cuboid is 1.5m and 1.15m respectively. Find its length.
Solution:
Given, volume of cuboid = 34.5 m3
b = 1.5 m, h = 1.15m
We know, volume of cuboid = l × b × h
=) 34.5 = l × 1.5 × 1.15
= l = 20m
(6) What will be the volume of a cube having length of edge 7.5 cm?
Solution:
Given, l = 7.5 cm
Volume of cube = l3 = (7.5)3 cm3
= 421.875 cm3
(7) Radius of base of a cylinder is 20cm and its height is 13cm, find its curved surface area and total surface area. (π = 3.14)
Solution:
Given, r = 20cm, h = 13cm
Curved surface area = 2 × 314 × 20 × 13
= 1632.8 cm2
Total surface are = 2πr2 + 2πrh
= 2 × 3.14 × (20)2 + 1632.8
= 4144.8 cm2
(8) Curved surface area of a cylinder is 1980 cm2 and radius of its base is 15cm. Find the height of the cylinder. (π = 22/7)
Solution:
Given, curved surface area of cylinder = 1980 cm2
r = 15cm
We know,
CSA of cylinder = 2πrh
=) 1980 = 2 × 22/7 × 15 × h
=) h = 1980 ×7/2×22×15
= 13860/660
= 21cm
Practice Set 9.2
(1) Perpendicular height of a cone is 12 cm and its slant height is 13 cm. Find the radius of the base of the cone.
Solution:
Given, h = 12 cm, l = 13cm
We know, l2 = r2 + h2
=) r2 = 132 – 122
=) 169 – 144 = 25
=) r = 5cm
(2) Find the volume of a cone, if its total surface area is 7128 sq.cm and radius of base is 28 cm. (π = 22/7)
Solution:
Total surface area of cone = 7128 cm2
r = 28 cm
∴ TSA of cone = π r (l + r)
=) 7128 = 22/7 × 28 (l + 28)
=) l + 28 = 7128×7/22×28
=) l = 81 – 28 = 53cm
∴ l2 = r2 + h2
= h = √l2 – r2
= √532 – 282
= 45 cm
∴ Volume of cone = 1/3 πr2h
= 1/3 × 22/7 × 28 × 28 × 45
= 36960 cm3
(3) Curved surface area of a cone is 251.2 cm2 and radius of its base is 8cm. Find its slant height and perpendicular height. (π = 3.14)
Solution:
Given CSA of cone = 251.2 cm2
r = 8 cm
We know, CSA of cone = πrl
=) 251.2 = 3.14 × 8 × l
=) l = 251.2/3.14 × 8
∴ l = 10cm
l2 = h2 + r2
=) h = √l2 – r2 = √100 – 64 = √36 = 6cm
(4) What will be the cost of making a closed cone of tin sheet having radius of base 6 m and slant height 8 m if the rate of making is Rs.10 per sq. m?
Solution:
Given,
r = 6 m, l = 8 m
∴ TSA of cone = πr (l + r)
=) 22/7 × 6 (6 + 8)
=) 22 × 12 = 264 m2
Cost of making the cone = ₹264 × 10
= ₹2640
(5) Volume of a cone is 6280 cubic cm and base radius of the cone is 30 cm. Find its perpendicular height. (π = 3.14)
Solution:
Given, volume of cone = 6280 cm3
r = 30 cm
We know, Volume of cone = 1/3 πr2h
=) 6280 = 1/3 × 3.14 × 30 × 30 × h
=) h = 6280 × 3/3.14 × 900
= 667 cm
(6) Surface area of a cone is 188.4 sq. cm and its slant height is 10cm. Find its perpendicular height (π = 3.14)
Solution:
Given, Surface area of cone = 188.4 cm2
l = 10cm
We know, CSA of cone = πrl
=) 188.4 = 3.14 × π × 10
= π = 1884/314 = 6 cm
∴ h = √l2 – r2
= √100 – 36 = 8cm
(7) Volume of a cone is 1212 cm3 and its height is 24cm. Find the surface area of the cone. (π = 22/7)
Solution:
Given volume of a cone = 1212 cm3
h = 24 cm
We know, volume of cone = 1/3 πr2h
=) 1212 = 1/3 × 22/7 × r2 × 24
=) r2 = 1212×21/22×24 = 48.02
= r = 6.94 cm = 7 cm (Approve)
∴ l = √h2 + r2 = √242 + 49
= 24 cm
CSA of cone = πrl
= 7 × 25 × 22/7
= 22 × 25
= 550 cm2
(8) The curved surface area of a cone is 2200 sq.cm and its slant height is 50 cm. Find the total surface area of cone. (π = 22/7)
Solution:
CSA of cone = 2200 cm2
l = 50cm
CSA o cone = πrl
=) 2200 = 22/7 × r × 50
=) r = 14cm
TSA of cone = 2200 + 22/7 × 14 × 14
= 2816 cm2
(9) There are 25 persons in a tent which is conical in shape. Every person needs an area of 4 sq. m. of the ground inside the tent. If height of the tent is 18m, find the volume of the tent.
Solution:
Given, h = 18 m
Are of each person on ground = 4 m2
∴ 25 person needs area = 25 × 4 = 100 m2
∴ Volume of cone = 1/3 πr2h
= 1/3 × 100 × 18
= 600 m3
(10) In a field, dry fodder for the cattle is heaped in a conical shape. The height of the cone is 2.1m. and diameter of base is 7.2 m. Find the volume of the fodder. if it is to be covered by polythin in rainy season then how much minimum polythin sheet is needed? (π = 22/7 and √17.37 = 4.17)
Solution:
Given, h = 2.1 m
l = 7.2 m
∴ r = 7.2/2 = 3.6 m
Volume of fodder = 1/3 πr2 h
= 1/3 × 22/7 × 3.6 × 3.6 × 2.1
= 28.512 m3
l = √h2 + r2 = √(2.1)2 + (3.6)2 = 4.17 m
∴ CSA of cone = πrl
= 22/7 × 3.6 × 4.1
= 47.18 m2
∴ Fodder is covered with 47.18m2 of polythene
Practice Set – 9.3
(1) Find the surface areas and volumes of spheres of the following radii.
(i) 4cm
(ii) 9cm
(iii) 3.5cm (π = 3.14)
Solution:
(i) Surface area = 4 π r2
= 4 × 3.14 × 42
= 200.96 cm2
Volume = 4/3 πr3
= 267 .95 cm3
(ii) Surface area = 4 π r2
= 4 × 3.14 × 92
= 1017.36 cm2
Volume = 4/3 π r3
= 3052.08 cm3
(iii) Surface area = 4 π r2
= 4 × 3.14 × (3.5)2
= 153.86 cm2
Volume = 4/3 π r3
= 179.5 cm3
(2) If the radius of a solid hemisphere is 5cm, then find its curved surface area and total surface area. (π = 3.14)
Solution:
Given, r = 50 m
CSA of hemisphere = 2 πr2
= 2 × 3.14 × 52
= 157 cm2
TSA of hemisphere = 3πr2
= 235.5 cm2
(3) If the surface area of a sphere is 2826 cm2 then find its volume. (π = 3.14)
Solution:
Surface area of sphere = 2826 cm2
=) 4 π r2 = 2826
=) r2 = 2826/4×3.14 = 225
=) r = 15 cm
Volume of sphere = 4/3 πr3
= 4/3 × 3.14 × 153
= 14130 cm3
(4) Find the surface area of a sphere, if its volume is 38808 cubic cm. (π = 22/7)
Solution:
Volume of sphere = 38808 cm3
=) 4/3 π r3 = 38808
=) r3 = 9261
=) r = 21cm
Surface area of sphere = 4 πr2
= 4 × 22/7 × 21 × 21
= 5544 cm2
(5) Volume of a hemisphere is 18000 π cubic cm. Find its diameter.
Solution:
Volume of hemisphere = 18000π cm3
=) 2/3 πr3 = 1800 π
=) r3 = 18000×3/2 = 27000
=) r = 30cm
∴ d = 2×30 = 60 cm
Problem Set – 9
(1) If diameter of a road roller is 0.9 m and its length is 1.4 m, how much area of a field will be pressed in its 500 rotations?
Solution:
Given, d = 0.9 m
r = 0.45 m
h = 1.4 cm
CSA of cylinder = 2 π rh
= 2 × 22/7 × 0.45 × 1.4
= 3.96 cm2
Area covered on 500 rotation = 500×3.96
= 1980 cm2
(2) To make an open fish tank, a glass sheet of 2 mm gauge is used. The outer length, breadth and height of the tank are 60.4 cm, 40.4 cm and 40.2 cm respectively. How much maximum volume of water will be contained in it?
Solution:
Thickness of glass = 2mm = 2/10 cm = 0.2 cm
Inner length = 60.4 – 0.2 – 0.2 = 60cm
Inner breadth = 40.4 – 0.2 – 0.2 = 40cm
Inner height = 40.2 – 0.2 = 40cm
Volume of water = l × b × h
= 60 × 40 × 40
= 96000 cm3
(3) If the ratio of radius of base and height of a cone is 5:12 and its volume is 314 cubic metre. Find its perpendicular height and slant height (π = 3.14)
Given, r/h = 5/12
Volume of cone = 314 m3
= 1/3 π r2 h = 314
= 1/3 × 3.14 × (5/12 h)2 × h = 314
=) h3 = 144×3×100/25 = 1728
=) h = 12 m
∴ r = 5m
∴ l = √h2 + r2 = √144 + 25 = 13m
(4) Find the radius of a sphere if its volume is 904.32 cubic cm. (π = 3.14)
Solution:
Volume of sphere = 904.32 cm3
= 4/3 π r3 = 904.32
= r3 = 904.32×3/4×3.14 = 216
= r = 6 cm
(5) Total surface area of a cube is 864 sq. cm. Find its volume.
Solution:
TSA of cube = 864 cm2
=) 6l2 = 864
=) l = 144
=) l = 12cm
Volume of cube = l3 = 123 = 1728 cm3
(6) Find the volume of a sphere, if its surface area is 154 sq. cm.
Solution:
Volume of sphere = 4/3 π r3
Surface area of sphere = 154 cm2
=) 4 π r2 = 154
=) r2 = 154 ×7/4×22 = 12.25
=) r = 3.5 cm
∴ Volume of sphere = 4/3 × 22/7 × (3.5)3
= 179.67 cm3
(7) Total surface area of a cone is 616 sq. cm. If the slant height of the cone is three times the radius of its base, find its slant height.
Solution:
TSA of cone = 616 cm2
l = 3r
TSA of cone = πr (l + r)
=) 616 = πr (3r + r)
= 4 πr2
= r2 = 616/4 π = 49.04
= 49 (Approx)
∴ r = 7cm
∴ l = 21cm
(8) The inner diameter of a well is 4.20 metre and its depth is 10 metre. Find the inner surface area of the well. Find the cost of plastering it from inside at the rate Rs.52 per sq. m.
Solution:
l = 4.20 m
r = 2.1 m
h = 10m
Surface area of well = 2 π rh
= 2 × 3.14 × 2.1 × 10
= 132 m2
Cost of plastering the above area
= 132 × 52
= ₹6864
(9) The length of a road roller is 2.1m and its diameter is 1.4m. For levelling a ground 500 rotations of the road roller were required. How much area of ground was leveled by the road roller? Find the cost of levelling at the rate of Rs. 7 per sq. m.
Solution:
h = 2.1 m
l = 1.4 m
r = 0.7 m
CSA of road and roller = 2 π rh
= 2 × 3.14 × 0.7 × 2.1
= 9.24 m2
Area of ground leveled = 500 × 9.24
= 4620 m2
Cost of leveling = 4620 × 7
= ₹ 32340
