\nThe curved surface area of a right circular cylinder of a height 14 cm is 88 cm2<\/sup>. Find the diameter of the base of the cylinder.
\nSolution:
\nLateral surface area of cylinder which has height 14 cm is 88 cm2<\/sup>.
\nDiameter of bottom of cylinder, d = ?
\nh = 14 cm, L.S.A. = 88 cm2<\/sup>, d = ?
\nLateral surface area of cylinder, L.S.A.
\nL.S.A.= 2\u03c0rh = 88
\n
![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-1.png\")
\n88 \u00d7 r = 88
\n\u2234 r = \\(\\frac{88}{88}\\) \u2234 r = 1 cm.
\n\u2234 Diameter, d = 2r =2 \u00d7 1 = 2 cm.<\/p>\n
Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.2 (If values are not given for ‘\u03c0’ Assume \u03c0 = ) Question 1. The curved surface area of a right circular cylinder of a height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. Solution: Lateral surface …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2266"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=2266"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2266\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=2266"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=2266"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=2266"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nIt is required to make a closed cylinderical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same ?
\nSolution:
\nHeight of cylindrical water tank, h = 1 m
\nDiameter of bottom, d = 140 cm.
\nh = 1 m, d = 140 cm,
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-2.png\")
\nr = 0.7 m.
\nArea of metal sheet required to prepare water tank, A
\nA = 2\u03c0r (r + h)
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-2.1.png\")
\n= 44 \u00d7 0.1 \u00d7 1.7
\n= 4.4 \u00d7 1.7
\nA = 7.48 m2<\/sup>.<\/p>\n
\nA metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see Fig. 13.11). Find its
\n(i) inner curved surface area,
\n(ii) outer curved surface area,
\n(iii) total surface area.
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-3.png\")
\nSolution:
\nLength of metal pipe, l = 77 cm,
\nh = 77 cm.
\nInner diameter = 4 cm, \u2234 r1<\/sub> = 2 cm.
\nOuter diameter= 4.4 cm. \u2234 r2<\/sub> = 2.2 cm.
\n(i) Inner surfce area of pipe (C.S.A) = 2\u03c0r1<\/sub>h
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-3.1.png\")
\n= 2 \u00d7 22 \u00d7 2 \u00d7 11
\n= 968 cm2<\/sup>.
\n(ii) Outer surface area of pipe (C.S.A) = 2\u03c0r2<\/sub>h
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-3.2.png\")
\n= 2 \u00d7 22 \u00d7 2.2 \u00d7 11
\n= 44 \u00d7 24.2
\n= 1064.8 cm2<\/sup>.
\n(iii) Total Surface area of Pipe (T.S.A.) = Inner CSA + Outer CSA + Area of circular ring.
\n= 2\u03c0r1<\/sub>h + 2\u03c0r2<\/sub>h + (2\u03c0r2<\/sub>2<\/sup>h – 2\u03c0r1<\/sub>2<\/sup>h)
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-3.3.png\")
\n= 2032.8 + 5.28
\n\u2234 T.S.A. = 2038.08 cm2<\/sup>.<\/p>\n
\nThe diameter of a roller is 84 cm and its length is 120 m. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2.
\nSolution:
\nRoller is in the form of cylinder.
\nDiameter, d = 84 cm,
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-4-300x44.png\")
\nl = h = 120 cm \u21d2 1.2 m.
\n\u2234 Curved surface area of Roller = 2 \u03c0rh
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-4.1.png\")
\n= 44 \u00d7 0.072
\n= 3.168 m2<\/sup>
\n\u2234 Area of 1 round of roller = 3.168 m2<\/sup>.
\nArea of 500 rounds … ? …
\n= 3.168 \u00d7 500
\n= 1584 m2<\/sup>.
\n\u2234 Total area of a playground is 1584 m2<\/sup>.<\/p>\n
\nA cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2<\/sup>.
\nSolution:
\nDiameter of cylindrical pole, d = 50 cm.
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-5.png\")
\n\u2234 r = 0.25 m
\nheight of pole, h = 3.5 m.
\nCurved surface area of pole = 2\u03c0rh
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-5.1.png\")
\n= 44 \u00d7 0.25 \u00d7 0.5
\n= 5.5 m2<\/sup>.
\nCost of painting for 1 m2 is Rs. 12.50,
\nCost of painting for 5.5 m2<\/sup> is … ? …
\n= 5.5 \u00d7 12.50
\n= Rs.68.75.<\/p>\n
\nCurved surface area of a right circular cylinder is 4.4 m2<\/sup>. If the radius of the base of the cylinder is 0.7 m, find its height.
\nSolution:
\nCurved surface arc of cylinder = 4.4 m2<\/sup>
\nradius, r = 0.7 m height, h = ?
\nC.S.A. of cylinder, A= 2\u03c0rh
\n2\u03c0rh = 4.4
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-6.png\")
\n44 \u00d7 0.1 \u00d7 h = 4.4
\n4.4 h = 4.4
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-6.1.png\")
\n\u2234 h = 1 m.<\/p>\n
\nThe inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
\n(i) its inner curved surface area.
\n(ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2<\/sup>.
\nSolution:
\nInner diameter of circular well, d = 3.5 m.
\nRadius, r = \\(\\frac{3.5}{2}\\) m
\nDepth of the well – height, h = 10 m.
\n(i) Inner surfacer area of circular well, C.S.A. = 2\u03c0rh
\nC.S.A. = 2\u03c0rh
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-7.png\")
\n= 22 \u00d7 0.5 \u00d7 10
\n= 110 m2<\/sup>
\n(ii) Cost of plastering of area 1 m2<\/sup> is Rs. 40
\nCost of plastering 110 m2<\/sup>… ? …
\n= 110 \u00d7 40
\n= Rs. 4400.<\/p>\n
\nIn a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
\nSolution:
\nDiameter of cylindrical pipe, d = 5 cm
\n= 0.5 m
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-8.png\")
\nLength of pipe = height, h = 28 m.
\nCurved surface area of pipe = ?
\nC.S.A.= 2\u03c0rh
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-8.1.png\")
<\/p>\n
\nFind
\n(i) The lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
\n(ii) how much steel was actually used, if \\(\\frac{1}{12}\\) of the steel actually used was wasted in making the tank.
\nSolution:
\n(i) Petrol tank is in the form of cylinder.
\nDiameter, d = 4.2 m
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-9.png\")
\nh = 4.5 m
\nTSA of petrol tank = 2\u03c0r(r + h)
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-9.1.png\")
\n= 44 \u00d7 0.3 \u00d7 6.6
\n= 87.12 m2<\/sup>
\n(ii) \u2234 Area of iron required for manufacturing tanker = 87.12 m2<\/sup>.
\nIf iron used is 1 sq. m then there is a loss of \\(\\frac{1}{12}\\).
\nIron required except loss = \\(1-\\frac{1}{12}\\)
\n= \\(\\frac{11}{12}\\) m2<\/sup>
\nArea of \\(\\frac{11}{12}\\) m2<\/sup> is 87.12 m2<\/sup>.
\nArea of 1 m2<\/sup> … ? …
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-9.2.png\")
\n= 95.04 m2<\/sup>.<\/p>\n
\nIn Fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-10.png\")
\nSolution:
\nDiameter of cylindrical frame, d= 20 cm,
\n\u2234 r = 10 cm.
\nheight, h = 30 cm.
\nFor folding 2.5 cm is required.
\n\u2234 Total height = 30 + 2.5 + 2.5
\n= 35 cm.
\nCurved surface area of cloth covered, A
\nA= 2\u03c0rh
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-10.1.png\")
\n= 44 \u00d7 50
\n= 2200 cm2<\/sup>.<\/p>\n
\nThe students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using card-board. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?
\nSolution:
\nPenholder is in the shape of a cylinder, which one of the edges is open and another is closed.
\nradius, r = 3 cm, height, h = 10.5 cm.
\nTotal cardboard required by one Competitor: = Curved Surface Area + Area of bottom = 2\u03c0rh + \u03c0r2<\/sup>
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-12.png\")
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-11.png\")
\n\u2234 Total cardboard required for 35 competitors is :
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.2-Q-11.1.png\")
<\/p>\nKSEEB Solutions for Class 9 Maths<\/a><\/h5>\n","protected":false},"excerpt":{"rendered":"