\nA wooden bookshelf has external dimensions as follows:
\nHeight = 110 cm, Depth = 25 cm, Breadth = 85 cm.
\n
![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Areas-and-Volumes-Ex-13.9-1.png\")
\nThe thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm2<\/sup> and the rate of painting is 10 paise per cm2<\/sup>, find the total expenses required for polishing and painting the surface of the bookshelf.
\nSolution:
\nOuter length, l = 25 cm.
\nouter breadth, b = 85 cm.
\nouter height, h = 110 cm.
\nOuter surface area = lh + 2lb + 2bh = lh + 2(lb + bh)
\n= 85 \u00d7 110 + 2(85 \u00d7 25 + 25 \u00d7 110)
\n= 9350 + 9750
\n= 19100 cm2<\/sup>.
\nArea of front face,
\n= [85 \u00d7 110 – 75 \u00d7 100 + 2(75 \u00d7 5)]
\n= (9350 – 7500 + 2(375)]
\n= 9350 – 7500 + 750
\n= 11000 – 7500
\n= 3500 cm2<\/sup>.
\nArea to be polished,
\n= 19100 + 3500
\n= 22600 cm2<\/sup>.
\nCost of polishing 1 cm3<\/sup> is Rs. 0.20.
\nCost of polishing 22600 cm3 … ? …
\n= 22600 \u00d7 0.20 = Rs. 4520.
\nLength of horizontal shelf, l = 75 cm.
\nbreadth, b = 20 cm.
\nheight, h = 30 cm.
\nArea of horizontal shelf
\n= 2(l + h)b + lh
\n= [2(75 + 30) \u00d7 20 + 75 \u00d7 30]
\n= (4200 + 2250] cm2<\/sup>.
\n= 6450 cm2<\/sup>.
\n\u2234 Area of painting 3 horizontal rows
\n= 3 \u00d7 6450
\n= 19350 cm2<\/sup>.
\nCost of painting for 1 cm3<\/sup> is Rs. 0.10.
\n\u2234 Cost of painting 19350 cm3 … ?
\n= 19350 \u00d7 0.10 = Rs. 1935.
\n\u2234 Total cost of polish and painting
\n= Rs. 4520 + Rs. 1935
\n= Rs. 6455.<\/p>\n
Question 2. Question 3. <\/p>\n","protected":false},"excerpt":{"rendered":" Karnataka Board Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 Question 1. A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm. The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces …<\/p>\n","protected":false},"author":6,"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\/2651"}],"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\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=2651"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2651\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=2651"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=2651"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=2651"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nThe front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Fig.
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Areas-and-Volumes-Ex-13.9-2.png\")
\nEight such spheres are used for this purpose and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2<\/sup> and black paint costs 5 paise per cm2<\/sup>.
\nSolution:
\nDiameter of a wooden frame, d = 21 cm.
\nRadius r = \\(\\frac{21}{2}\\)
\nOuter surface area of wooden spheres,
\nA = 4\u03c0r2<\/sup>
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Areas-and-Volumes-Ex-13.9-3.png\")
\n\u2234 A = 1386cm2<\/sup>
\nRadius of cylinder, r1<\/sub> = 1.5 cm
\nheight, h = 7 cm
\nThe curved surface area of cylinder support,
\nA = 2\u03c0rh
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Areas-and-Volumes-Ex-13.9-4.png\")
\nArea of silver painted
\n= [8 \u00d7 (1386 – 7.07)] cm2<\/sup>
\n= 8 \u00d7 1378.93
\n= 11031.44 cm2<\/sup>
\nCost of silver paint = 11031.44 \u00d7 0.25
\n= Rs. 2757.86
\nArea of black paint = (8 \u00d7 66) cm2<\/sup>
\n= 528 cm2<\/sup>
\nCost of black paint = 528 \u00d7 0.05
\n= Rs. 26.40
\nTotal cost of silver and black paint.
\n= Rs. 2757.76 + Rs. 26.40
\n= Rs. 2784.26.<\/p>\n
\nThe diameter of a sphere is decreased by 25%. By what percent does its curved surface area decrease?
\nSolution:
\nLet the diametrer of sphere be \u2018d\u2019
\nRadius of sphere, r1<\/sub> = \\(\\frac{\\mathrm{d}}{2}\\)
\nRadius of outer sphere, r2<\/sub> = \\(\\frac{\\mathrm{d}}{2}\\left(1-\\frac{25}{100}\\right)\\)
\n\u2234 r2<\/sup> = \\(\\frac{3}{8}\\)d
\nOuter Area of Sphere, S1<\/sub> = 4\u03c0r1<\/sub>2<\/sup>
\n= 4\u03c0\\(\\left(\\frac{\\mathrm{d}}{2}\\right)^{2}\\)
\nS1<\/sub> = \u03c0d2<\/sup>
\nThe diameter of sphere is decreased by 25%. Then its outer surface area,
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Areas-and-Volumes-Ex-13.9-5.png\")
<\/p>\nKSEEB Solutions for Class 9 Maths<\/a><\/h5>\n