{"id":26530,"date":"2021-06-26T11:03:36","date_gmt":"2021-06-26T05:33:36","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=26530"},"modified":"2021-06-26T11:03:36","modified_gmt":"2021-06-26T05:33:36","slug":"ncert-solutions-for-class-9-maths-chapter-13-ex-13-9","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-13-ex-13-9\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 13 Surface Areas and Volumes Ex 13.9 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.9<\/h2>\n

Question 1.
\nA wooden bookshelf has external dimension as follows: Height = 110 cm, Depth = 25 cm, breadth = 85 cm (see Fig. 13.31). The 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<\/span><\/sup><\/span> and the rate of painting is 10 paise per cm2<\/span><\/sup><\/span>, find the total expenses required for polishing and painting the surface of the bookshelf.
\n\"NCERT
\nSolution:
\nWe have given that
\nHeight of bookshelf = 110 cm
\nDepth of bookshelf = 25 cm
\nand Breadth of bookshelf = 85 cm
\nThe thickness of the plank = 5 cm
\nTotal surface area of external face (Shading portion) = 2 (25 \u00d7 110) + 2 (85 \u00d7 25) + 110 \u00d7 85 + 4(5 \u00d7 85) + 2(90 \u00d7 5)
\n= 5500 + 4250 + 9350 + 1700 + 900
\n= 21,700 cm2<\/span><\/sup><\/span>
\nTotal surface area of internal faces (without shading portion)
\n= 6(20 \u00d7 75) + 2(90 \u00d7 20) + 90 \u00d7 75
\n= 9000 + 3600 + 6750
\n= 19350 cm2<\/span><\/sup><\/span>
\nNow, cost of polishing the external faces at the rate of 20 paise\/cm2<\/span><\/sup><\/span> is = 21700 \u00d7 0.20 = Rs. 4340
\nand cost of painting the internal faces at tire rate of 10 paise\/cm2<\/span><\/sup><\/span> is, the bookself = 4340 + 1935 = Rs. 6275<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nThe front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Fig. 13.32. Eight 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<\/span><\/sup><\/span> and black paint costs 5 paise per cm2<\/span><\/sup><\/span>.
\n\"NCERT
\nSolution:
\nThe surface area of one wooden sphere = 4\u03c0r2<\/span><\/sup><\/span>
\n= \\(4 \\times \\frac{22}{7} \\times\\left(\\frac{21}{2}\\right)^{2}\\)
\nSo, total surface area of all the 8.
\nWooden sphere = \\(8 \\times 4 \\times \\frac{22}{7} \\times\\left(\\frac{21}{2}\\right)^{2}\\) = 11088 cm2<\/span><\/sup><\/span>
\nNow, some areas of the sphere can’t be silver printed due to the support.
\nSupport area = \\(8 \\times \\frac{22}{7} \\times(1.5)^{2}\\) = 56.57 cm2<\/span><\/sup><\/span>
\nRequired area for silver painting = 11088 – 56.57 = 11031.43 cm2<\/span><\/sup><\/span>
\nCost of silver paint = 11031.43 \u00d7 0.25 = Rs. 2757.85
\nNow, support is to be painted in black colour.
\nSo, curved surface of cylindrical support of:
\n= 8 \u00d7 2 \u00d7 \\(\\frac {22}{7}\\) \u00d7 7 \u00d7 15
\n= 528 cm2<\/span><\/sup><\/span>
\nCost to paint the support in black = 528 \u00d7 0.05 = Rs. 26.40
\nTotal cost of paint = 2759.85 + 26.40 = Rs. 2784.25<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nThe diameter of a sphere is decreased by 25%. By what percent does its curved surface area decrease?
\nSolution:
\nLet the diameter of sphere = d
\n\"NCERT
\n\u2234 Radius of sphere = \\(\\frac{d}{2}\\)
\n\u2234 Curved surface area of sphere = 4\u03c0r2<\/span><\/sup><\/span>
\n= \\(4 \\times \\pi \\times \\frac{d}{2} \\times \\frac{d}{2}\\)
\n= \u03c0d2<\/span><\/sup><\/span>
\nIn second case,
\ndiameter of new sphere = d – 25% of d
\n= d – \\(\\frac{25}{100} \\times d\\)
\n= \\(\\frac{3 d}{4}\\)
\n\u2234 Radius of new sphere = \\(\\frac{3 d}{8}\\)
\n\"NCERT
\n\u2234 Curved surface area of new sphere = 4\u03c0r2<\/span><\/sup><\/span>
\n= \\(4 \\times \\pi \\times \\frac{3 d}{8} \\times \\frac{3 d}{8}\\)
\n= \\(\\frac{9 \\pi d^{2}}{16}\\)
\nTherefore, surface area decrease in second case is = \u03c0d2<\/span><\/sup><\/span> – \\(\\frac{9 \\pi d^{2}}{16}\\)
\n= \\(\\frac{16 \\pi d^{2}-9 \\pi d^{2}}{16}\\)
\n= \\(\\frac{7 \\pi d^{2}}{16}\\)
\nSo, Percent decreasing = \\(\\frac{7 \\pi d^{2}}{\\frac{16}{\\pi d^{2}} \\times 100}\\)
\n= \\(\\frac{7 \\pi d^{2}}{16 \\pi d^{2}} \\times 100\\)
\n= 43.75%
\nSo, the curved surface area decrease in the second case is 43.75%.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.9 Question 1. A wooden bookshelf has external dimension as follows: Height = 110 cm, Depth …<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-13-ex-13-9\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 Questions and Answers are prepared by our highly skilled subject experts. 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