{"id":26199,"date":"2021-06-25T11:58:26","date_gmt":"2021-06-25T06:28:26","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=26199"},"modified":"2021-06-25T11:58:26","modified_gmt":"2021-06-25T06:28:26","slug":"ncert-solutions-for-class-9-maths-chapter-13-ex-13-3","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-13-ex-13-3\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 13 Surface Areas and Volumes Ex 13.3 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 13.3<\/h2>\n

Question 1.
\nThe diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find the curved surface area.
\nSolution:
\n\"NCERT
\nWe have given that,
\nDiameter of the base of a cone = 10.5 cm
\nThe radius of the base of a cone = 5.25 cm
\nand slant height of the cone (l) = 10 cm
\n\u2234 The curved surface area of cone = \u03c0rl
\n= \\(\\frac {22}{7}\\) \u00d7 5.25 \u00d7 10
\n= 165 cm2<\/sup><\/p>\n

\"NCERT<\/p>\n

Question 2.
\nFind the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
\nSolution:
\n\"NCERT
\nWe have given,
\nSlant height of the cone = 21 m
\nand diameter of its base = 24 m
\n\u2234 radius of its base = 12 m
\n\u2234 Total surface area of a cone = \u03c0r(l + r)
\n= \\(\\frac {22}{7}\\) \u00d7 12 (21 + 12)
\n= \\(\\frac {22}{7}\\) \u00d7 12 \u00d7 33
\n= 1244.57 m2<\/sup> (approx)<\/p>\n

Question 3.
\nThe curved surface area of a cone is 308 cm2<\/sup> and its slant height is 14 cm. Find
\n(i) radius of the base
\n(ii) Total surface area of the cone.
\nSolution:
\n(i) We have given that
\nThe curved surface area of cone = 308 cm2<\/sup>
\n\"NCERT
\nand slant height of the cone = 14 cm.
\nWe know that,
\nCurved surface area of cone = \u03c0rl
\n308 = \\(\\frac {22}{7}\\) \u00d7 r \u00d7 14
\nr = 7 cm
\nTherefore, radious of base (r) = 7 cm<\/p>\n

(ii) Total surface area of cone = \u03c0r(l + r)
\n= \\(\\frac {22}{7}\\) \u00d7 7(14 + 7)
\n= 462 cm2<\/sup><\/p>\n

\"NCERT<\/p>\n

Question 4.
\nA conical tent is 10 m high and the radius of its base is 24 m. Find.
\n(i) Slant height of the tent
\n(ii) Cost of the canvas required to make the tent, if the cost of 1 m2<\/sup> canvas is Rs. 70.
\nSolution:
\n(i) We have given that
\nheight of the cone = 10 m
\nand radious of the cone = 24 m
\nBy Pythagoras theorem we know that
\nh2<\/sup> = p2<\/sup> + b2<\/sup>
\n\u21d2 AC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n\u21d2 AC2<\/sup> = (10)2<\/sup> + (24)2<\/sup>
\n\u21d2 AC2<\/sup> = 100 + 576
\n\u21d2 AC2<\/sup> = 676
\n\u21d2 AC = \u221a676 = 26 m
\n\"NCERT
\nTherefore slant height of the cone = 26 m.
\n(ii) Canvas required to make the tent = curved surface area of cone
\nor, Curved surface area of cone = \u03c0rl
\n= \\(\\frac {22}{7}\\) \u00d7 24 \u00d7 26
\n= \\(\\frac{13728}{7}\\) m2<\/sup>
\nCost of canvas = Rs. 70 per m2<\/sup>
\n\u2234 Total cost of canvas required to make the tent = \\(\\frac{13728}{7} \\times 70\\) = Rs. 1,37,280.<\/p>\n

Question 5.
\nWhat length of tarpaulin 3 m wide will be required to make a conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use \u03c0 = 3.14)
\nSolution:
\nWe have given that
\nThe radius of base of cone = 6 m
\nHeight of conical tent = 8 m
\n\"NCERT
\nBy Pythagoras theorem,
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n\u21d2 AC2<\/sup> = (9)2<\/sup> + (6)2<\/sup>
\n\u21d2 AC2<\/sup> = 64 + 36
\n\u21d2 AC2<\/sup> = 100
\n\u21d2 AC = 10 cm
\n\u2234 Curved surface area of cone = \u03c0rl
\n= \\(\\frac {22}{7}\\) \u00d7 6 \u00d7 10
\n= \\(\\frac{1320}{7}\\) m2<\/sup>
\nAccording to question
\nCurved surface area of conical tent = Area of rectangular tarpaulin.
\n\\(\\frac{1320}{7}\\) = x \u00d7 3
\nx = \\(\\frac{1320}{7 \\times 3}\\) = 62.87 m (approx)
\nAgain, stitching margins and wastage in cutting is 20 cm approximately.
\nTotal length of tarpaulin = 62.87 + 0.20 = 63m (approx).<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nThe slant height and base diameter of a Conical tomb is 25 m and 14 m respectively. Find the white-washing its curved surface at the rate of Rs. 210 per 100 m2<\/sup>.
\nSolution:
\nWe have given that
\nThe slant height of cone = 25 m
\nDiameter of its base = 14 m
\nThe radius of its base = 7 m
\n\"NCERT
\n\u2234 Curved surface area of cone = \u03c0rl
\n= \\(\\frac {22}{7}\\) \u00d7 7 \u00d7 25
\n= 550 m2<\/sup>
\nNow, rate of white washing = Rs. 210 per 100 m2<\/sup>
\n\u2234 Total cost of white washing the conical tomb = 550 \u00d7 \\(\\frac {210}{100}\\) = Rs. 1155.<\/p>\n

Question 7.
\nA joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
\nSolution:
\nWe have given that
\nRadius of conical cap = 7 cm
\nand height of conical cap = 24 cm
\nBy Pythagoras theorem,
\nAC = \\(\\sqrt{(\\mathrm{AB})^{2}+(\\mathrm{BC})^{2}}\\)
\n\u21d2 AC = \\(\\sqrt{(24)^{2}+(7)^{2}}\\)
\n\u21d2 AC = \\(\\sqrt{576+49}\\)
\n\u21d2 AC = \u221a625
\n\u21d2 AC = 25 cm
\n\"NCERT
\n\u2234 Curved surface area of one cap = \u03c0rl
\n= \\(\\frac {22}{7}\\) \u00d7 7 \u00d7 25
\n= 550 cm2<\/sup>
\nSheet required to make one jocker cap = 550 cm2<\/sup>
\nSheet required to make ten such caps = 550 \u00d7 10 = 5500 cm2<\/sup>.<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nA bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each one has a base diameter of 40 cm and height lm. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per m2<\/sup>, what will be the cost of painting all these cones?
\n(Use \u03c0 = 3.14 and take \u221a1.04 = 1.02)
\nSolution:
\nWe have given that
\nHeight of the cone = 1 m
\nand diameter of its base = 40 cm = 0.40 m
\nradius = \\(\\frac{0.40}{2}\\) = 0.20 m
\n\"NCERT
\nBy Pythagoras theorem
\nAC = \\(\\sqrt{(\\mathrm{AB})^{2}+(\\mathrm{BC})^{2}}\\)
\n= \\(\\sqrt{(1)^{2}+(0.20)^{2}}\\)
\n= \\(\\sqrt{100400}\\) (\u2234 \u221a1.04 = 1.02 Given)
\n= 1.02m
\nNow, curved surface area of one cone = \u03c0rl
\n= 3.14 \u00d7 0.20 \u00d7 1.02
\n= 0.640 m2<\/sup> (approx)
\nTherefore, the curved surface area of 50 such cones = 50 \u00d7 0.640
\n= 32.02 m2<\/sup>
\nRate of painting the cone = Rs. 12 per m2<\/sup>
\nTotal cost for painting fifty cones is 32.02 \u00d7 12 = Rs. 384.24 (approx)<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.3 Question 1. The diameter of the base of a cone is 10.5 cm and its …<\/p>\n

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NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.3 Question 1. 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