NCERT Solutions for Class 9 Maths<\/a> Chapter 13 Surface Areas and Volumes Ex 13.7 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.7<\/h2>\n Question 1. \nFind the volume of the right circular cone with \n(i) radius 6 cm, height 7 cm. \n(ii) radius 3.5 cm, height 12 cm. \nSolution: \n(i) We have given \nradius of right circular cone = 6 cm \nand height of right circular cone = 7 cm. \n \nVolume of right circular cone = \\(\\frac{1}{3} \\pi r^{2} h\\) \n= \\(\\frac{1}{3} \\times \\frac{22}{7} 6 \\times 6 \\times 7\\) \n= 264 cm3<\/sup><\/p>\n(ii) We have given \nRadius of cone = 3.5 cm \nand height of cone = 12 cm. \n \nVolume of cone = \\(\\frac{1}{3} \\pi r^{2} h\\) \n= \\(\\frac{1}{3} \\times \\frac{22}{7}\\) \u00d7 3.5 \u00d7 3.5 \u00d7 12 \n= 154 cm3<\/sup><\/p>\n <\/p>\n
Question 2. \nFind the capacity in litres of a conical vessel with \n(i) radius 7 cm, slant height 25 cm. \n(ii) height 12 cm and slant height 13 cm. \nSolution: \n(i) We have given that, \nRadius of conical vessel = 7 cm \nand slant height = 25 cm \nBy Pythagoras theorem, \n \n\u2234 Volume of conical vessel = \\(\\frac{1}{3} \\pi r^{2} h\\) \n= \\(\\frac{1}{3} \\times \\frac{22}{7}\\) \u00d7 7 \u00d7 7 \u00d7 24 \n= 1232 cm3<\/sup> \nWe know that \n1000 cm3<\/sup> = 1 l \n\u2234 1232 cm3<\/sup> = 1.232 l \nTherefore, capacity of conical versel is 1.232 litres.<\/p>\n(ii) We have given that \nHeight of conical vessel (h) = 12 cm \nSlant height of conical vessel (l) = 13 cm \nBy Pythagoras theorem \n \n\u2234 Volume of conical vessel = \\(\\frac{1}{3} \\pi r^{2} h\\) \n= \\(\\frac{1}{3} \\times \\frac{22}{7}\\) \u00d7 5 \u00d7 5 \u00d7 12 \n= \\(\\frac{2200}{7}\\) cm3<\/sup> \nSince, 1000 cm3<\/sup> = 1 l \n\\(\\frac{2200}{7}\\) cm3<\/sup> = \\(\\frac{2.2}{7}\\) l \nTherefore, the capacity of conical vessel is \\(\\frac{2.2}{7}\\) litres.<\/p>\n <\/p>\n
Question 3. \nThe height of a cone is 15 cm. If its volume is 1570 cm3<\/sup>, find the radius of the base. \nSolution: \nWe have given \nHeight of cone (h) = 15 cm \nand volume of cone = 1570 cm3<\/sup> \nWe know that \nVolume of cone = \\(\\frac{1}{3} \\pi r^{2} h\\) \n \nTherfore, radius of required cone = 10 cm.<\/p>\nQuestion 4. \nIf the volume of a right circular cone of height 9 cm is 48\u03c0 cm3<\/sup>, find the diameter of its base. \nSolution: \nWe have given that \nHeight of cone = 9 cm \nand volume of cone = 48\u03c0 \nBut we know that volume of cone = \\(\\frac{1}{3} \\pi r^{2} h\\) \n \nTherefore diameter of required right circular cone is 2 \u00d7 4 = 8 cm.<\/p>\nQuestion 5. \nA conical pit top diameter 3.5 is 12 m deep. What is its capacity in kilo liters. \nSolution: \nWe have given that \ndiameter of conical pit = 3.5m. \nradius of conical pit = \\(\\frac{3.5}{2}\\) m = 1.75 m \nand height of conical pit = 12 m \n \n\u2234 Volume of conical pit = \\(\\frac{1}{3} \\pi r^{2} h\\) \n= \\(\\frac{1}{3} \\times \\frac{22}{7}\\) \u00d7 1.75 \u00d7 1.75 \u00d7 12 \n= 38.5 m3<\/sup> \nWe know that \n1 m3<\/sup> = 10001 \n38.5 m3<\/sup> = 1000 \u00d7 38.5 \n= 38500 litres \n= 38.5 kilolitres.<\/p>\n <\/p>\n
Question 6. \nThe volume of a right circular cone is 9856 cm3<\/sup>. If the diameter of the base is 28 cm, find. \n(i) height of the cone \n(ii) slant height of the cone \n(iii) curved surface area of the cone. \nSolution: \n(i) We have given that \ndiameter of cone = 28 cm \nRadius of cone = 14 cm \nand volume of cone = 9856 cm3<\/sup> \n \nWe know that, \nVolume of cone = \\(\\frac{1}{3} \\pi r^{2} h\\) \n9856 = \\(\\frac{1}{3} \\times \\frac{22}{7}\\) \u00d7 14 \u00d7 14 \u00d7 h \nh = \\(\\frac{9856 \\times 3 \\times 7}{22 \\times 14 \\times 14}\\) = 48 cm<\/p>\n(ii) We have, \nheight of the cone = 48 cm \nand radius of cone = 14 cm \nBy Pathagoras theorem, \nAC = \\(\\sqrt{\\mathrm{AB}^{2}+\\mathrm{BC}^{2}}\\) \n= \\(\\sqrt{(48)^{2}+(14)^{2}}\\) \n= \\(\\sqrt{2304+196}\\) \n= 50 cm. \nTherefore slant height of the cone = 50 cm<\/p>\n
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(iii) We know that, \nCurved surface area of cone = \u03c0rl \n= \\(\\frac {22}{7}\\) \u00d7 14 \u00d7 50 \n= 2200 cm2<\/sup> \nHence, curved surface area of the cone is 2200 cm2<\/sup><\/p>\nQuestion 7. \nA right triangle ABC with side 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained. \nSolution: \nWhen we revolved the right \u2206ABC about the side 12 cm, then it form a right circular cone of radius 6 cm and height 12 cm. \n \nSo, volume of such cone = \\(\\frac{1}{3} \\pi r^{2} h\\) \n= \\(\\frac {1}{3}\\) \u00d7 \u03c0 \u00d7 5 \u00d7 5 \u00d7 12 \n= 100\u03c0 cm3<\/sup> \nTherefore, volume of the solid formed by the revolved a right angle triangle about the side 12 cm is 100\u03c0 cm3<\/sup>.<\/p>\nQuestion 8. \nIf the triangle ABC in the question 7 is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solid obtained. \nSolution: \nIf we revolved the right angle triangle ABC about the side 5 cm is formed a right circular cone. \n \nSo, volume of the such cone = \\(\\frac{1}{3} \\pi r^{2} h\\) \n= \\(\\frac {1}{3}\\) \u00d7 \u03c0 \u00d7 12 \u00d7 12 \u00d7 5 \n= 240\u03c0 \nTherefore, volume of the solid formed by the revolved a right angle triangle about the side 5 cm is 240\u03c0 cm3<\/sup>. \nRatio of the volumes of the two solid obtained by the revolved a right angle triangle is = \\(\\frac{100 \\pi}{240 \\pi}\\) = 5 : 12<\/p>\n <\/p>\n
Question 9. \nA heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3m. Find its volume. The heap is to be covered by canvas to protect it form rain. Find the area of the cavas required. \nSolution: \nWe have given that \nDiameter of right circular cone = 10.5 cm \nRadius of right circular cone = 3.25 m \nand height of the cone = 3m \n \n\u2234 Volume of the cone = \\(\\frac{1}{3} \\pi r^{2} h\\) \n= \\(\\frac{1}{3} \\times \\frac{22}{7}\\) \u00d7 5.25 \u00d7 5.25 \u00d7 3 \n= 86.625 m3<\/sup> \nAgain, \nBy Pythagoars theorem. \nAC = \\(\\sqrt{\\mathrm{AB}^{2}+\\mathrm{BC}^{2}}\\) \n= \\(\\sqrt{(3)^{3}+(5.25)^{2}}\\) \n= \\(\\sqrt{9+27.5625}\\) \n= 6.04 m (approx) \nTherefore, \nCanvas required to covers it = Curved surface area of cone \nNow, Curved surface area of cone = \u03c0rl \n= \\(\\frac {22}{7}\\) \u00d7 5.25 \u00d7 6.04 \n= 99.66 m2<\/sup> \nHence, canvas required to cover the conical shape heap of wheat is 99.66 m2<\/sup>.<\/p>\n","protected":false},"excerpt":{"rendered":"These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.7 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.7 Question 1. Find the volume of the right circular cone with (i) radius 6 cm, …<\/p>\n
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.7<\/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":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.7 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n