{"id":26399,"date":"2021-06-25T17:49:07","date_gmt":"2021-06-25T12:19:07","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=26399"},"modified":"2021-06-25T17:51:27","modified_gmt":"2021-06-25T12:21:27","slug":"ncert-solutions-for-class-9-maths-chapter-13-ex-13-7","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-13-ex-13-7\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.7"},"content":{"rendered":"

These 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>\n

NCERT 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\"NCERT
\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\"NCERT
\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

\"NCERT<\/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\"NCERT
\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\"NCERT
\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

\"NCERT<\/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\"NCERT
\nTherfore, radius of required cone = 10 cm.<\/p>\n

Question 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\"NCERT
\nTherefore diameter of required right circular cone is 2 \u00d7 4 = 8 cm.<\/p>\n

Question 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\"NCERT
\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

\"NCERT<\/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\"NCERT
\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

\"NCERT<\/p>\n

(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>\n

Question 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\"NCERT
\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>\n

Question 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\"NCERT
\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

\"NCERT<\/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\"NCERT
\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

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NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.7 Question 1. 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