\nFind the volume of the right circular cone with
\n(i) radius 6 cm, height 7 cm.
\n(ii) radius 3.5 cm, slant height 12 cm.
\nSolution:
\n(i) Volume of Cone, V = \\(\\frac{1}{3} \\pi r^{2} h\\)
\n
![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-1.png\")
\n\u2234 V = 264 cm3<\/sup>.
\n(ii) r = 3.5 cm, h = 12 cm.
\nVolume of Cone, V = \\(\\frac{1}{3} \\pi r^{2} h\\)
\n
![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-1.1.png\")
\n= 11 \u00d7 14
\n\u2234 V = 154 cm3<\/sup>.<\/p>\n
Question 2. Question 3. Question 4. Question 5. Question 6. (ii) l2<\/sup> = h2<\/sup> + r2<\/sup> (iii) Curved surface area of the cone, A Question 7. Question 8. Question 9. Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.7 (Assume \u03c0 = unless stated otherwise) Question 1. Find the volume of the right circular cone with (i) radius 6 cm, height 7 cm. (ii) radius 3.5 cm, slant height 12 cm. Solution: (i) Volume of Cone, V = \u2234 V = …<\/p>\n","protected":false},"author":5,"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\/2666"}],"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\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=2666"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2666\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=2666"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=2666"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=2666"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nFind the capacity in litres of a conical vessel with
\n(i) radius 7 cm, slant height 25 cm.
\n(ii) height 12 cm, slant height 13 cm.
\nSolution:
\n(i) radius, r = 7 cm, slant height, l = 25 cm
\nVolume of conical vessel, V = \\(\\frac{1}{3} \\pi r^{2} h\\)
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-2.png\")
\n\u2234 h = 24 cm.
\nV = \\(\\frac{1}{3} \\pi r^{2} h\\)
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-2.1.png\")
\n= 22 \u00d7 7 \u00d7 8
\n\u2234 V = 1232 cm3<\/sup>
\n\u2234 Volume of conical vessel
\n= \\(\\frac{1232}{1000}\\) = 1.232 litres
\n(ii) height, h =12 cm, slant height, l = 13cm. r = ?, V = ? .
\nh2<\/sup> = l2<\/sup> – r2<\/sup>
\n\u2234 r2<\/sup> = l2<\/sup> – h2<\/sup>
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-2.2.png\")
\n\u2234 r = 5 cm.
\nVolume of vessel, V = \\(\\frac{1}{3} \\pi r^{2} h\\)
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-2.3.png\")
\n\u2234 V = 314.28 cm3<\/sup>.
\nIn litres, = \\(\\frac{314.28}{1000}\\) = 0.31428 litres.<\/p>\n
\nThe height of a cone is 15 cm. If its volume is 1570 cm3<\/sup>, find the radius of the base. (Use \u03c0 = 3.14)
\nSolution:
\nHeight of the cone, h = 15 cm.
\nVolume, V = 1570 cm3<\/sup>, radius, r = ?
\nVolume of a cone. V = \\(\\frac{1}{3} \\pi r^{2} h\\)
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-3.png\")
\nr2<\/sup> = 100
\n\u2234 r = 10 cm.<\/p>\n
\nIf the volume of a right circular cone of height 9 cm is 48\u03c0 cm3<\/sup>, find the diameter of its base.
\nSolution:
\nVolume of a right circular cone,V= 48\u03c0 cm3<\/sup>,
\nheight, h = 9 cm.
\ndiameter, d = ?
\nVolume of cone, V = \\(\\frac{1}{3} \\pi r^{2} h\\)
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-4.png\")
\n\u2234 r2<\/sup> = 16
\n\u2234 r = 4 cm
\n\u2234 diameter, d = 2r
\n= 2 \u00d7 4
\n\u2234 d = 8 cm.<\/p>\n
\nA conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
\nSolution:
\nd = 3.5m., h = 12 m., V = ?
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-5.png\")
\nVolume of Conical pit, V
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-5.1.png\")
\n= 22 \u00d7 0.5 \u00d7 3.5
\n\u2234 V = 38.5 m3<\/sup>
\nV = 38.5 kilolitres<\/p>\n
\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:
\nVolume of Cone, V = 9856 cm3<\/sup>
\ndiameter of the base, d = 28 cm.
\n(i) h = ?,
\n(ii) I = ?,
\n(iii) C.S.A. = ?
\n(i) V = 9856 cm3<\/sup>.
\nd = 28 cm. \u2234 r = \\(\\frac{28}{2}\\) = 14 cm.
\nVolume of cone, V= \\(\\frac{1}{3} \\pi r^{2} h\\)
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-6.png\")
\nh = \\(\\frac{2688}{56}\\)
\nh = 48 cm.<\/p>\n
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-6.1.png\")
\nl = 50 cm.<\/p>\n
\nA = \u03c0rl
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-6.2.png\")
\n= 22 \u00d7 100
\nA= 2200 cm2<\/sup>.<\/p>\n
\nA right triangle ABC with sides 5 cm, 12 cm, and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
\nSolution:
\nCone is formed \u2206ABC is revolved about the side 12 cm.
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-7.png\")
\nr = 5 cm,
\nh = 12 cm,
\nV = ?
\nVolume of Cone, V
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-7.1.png\")
\n\u2234 V = 314.29 cm3<\/sup>.<\/p>\n
\nIf the triangle ABC in Question 7 above 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 solids obtained in Questions 7 and 8.
\nSolution:
\nCone is formed when \u2206ABC is revolved about 5 cm.
\nr = 12 cm,
\nh = 5 cm.
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-8.png\")
\nVolume of Cone, V
\nv = \\(\\frac{1}{3} \\pi r^{2} h\\)
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-8.1.png\")
\nV = 754.28 cm3<\/sup>.
\nVolume of Cone in problem 7 = 314.29 cm3<\/sup>
\nVolume of Cone in problem 8 = 754.28 cm3<\/sup>
\n\u2234 ratio of both Volumes = \\(\\frac{314.29}{754.28}\\)
\n= \\(\\frac{5}{2}\\) = 5 : 12<\/p>\n
\nA heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find ifs volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
\nSolution:
\nDiameter of heap of wheat in the form of Cone,
\ndiameter, d = 10.5 m = \\(\\frac{21}{2}\\)
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-9.png\")
\nheight, h = 3 m. V = ?
\nVolume of Canvas Cover, V = \\(\\frac{1}{3} \\pi r^{2} h\\)
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-9.1.png\")
\n= 86.625 m3<\/sup>.
\nArea of Canvas required to cover a heap of wheat, C.S.A.
\nC.S.A.= \u03c0rl
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-9.2.png\")
\n\u2234 Area of Canvas, A = \u03c0rl
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.7-Q-9.3.png\")
\n\u2234 A = 99.83 m2<\/sup>.<\/p>\nKSEEB Solutions for Class 9 Maths<\/a><\/h5>\n","protected":false},"excerpt":{"rendered":"