{"id":4879,"date":"2023-05-17T16:42:51","date_gmt":"2023-05-17T11:12:51","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=4879"},"modified":"2023-05-18T17:58:48","modified_gmt":"2023-05-18T12:28:48","slug":"ap-board-9th-class-maths-solutions-chapter-10-ex-10-3","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-board-9th-class-maths-solutions-chapter-10-ex-10-3\/","title":{"rendered":"AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.3"},"content":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions<\/a> Chapter 10 Surface Areas and Volumes Ex 10.3 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 9th Class Maths Solutions 10th Lesson Surface Areas and Volumes Exercise 10.3<\/h2>\n

\"AP<\/p>\n

Question 1.
\nThe base area of a cone is 38.5 cm Its volume is 77 cm3<\/sup>. Find its height.
\nSolution:
\nBase area of the cone, \u03c0r2<\/sup> = 38.5 cm2<\/sup>
\nVolume of the cone, V = \\(\\frac{1}{3}\\) \u03c0r2<\/sup> h = 77
\n\u03c0r2<\/sup> = 38.5
\n\\(\\frac{22}{7}\\) r2<\/sup> = 38.5
\nr2<\/sup> = 38.5 x \\(\\frac{7}{22}\\)
\nr2<\/sup> = 12.25
\nr = \\(\\sqrt{12.25}\\) = 3.5
\nV= \\(\\frac{1}{22}\\) x \\(\\frac{22}{7}\\) x 3.5 x 3.5 x h = 77
\n\u2234 h = \\(\\frac{77 \\times 3 \\times 7}{22 \\times 12.25}\\) = 6
\n\u2234 Height of the cone = 6 cm<\/p>\n

Question 2.
\nThe volume of a cone is 462 m3<\/sup>. Its base radius is 7 m. Find its height.
\nSolution:
\nThe volume of a cone”V’= \\(\\frac{1}{3}\\) \u03c0r2<\/sup> h = 462
\nRadius ‘r’ = 7 m
\nHeight = h (say)
\n\\(\\frac{1}{3}\\) x \\(\\frac{22}{7}\\) x 7 x h = 462
\nh = \\(\\frac{462 \\times 3}{22 \\times 7}\\) = 9
\n\u2234 Height = 9m<\/p>\n

\"AP<\/p>\n

Question 3.
\nCurved surface area of a cone is 308 cm2<\/sup> and its slant height is 14 cm. Find (1) radius of the base (ii) total surface area of the cone.
\nSolution:
\nC.S.A. of the cone, \u03c0rl = 308
\nSlant height, l = 14 cm
\ni) \u03c0rl = 308; l = 14 cm
\n\\(\\frac{22}{7}\\) x r x 14 = 308
\nr = \\(\\frac{308}{44}\\) = 7cm<\/p>\n

ii) T.S.A. = \u03c0rl + \u03c0r2<\/sup>
\n= \u03c0r (r + l) = \\(\\frac{22}{7}\\) x 7 x (7 + 14)
\n= 22 x 21 = 462 cm3<\/sup><\/p>\n

Question 4.
\nThe cost of painting the total surface area of a cone at 25 paise per cm2 is \u20b9176. Find the volume of the cone, if its slant height is 25 cm.
\nSolution:
\nSlant height of the cone, l = 25 cm
\nTotal cost at the rate of 25 p\/cm2<\/sup>
\n= \u20b9176
\n\u2234 Total surface area of the cone
\n= \\(\\frac{176}{25}\\) x 100 = 176 x 4 = 704cm2<\/sup>
\nBut T.S.A. of the cone = \u03c0r (r + l) = 704
\nThus \\(\\frac{22}{7}\\)r(r + 25) = 704
\nr(r + 25) = \\(\\frac{704 \\times 7}{22}\\) = 224
\nr2<\/sup> + 25r = 224
\n\u21d2 r2<\/sup> + 32r – 7r – 224 = 0
\n\u21d2 r (r + 32) – 7 (r + 32) = 0
\n\u21d2 (r + 32) (r – 7) = 0
\n\u21d2 r = 7 (\u2235 \u2019r’ can’t be negative)<\/p>\n

\"AP
\n\u2234 Volume of the cone = \\(\\frac{1}{3}\\) \u03c0r2<\/sup>h
\n= \\(\\frac{1}{3}\\) x \\(\\frac{22}{7}\\) x 7 x 7 24
\n= 22 x 7 x 8 = 1232cm3<\/sup><\/p>\n

\"AP<\/p>\n

Question 5.
\nFrom a circle of radius 15 cm, a sector with angle 216\u00b0 is cut out and its bounding radii are bent so as to form a cone. Find its volume.
\nSolution:
\nRadius of the sector, ‘r’ = 15 cm
\nAngle of the sector, ‘x’ = 216\u00b0
\n\u2234 Length of the arc, l = \\(\\frac{x}{360}\\) x 2\u03c0r
\n\\(=\\frac{216}{360} 2 \\pi r=\\frac{3}{5}(2 \\pi r)\\)
\nPerimeter of the base of the cone = Length of the arc
\n2\u03c0r of cone = \\(=\\frac{6}{5}\\) \u03c0r of the circle
\nRadius of the cone ‘r’ = \\(\\frac{3}{5}\\) x 15 = 9
\nRadius ‘r’ of the circle = slant height l of the cone = 9 cm
\n\"AP
\n= 1018.3 cm3<\/sup><\/p>\n

\"AP<\/p>\n

Question 6.
\nThe height of a tent is 9 m. Its base diameter is 24 m. What is its slant height ? Find the cost of canvas cloth required if it costs \u20b914 per sq.m.
\nSolution:
\nHeight of a conical tent ‘h’ = 9 m
\nBase diameter = 24 m
\nThus base radius \u2019r’= \\(\\frac{d}{2}=\\frac{24}{2}\\) = 12m
\nCost of canvas = \u20b9 14 per sq.m.
\nC.S.A. of the cone = \u03c0rl
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 7.
\nThe curved surface area of a cone is 1159\\(\\frac { 5 }{ 7 }\\) cm2<\/sup>. Area of its base is 254 \\(\\frac { 4 }{ 7 }\\) cm2<\/sup>. Find its volume.
\nSolution:
\nC.S.A. of the cone = \u03c0rl
\n\"AP
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 8.
\nA tent is cylindrical to a height of 4.8 m and conical above it. The ra\u00acdius of the base is 4.5 m and total height of the tent is 10.8 m. Find the canvas required for the tent in square
\nmeters.
\nSolution:
\n\"AP
\nRadius of cylinder, l = 4.5 m
\nHeight of the cylinder = h = 4.8 m
\n\u2234 C.S.A. of the cylinder = 2\u03c0rh
\n= 2 x \\(\\frac{22}{7}\\) x 4.5 x 4.8
\n= 135.771 m2<\/sup>
\nRadius of the cone ‘r’ =
\nRadius of the cylinder = 4.5 m
\nHeight of the cone \u2019h’ = 10.8 – 4.8 = 6 m
\n\u2234 Slant height of the cone
\n\"AP
\n\u2234 C.S.A. of the cone = \u03c0rl
\n= \\(\\frac{22}{7}\\) x 4.5 x 7.5
\n= \\(\\frac{742.5}{7}\\) = 106.071m2<\/sup>
\n\u2234 Total canvas required
\n= C.S.A of cylinder + C.S.A. of cone
\n= 135.771 + 106.071
\n= 241.842 m2<\/sup><\/p>\n

\"AP<\/p>\n

Question 9.
\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 extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (use \u03c0 = 3.14) [Note : Take 20 cm as 0.6 m2<\/sup>]
\nSolution:
\nRadius of the cone, r = 6 m
\nHeight of the cone, h = 8 m
\n\"AP
\n\u2234 C.S.A. = \u03c0rl = 3.14 x 6 x 10 = 188.4 m2<\/sup>
\nLet the length of the tarpaulin = l
\n\u2234 Area of the tarpaulin, lb = 188.4 + 0.6
\n= 189 m2<\/sup>
\n\u21d2 3l = 189
\n\u21d2 l = \\(\\frac{189}{3}\\) = 63m<\/p>\n

Question 10.
\nA joker’s cap is in the form of a right circular cone of base radius 7 cm and height 27 cm. Find the area of the sheet required to make 10 such caps.
\nSolution:
\nRadius of the cone, r = 7 cm
\nHeight of the cone, h = 27 cm
\nSlant height of the cone (l)
\n\"AP
\n\u2234 Total area of the sheet required for
\n10 caps = 10 x 22 \\(\\sqrt{778}\\)
\n= 6136.383 cm2<\/sup><\/p>\n

\"AP<\/p>\n

Question 11.
\nWater is pouring into a conical vessel (as shown in the given figure), at the rate of 1.8 m3<\/sup> per minute. How long will it take to fill the vessel?
\n\"AP
\nSolution:
\nFrom the figure, diameter of the cone 5.2 m
\nThus its radius \u2019r\u2019 = \\(\\frac{5.2}{2}\\) = 2.6 m
\n\u2234 Height of the cone = h = 6.8 m
\nVolume of the cone = \\(\\frac{1}{3}\\) \u03c0r2<\/sup> h
\n= \\(\\frac{1}{3} \\times \\frac{22}{7}\\) x 2.6 x 2.6 x 6.8
\n= \\(\\frac{1011.296}{21}\\)
\n= 48.156 m3<\/sup>
\nQuantity of water that flows per minute
\n= 1.8 m3<\/sup>
\n\u2234 Total time required = \\(\\frac{\\text { Total volume }}{1.8}\\)
\n= \\(\\frac{48.156}{1.8}\\) = 26.753
\n27 minutes.<\/p>\n

\"AP<\/p>\n

Question 12.
\nTwo similar cones have volumes 12\u03c0 CU. units and 96\u03c0 CU. units. If the curved surface area of smaller cone is 15\u03c0 sq.units, what is the curved surface area of the larger one?
\nSolution:
\n\"AP
\n\u03c0rl = 15 \u03c0
\n\\(r \\sqrt{\\left(r^{2}+h^{2}\\right)}=15\\)
\nSquaring on both sides
\nr2<\/sup> (r2<\/sup> + h2<\/sup>) = 15 x 15
\n= 3 x 5 x 3 x 5
\n= 3 x 3 x 25
\nr2<\/sup>(r2<\/sup> + h2<\/sup>) = 32<\/sup>(32<\/sup> + 42<\/sup>)
\n\u2234 r = 3 cm, h = 4 cm
\nC.S.A. = \u03c0 x 3 x \\(\\left(\\sqrt{3^{2}+4^{2}}\\right)\\) = 15\u03c0
\n\\(\\frac{1}{3}\\)\u03c0r2<\/sup>H = 96\u03c0
\n\\(\\frac{3 \\times 3 \\times \\mathrm{H}}{3}\\) = 96
\n\u2234 H = \\(\\frac{96}{3}\\) = 32 units
\n\"AP<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.3 Textbook Questions and Answers. AP State Syllabus 9th Class Maths Solutions 10th Lesson Surface Areas and Volumes Exercise 10.3 Question 1. The base area of a cone is 38.5 cm Its volume is 77 cm3. Find its height. Solution: … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/4879"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=4879"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/4879\/revisions"}],"predecessor-version":[{"id":40558,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/4879\/revisions\/40558"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=4879"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=4879"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=4879"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}