{"id":756,"date":"2020-09-15T08:59:29","date_gmt":"2020-09-15T03:29:29","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=756"},"modified":"2021-07-02T15:01:51","modified_gmt":"2021-07-02T09:31:51","slug":"kseeb-sslc-class-10-maths-solutions-chapter-15-ex-15-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-15-ex-15-1\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1<\/strong>\u00a0are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.1.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.1<\/h2>\n

(Unless stated otherwise, take \u03c0 \\(=\\frac{22}{7}\\))<\/p>\n

Question 1.
\n2 cubes each of volume 64cm3<\/sup> are joined end to end. Find the surface area of the resulting cuboid.
\nSolution:
\nIf measurement of each side of each cube be l cm, then
\n\"KSEEB
\n(l)3<\/sup> \\(=\\sqrt[3]{64}\\)
\n\u2234 l = 4 cm.
\nLength of cubes if both joined together,
\nLength, l = 4 + 4 = 8 cm.
\nbreadth, b = 4 cm.
\nheight, h = 4 cm.
\nSurface area of the cuboid,
\n= 2lb + 2lh + 2bh
\n= 2(8 \u00d7 4) + 2(8 \u00d7 4) + 2(4 \u00d7 4)
\n= 2 \u00d7 32 + 2 \u00d7 32 + 2 \u00d7 16 = 64 + 64 + 32
\n= 160 sq.cm.<\/p>\n

Question 2.
\nA vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
\nSolution:
\n(i) Diameter of a hollo hemisphere = 14 cm.
\n\u2234 Radius of a hollow hemisphere, r
\nr = 7 cm.
\nInner curved surface of a hemisphere = 2\u03c02<\/sup>
\n\"KSEEB
\n\"KSEEB
\n= 14 \u00d7 22
\n= 308 sq. cm.
\n(ii) Radius of hollow cylinder, r = 7 cm.
\nHeight, h = 13 – 7 = 6 cm.
\n\u2234 Inner Surface area of the Cylinder,
\n= 2\u03c0rh
\n\\(=2 \\times \\frac{22}{7} \\times 7 \\times 6\\)
\n= 44 \u00d7 6
\n= 264 sq.cm.
\n\u2234 Inner surface area of the vessel = Area of hemisphere + Area of cylinder
\n= 308 + 264
\n= 572 sq. cm.<\/p>\n

Question 3.
\nA toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
\nSolution:
\n(i) Radius of base of circular toy,
\nr = 3.5 cm.
\n\u2234 Surface Area of hemisphere, = 2\u03c0r2<\/sup>
\n\"KSEEB
\n\"KSEEB
\n= 77 sq.cm
\n(ii) Let slant height of cone is l cm. then
\n\"KSEEB
\n\u2234 Curved surface area of a cone = \u03c0rl
\n\"KSEEB
\n\"KSEEB
\n= 137.5 sq.cm.
\n\u2234 Total surface area of a toy = Surface area of hemisphere + Curved surface area of a cone
\n= 77.0 + 137.5
\n= 214.5 sq.cm.<\/p>\n

Question 4.
\nA cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have ? Find the surface area of the solid.
\nSolution:
\n(i) A cubical block of side 7 cm is surmounted by a hemisphere.
\n\u2234 Diameter of hemisphere is = 7 cm.
\nRadius of hemisphere, r = 3.5 cm.
\n\u2234 Surface area of hemisphere = Curved surface area – Area of base of hemisphere
\n\"KSEEB
\n\"KSEEB
\n= 38.5
\n(ii) The curved surface area of square,
\n=6 \u00d7 l2<\/sup>
\n= 6 \u00d7 (7)2<\/sup>
\n= 6 \u00d7 49
\n= 294 sq. cm.
\n\u2234 Total surface area of cubical block, = Curved surface area of cubical block + Surface area of hemisphere.
\n= 294 + 38.5
\n=332.5sq.cm.<\/p>\n

Question 5.
\nA hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the cube. Determine the surface area of the remaining solid.
\nSolution:
\nLet, Length of each edge of cubical block = Diameter of hemisphere = l
\n\u2234 Radius of hemisphere, \\(r=\\frac{l}{2} unit\\).
\n\"KSEEB
\nTotal surface area of newly formed cube = Curved surface area of cube – Upper part of Cube + Area of hemisphere which is depressed
\n\"KSEEB<\/p>\n

Question 6.
\nA medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Figure given below). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
\n\"KSEEB
\nSolution:
\n\"KSEEB
\n(i) Surface area of cylindrical capsule, = 2\u03c0rh
\n\\(=2 \\pi \\times\\left(\\frac{5}{2}\\right) \\times 9\\)
\n= 45\u03c0 mm2<\/sup>,
\n(ii) Surface area of both hemispheres,
\n\"KSEEB
\n= 25\u03c0 mm2<\/sup>
\n\u2234 Total surface area of capsule, = 45\u03c0 + 25\u03c0 mm2<\/sup>
\n= 70\u03c0 mm2<\/sup>
\n\\(=70 \\times \\frac{22}{7}\\)
\n= 220 mm2<\/sup><\/p>\n

Question 7.
\nA tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1m and 4m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs. 500 per m2<\/sup>. (note that the base of the tent will not be covered with canvas.)
\nSolution:
\n(i) Radius of base of cylinder, r= 2 mm.
\nHeight of base of cylinder, h= 2.1 m.
\n\u2234 Curved Surface area of cylinder = 2\u03c0rh
\n= 2\u03c0 \u00d7 2 \u00d7 (2.1)
\n\\(=4 \\times \\frac{22}{7} \\times 2.1\\)
\n= 26.4 m2<\/sup>.
\n\"KSEEB
\n(ii) Radius of conical tent,
\nr = 2 m.
\nlength, l = 2.8 m.
\n\u2234 Curved Surface area of cylinder = \u03c0rl
\n\\(=\\frac{22}{7} \\times 2 \\times 2.8\\)
\n= 17.6 m2<\/sup>.
\n\u2234 Total area of the tent = 26.4 + 17.6
\n= 44m2<\/sup>
\nCost of 1 aq.m. of canvas is Rs. 500.
\nCost of 44 sq. m. canvas … ??
\n500 \u00d7 44
\n= Rs. 22,000.<\/p>\n

Question 8.
\nFrom a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2<\/sup>.
\nSolution:
\nHeight of the cylinder, h = 2.4 m.
\nDiameter of the cylinder, d = 1.4 m.
\n\u2234 Radius, r = 0.7 m.
\n(i) Outer Curved surface area of cylinder = 2\u03c0rh
\n\"KSEEB
\n\\(=2 \\times \\frac{22}{7} \\times(0.7) \\times 2.4\\)
\n= 44. \u00d7 0.24
\n= 10.56 cm2<\/sup>.
\n(ii) Area of base of = \u03c0r2<\/sup>
\n\"KSEEB
\n= 1.54 cm2<\/sup>.
\n(iii) Inner surface area of cylinder, = \u03c0rl
\n\"KSEEB
\n= 5.5 cm2<\/sup>
\n\u2234 Total surface area of newly formed cube = 10.56 + 1.54 + 5.5 = 17.6 cm2<\/sup>.
\n\u2234 Nearest value is 18 sq.cm.<\/p>\n

Question 9.
\nA wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm. and its base is of radius 3.5 cm. find the total surface area of the article.
\n\"KSEEB
\nSolution:
\nRadius of base of cylinder, r= 3.5 cm
\nHeight, h = 10 cm.
\n\"KSEEB
\nTotal area of article, = Curved Surface area of Cylinder + 2 \u00d7 Area of Hemisphere
\n= 2\u03c0rh + 2 \u00d7 2\u03c0r2<\/sup>
\n\"KSEEB
\n= 374 sq.cm.<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.1, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1\u00a0are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.1. Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.1 (Unless …<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/756"}],"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\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=756"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/756\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=756"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=756"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=756"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}