{"id":28886,"date":"2021-07-13T17:08:54","date_gmt":"2021-07-13T11:38:54","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=28886"},"modified":"2021-07-13T17:08:54","modified_gmt":"2021-07-13T11:38:54","slug":"ncert-solutions-for-class-10-maths-chapter-13-ex-13-1","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-13-ex-13-1\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 13 Surface Areas and Volumes Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\n2 cubes each of volume 64 cm3<\/sup> are joined end to end. Find the surface area of the resulting cuboid.
\n\"NCERT
\nSolution:
\nVolume of one cube = 64 cm3<\/sup>
\nLet edge of one cube = a
\nVolume of the cube = (edge)3<\/sup>
\na3<\/sup> = 64 \u21d2 a = 4 cm
\nSimilarly, edge of the another cube = 4 cm.
\nNow, both cubes are joined together and a cuboid is formed as shown in the figure.
\nNow, length of the cuboid (l) = 8 cm
\nbreadth of the cuboid (b) = 4 cm
\nheight of the cuboid (h) = 4 cm
\nSurface area of the cuboid so formed = 2 (lb + bh + hl)
\n= 2(8 x 4 + 4 x 4 + 4 x 8)
\n= 2(32 + 16 + 32) = 160 cm\u00b2<\/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.
\n\"NCERT
\nSolution:
\nGiven: diameter of the hemisphere = 14 cm
\nRadius = \\(\\frac { 14 }{ 2 }\\) = 7 cm
\nCurved surface area of the hemisphere = 2\u03c0r\u00b2 = 2 x \\(\\frac { 22 }{ 7 }\\) x 7 x 7 cm\u00b2
\n= 14 x 22 cm\u00b2 = 308 cm\u00b2
\nHere, total height of the vessel = 13 cm
\nHeight of the cylinder = Total height – Height of the hemisphere = 13 cm – 7 cm = 6 cm
\nand radius of the cylinder = radius of the hemisphere = 7 cm
\nInner surface area of the cylinder = 2\u03c0rh = 2 x \\(\\frac { 22 }{ 7 }\\) x 7 x 6
\n= 2 x 22 x 6 = 264 cm\u00b2
\nInner surface area of the vessel = Inner surface area of the cylinder + curved surface area of the hemisphere
\n= 264 cm\u00b2 + 308 cm\u00b2 = 572 cm\u00b2.<\/p>\n

\"NCERT<\/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:
\nRadius of the hemispheree, r = 3.5 cm
\nRadius of the base of the cone, r = 5 cm
\nHeight of the cone, h = (15.5 – 3.5) cm = 12 cm
\n\"NCERT
\n\u2234 total surface area of the toy = (curved surface area of the hemisphere + (curved surface area of the cone)
\n\"NCERT
\nHence, the total surface area of the toy is 214.5 cm\u00b2.<\/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:
\nGreatest diameter of hemisphere is = 7 cm (side of cube)
\nRadius of hemisphere \\(\\frac { 7 }{ 2 }\\) = cm.
\nSurface area of solid
\n\"NCERT<\/p>\n

\"NCERT<\/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 remaining solid.
\nSolution:
\nDiameter of hemisphere = l
\nSo radius of hemisphere = \\(\\frac { l }{ 2 }\\)
\n\"NCERT
\nSurface area of cube = 6 (edge)\u00b2 – 6l\u00b2.
\nSurface area of remaining solid = Total surface area of cube-back area of hemisphere + Curved surface area of hemisphere.
\n\"NCERT<\/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). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
\n\"NCERT
\nSolution:
\nRadius of the cylinder = radius of the
\nhemispherical ends = r = \\(\\frac { 5 }{ 2 }\\) mm
\nHeight of the cylinder = h = 14 – 2 x \\(\\frac { 5 }{ 2 }\\) = 9 mm
\n\u2234 Total surface area = curved surface area of the cylinder + surface area of two hemispherical ends
\n= (2\u03c0rh + 2 x \u03c0r\u00b2) mm\u00b2
\n= 2\u03c0r (h + 2r) cm\u00b2
\n= 2 x \\(\\frac { 22 }{ 7 }\\) x \\(\\frac { 5 }{ 2 }\\)(9 + 2 x \\(\\frac { 5 }{ 2 }\\))
\n= \\(\\frac { 110 }{ 7 }\\)(9 + 5)
\n\u21d2 = \\(\\frac { 110 }{ 7 }\\) x 14
\n\u2234 Surface area = 220mm\u00b2<\/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 2.1 m and 4 m 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 \u20b9 500 per m\u00b2. (Note that the base of the tent will not be covered with canvas.)
\nSolution:
\nArea of the canvas used = Curved surface area of the cylinder + (Curved surface area of the cone)
\n\"NCERT
\nCost of the canvas of the tent at the rate \u20b9 500 per m\u00b2.
\n= \u20b9 (44 x 500) = \u20b9 22,000.<\/p>\n

\"NCERT<\/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 cm\u00b2.
\nSolution:
\nHeight of solid cylinder is 2.4 cm
\nDiameter of cylinder is 1.4 cm
\n\u2234 Radius = \\(\\frac { 1.4 }{ 2 }\\) cm = 0.7 cm
\n\"NCERT
\nTotal surface area of remaining solid
\n= 2\u03c0rh + \u03c0rl\u00b2 + nrl … (i)
\nwhere \u03c0rl curved surface area of cone
\n\"NCERT<\/p>\n

Question 9.
\nA wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in 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\"NCERT
\nSolution:
\nRadius of the cylinder = Radius of the hemispherical ends = r = 3.5 cm
\nHeight of the cylinder = 10 cm.
\nTotal surface area of the article
\n= Curved surface area of the cylinder + Surface area of the two hemispherical ends.
\n= (2\u03c0rh + 2 x 2\u03c0r\u00b2) cm\u00b2.
\n= 2\u03c0r (h + 2r) cm\u00b2.
\n= 2 x \\(\\frac { 22 }{ 7 }\\) x 3.5 (10 + 7)
\n= 22 (14) = 374 cm\u00b2.<\/p>\n

\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1 Question 1. 2 cubes each of volume 64 cm3 are joined end to end. Find …<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1<\/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":{"spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nNCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-13-ex-13-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts. 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