{"id":3447,"date":"2021-02-18T16:11:18","date_gmt":"2021-02-18T10:41:18","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=3447"},"modified":"2021-09-15T11:40:43","modified_gmt":"2021-09-15T06:10:43","slug":"ap-ssc-10th-class-maths-solutions-chapter-10-ex-10-2","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-ssc-10th-class-maths-solutions-chapter-10-ex-10-2\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2"},"content":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 10 Mensuration Ex 10.2 Textbook Questions and Answers.<\/p>\n

AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.2<\/h2>\n

10th Class Maths 10th Lesson Mensuration Ex 10.2 Textbook Questions and Answers<\/h3>\n

Question 1.
\nA toy is in the form of a cone mounted on a hemisphere. The diameter of the base and the height of the cone are 6 cm and 4 cm respectively. Determine the surface area of the toy. (Use \u03c0 = 3.14)
\nAnswer:
\n\"AP
\nDiameter of the base of the cone d = 6 cm.
\n\u2234 Radius of the base of the cone
\nr = \\(\\frac{d}{2}\\) = \\(\\frac{6}{2}\\) = 3 cm
\nHeight of the cone = h = 4 cm
\nSlant height of the cone l = \\(\\sqrt{r^{2}+h^{2}}\\)
\n= \\(\\sqrt{3^{2}+4^{2}}\\)
\n= \\(\\sqrt{9+16}\\)
\n= \u221a25
\n= 5 cm
\n\u2234 C.S.A of the cone = \u03c0rl
\n= \\(\\frac{22}{7}\\) \u00d7 3 \u00d7 5
\n= \\(\\frac{330}{7}\\) cm2<\/sup>
\nRadius of the hemisphere = \\(\\frac{d}{2}\\) = \\(\\frac{6}{2}\\) = 3 cm
\nC.S.A. of the hemisphere = 2\u03c0r2<\/sup>
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 3 \u00d7 3
\n= \\(\\frac{396}{7}\\)
\nHence the surface area of the toy = C.S.A. of cone + C.S.A. of hemisphere
\n= \\(\\frac{330}{7}\\) + \\(\\frac{396}{7}\\)
\n= \\(\\frac{726}{7}\\) \u2243 103.71 cm2<\/sup>.<\/p>\n

\"AP<\/p>\n

Question 2.
\nA solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. The radius of the common base is 8 cm and the heights of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the total surface area of the solid. [Use \u03c0 = 3.14]
\nAnswer:
\nTotal surface area = C.S.A. of the cone + C.S.A. of cylinder + C.S.A of the hemisphere.
\n\"AP
\nCone:
\nRadius (r) = 8 cm
\nHeight (h) = 6 cm
\nSlant height l = \\(\\sqrt{r^{2}+h^{2}}\\)
\n= \\(\\sqrt{8^{2}+6^{2}}\\)
\n= \\(\\sqrt{64+36}\\)
\n= \u221a100
\n= 10 cm
\nC.S.A. = \u03c0rl
\n= \\(\\frac{22}{7}\\) \u00d7 8 \u00d7 10
\n= \\(\\frac{1760}{7}\\) cm2<\/sup>
\nCylinder:
\nRadius (r) = 8 cm;
\nHeight (h) = 10 cm
\nC.S.A. = 2\u03c0rh
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 8 \u00d7 10
\n= \\(\\frac{3520}{7}\\) cm2<\/sup>
\nHemisphere:
\nRadius (r) = 8 cm
\nC.S.A. = 2\u03c0r2<\/sup>
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 8 \u00d7 8
\n= \\(\\frac{2816}{7}\\) cm2<\/sup>
\n\u2234 Total surface area of the given solid
\n= \\(\\frac{1760}{7}\\) + \\(\\frac{3520}{7}\\) + \\(\\frac{2816}{7}\\)
\nT.S.A. = \\(\\frac{8096}{7}\\) = 1156.57 cm2<\/sup>.<\/p>\n

\"AP<\/p>\n

Question 3.
\nA medicine capsule is ih the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the capsule is 14 mm. and the width is 5 mm. Find its surface area.
\nAnswer:
\n\"AP
\nSurface area of the capsule = C.S.A. of 2 hemispheres + C.S.A. of the cylinder
\ni) Now for Hemisphere:
\nRadius (r) = \\(\\frac{d}{2}\\) = \\(\\frac{5}{2}\\) = 2.5 mm
\nC.S.A of each hemisphere = 2\u03c0r2<\/sup>
\nC.S.A of two hemispheres
\n= 2 \u00d7 2\u03c0r2<\/sup> = 4\u03c0r2<\/sup>
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 \\(\\frac{5}{2}\\) \u00d7 \\(\\frac{5}{2}\\)
\n= \\(\\frac{550}{7}\\)
\n= 78.57 mm2<\/sup>.<\/p>\n

ii) Now for Cylinder:
\nLength of capsule = AB =14 mm
\nThen height (length) cylinder part = 14 – 2(2.5)
\nh = 14 – 5 = 9 mm
\nRadius of cylinder part (r) = \\(\\frac{5}{2}\\)
\nNow C.S.A of cylinder part = 2\u03c0rh
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 \\(\\frac{5}{2}\\) \u00d7 9
\n= \\(\\frac{900}{7}\\)
\n= 141.428 mm2<\/sup>
\nNow total surface area of capsule
\n= 78.57 + 141.43 = 220 mm2<\/sup><\/p>\n

Question 4.
\nTwo cubes each of volume 64 cm3<\/sup> are joined end to end together. Find the surface area of the resulting cuboid.
\nAnswer:
\nGiven, volume of the cube.
\nV = a3<\/sup> = 64 cm3<\/sup>
\n\u2234 a3<\/sup> = 4 \u00d7 4 \u00d7 4 = 43<\/sup> , Hence a = 4 cm
\nWhen two cubes are added, the length of cuboid = 2a = 2 \u00d7 4 = 8 cm,
\nbreadth = a = 4 cm.
\nheight = a = 4 cm is formed.
\n\"AP
\n\u2234 T.S.A. of the cuboid
\n= 2 (lb + bh + lh)
\n= 2(8 \u00d7 4 + 4 \u00d7 4 + 8 \u00d7 4)
\n= 2(32 + 16 + 32)
\n= 2 \u00d7 80
\n= 160 cm2<\/sup>
\n\u2234 The surface area of resulting cuboid is 160 cm2<\/sup>.<\/p>\n

\"AP<\/p>\n

Question 5.
\nA storage tank consists of a circular cylinder with a hemisphere stuck on either end. If the external diameter of the cylinder be 1.4 m. and its length be 8 m. Find the cost of painting it on the outside at rate of Rs. 20 per m2<\/sup>.
\nAnswer:
\nTotal surface area of the tank = 2 \u00d7 C.S.A. of hemisphere + C.S.A. of cylinder.
\n\"AP
\nHemisphere:
\nRadius (r) = \\(\\frac{d}{2}\\) = \\(\\frac{1.4}{2}\\) = 0.7 m
\nC.S.A. of hemisphere = 2\u03c0r2<\/sup>
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 0.7 \u00d7 0.7
\n= 3.08 m2<\/sup>.
\n2 \u00d7 C.S.A. = 2 \u00d7 3.08 m2<\/sup> = 6.16 m2<\/sup>
\nCylinder:
\nRadius (r) = \\(\\frac{d}{2}\\) = \\(\\frac{1.4}{2}\\) = 0.7 m
\nHeight (h) = 8 m
\nC.S.A. of the cylinder = 2\u03c0rh
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 0.7 \u00d7 8
\n= 35.2 m2<\/sup>
\n\u2234 Total surface area of the storage tank = 35.2 + 6.16 = 41.36 m2<\/sup>
\nCost of painting its surface area @ Rs. 20 per sq.m, is
\n= 41.36 \u00d7 20 = Rs. 827.2.<\/p>\n

Question 6.
\nA hemisphere is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the length of the cube. Determine the surface area of the remaining solid.
\nAnswer:
\nLet the length of the edge of the cube = a units
\n\"AP
\nT.S.A. of the given solid = 5 \u00d7 Area of each surface + Area of hemisphere
\nSquare surface:
\nSide = a units
\nArea = a2<\/sup> sq. units
\n5 \u00d7 square surface = 5a2<\/sup> sq. units
\nHemisphere:
\nDiameter = a units;
\nRadius = \\(\\frac{a}{2}\\)
\nC.S.A. = 2\u03c0r2<\/sup>
\n= 2\u03c0\\(\\left(\\frac{a}{2}\\right)^{2}\\)
\n= 2\u03c0\\(\\frac{a^{2}}{4}\\) = \\(\\frac{\\pi \\mathrm{a}^{2}}{2}\\) sq. units
\nTotal surface area = 5a2<\/sup> + \\(\\frac{\\pi \\mathrm{a}^{2}}{2}\\) = a2<\/sup>\\(\\left(5+\\frac{\\pi}{2}\\right)\\) sq. units.<\/p>\n

\"AP<\/p>\n

Question 7.
\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 radius is of 3.5 cm, find the total surface area of the article.
\n\"AP
\nAnswer:
\nSurface area of the given solid = C.S.A. of the cylinder + 2 \u00d7 C.S.A. of hemisphere.
\nIf we take base = radius
\nCylinder:
\nRadius (r) = 3.5 cm
\nHeight (h) = 10 cm
\nC.S.A. = 2\u03c0rh
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 3.5 \u00d7 10
\n= 220 cm2<\/sup>
\nHemisphere:
\nRadius (r) = 3.5 cm
\nC.S.A. = 2\u03c0r2<\/sup>
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 3.5 \u00d7 3.5
\n= 77 cm2<\/sup>
\n2 \u00d7 C.S.A. = 2 \u00d7 77 = 154 cm2<\/sup>
\n\u2234 T.S.A. = 220 + 154 = 374 cm2<\/sup>.<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Ex 10.2 Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.2 10th Class Maths 10th Lesson Mensuration Ex 10.2 Textbook Questions and Answers Question 1. A toy is in the form of a cone …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/3447"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=3447"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/3447\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=3447"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=3447"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=3447"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}