10th Class Maths 10th Lesson Mensuration Optional Exercise Textbook Questions and Answers<\/h3>\n
Question 1. Question 2. Question 3. Question 4. Question 5. AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Optional Exercise AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Optional Exercise Textbook Questions and Answers. 10th Class Maths 10th Lesson Mensuration Optional Exercise Textbook Questions and Answers Question 1. A golf ball has diameter equal to 4.1 cm. … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":35219,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2023\/02\/Chapter-10-Mensuration-Optional-Exercise.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/5921"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=5921"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/5921\/revisions"}],"predecessor-version":[{"id":34269,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/5921\/revisions\/34269"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/35219"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=5921"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=5921"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=5921"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nA golf ball has diameter equal to 4.1 cm. Its surface has 150 dimples each of radius 2 mm. Calculate total surface area which is exposed to the surroundings. (Assume that the dimples are all hemispherical) [\u03c0 = \\(\\frac{22}{7}\\)]
\nAnswer:
\nArea exposed = surface area of the ball – total area of 150 hemispherical with radius 2 mm
\n\u2234 C.S.A of hemisphere = 2\u03c0r2<\/sup>
\n![\"AP](\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-10-Mensuration-Optional-Exercise-1.png\")
\n= 52.831 – 37.71 = 15.12 cm2<\/sup>.<\/p>\n![\"AP](\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2020\/12\/AP-Board-Solutions.png\")
<\/p>\n
\nA cylinder of radius 12 cm. contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. [\u03c0 = \\(\\frac{22}{7}\\)]
\nAnswer:
\nRise in the water level is seen as a cylinder of radius ‘r’ = 12 cm
\nHeight, h = 6.75 cm.
\nVolume of the rise = Volume of the spherical iron ball dropped
\n![\"AP](\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-10-Mensuration-Optional-Exercise-3.png\")
\n= 9 \u00d7 12 \u00d7 6.75
\n= 108 \u00d7 6.75
\n= 729
\n![\"AP](\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-10-Mensuration-Optional-Exercise-2.png\")
\nr3<\/sup> = 9 \u00d7 9 \u00d7 9
\n\u2234 729 = (3 \u00d7 3) \u00d7 (3 \u00d7 3) \u00d7 (3 \u00d7 3)
\n\u2234 Radius of the ball r = 9 cm.<\/p>\n<\/p>\n
\nA solid toy is in the form of a right circular cylinder with a hemispheri\u00accal shape at one end and a cone at the other end. Their common diameter is 4.2 cm. and height of the cylindrical and conical portion are 12 cm. and 7 cm. respectively. Find the volume of the solid toy. [\u03c0 = \\(\\frac{22}{7}\\)]
\nAnswer:
\n![\"AP](\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-10-Mensuration-Optional-Exercise-4.png\")
\nVolume of the toy = volume of the hemisphere + volume of the cylinder + volume of the cone.
\n![\"AP](\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-10-Mensuration-Optional-Exercise-5.png\")
\nHemisphere:
\nRadius = \\(\\frac{d}{2}\\) = \\(\\frac{4.2}{2}\\) = 2.1 cm
\n![\"AP](\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-10-Mensuration-Optional-Exercise-6.png\")
\nCylinder:
\nRadius, r = \\(\\frac{d}{2}\\) = \\(\\frac{4.2}{2}\\) = 2.1 cm
\nheight, h = 12 cm
\nV = \u03c0r2<\/sup>h
\n= \\(\\frac{22}{7}\\) \u00d7 2.1 \u00d7 2.1 \u00d7 12
\n= 166.32 cm3<\/sup>
\nCone: Radius, r = \\(\\frac{d}{2}\\) = \\(\\frac{4.2}{2}\\) = 2.1 cm
\nHeight, h = 7 cm
\nVolume = \\(\\frac{1}{3}\\)\u03c0r2<\/sup>h
\n= \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 2.1 \u00d7 2.1 \u00d7 7
\n= 32.34 cm3<\/sup>
\n\u2234 Total volume = 19.404 + 166.32 + 32.34 = 218.064 cm3<\/sup>.<\/p>\n<\/p>\n
\nThree metal cubes with edges 15 cm., 12 cm. and 9 cm. respectively are melted together and formed into a simple cube. Find the diagonal of this cube.
\nAnswer:
\nEdges l1<\/sub> = 15 cm, l2<\/sub> = 12 cm, l3<\/sub> = 9 cm.
\n![\"AP](\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-10-Mensuration-Optional-Exercise-7.png\")
\nVolume of the resulting cube = Sum of the volumes of the three given cubes
\nL3<\/sup> = l1<\/sub>3<\/sup> + l2<\/sub>3<\/sup> + l3<\/sub>3<\/sup>
\nL3<\/sup> = 153<\/sup> + 123<\/sup> + 93<\/sup>
\nL3<\/sup> = 3375 + 1728 + 729
\nL3<\/sup> = 5832 = 18 \u00d7 18 \u00d7 18
\n\u2234 Edge of the new cube l = 18 cm
\n![\"AP](\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-10-Mensuration-Optional-Exercise-9.png\")
<\/p>\n
\nA hemispherical bowl of internal diameter 36 cm. contains a liquid. This liquid is to be filled in cylindrical bottles of radius 3 cm. and height 6 cm. How many bottles are required to empty the bowl?
\nAnswer:
\nLet the number of bottles required = n
\nThen total volume of n bottles = volume of the hemispherical bowl
\nn. \u03c0r1<\/sub>2<\/sup>h = \u03c0r2<\/sub>2<\/sup>h
\nBottle:
\nRadius, r = 3 cm; Height, h = 7 cm
\nVolume, V = \u03c0r2<\/sup>h
\n= \\(\\frac{22}{7}\\) \u00d7 3 \u00d7 3 \u00d7 7
\n= 198 cm3<\/sup>
\n\u2234 Total volume of n bottles = 198 . n cm3<\/sup>.
\nBowl:
\nRadius, r = \\(\\frac{d}{2}\\) = \\(\\frac{36}{2}\\) = 18 cm
\n![\"AP](\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-10-Mensuration-Optional-Exercise-8.png\")
\n\u2234 62 bottles are required to empty the bowl.<\/p>\n","protected":false},"excerpt":{"rendered":"