{"id":27689,"date":"2020-07-13T14:50:48","date_gmt":"2020-07-13T09:20:48","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=27689"},"modified":"2021-07-02T15:07:05","modified_gmt":"2021-07-02T09:37:05","slug":"kseeb-solutions-for-class-8-maths-chapter-16-additional-questions","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-8-maths-chapter-16-additional-questions\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 16 Mensuration Additional Questions"},"content":{"rendered":"

Students can Download Maths Chapter 16 Mensuration Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 8 Maths Chapter 16 Mensuration Additional Questions<\/h2>\n

Question 1.
\nThree metal cubes whose edges measure 3 cm, 4 cm and 5 cm respectively are melted to form a single cube find (1) side – length (2) total surface area of the new cube. What is the difference between total surface area of the new cube and the sum of total surface areas of the original cubes?
\nSolution:
\nVolume of a cube = l3<\/sup>
\nVolume of cube 1 = 33<\/sup> = 27 cm3<\/sup>
\nVolume of cube 2 = 43<\/sup> = 64 cm3<\/sup>
\nVolume of cube 3 = 53<\/sup> = 125 cm3<\/sup>
\nThe volume of 3 cubes = 216 cm3<\/sup>
\nVolume of the New cube formed = 216 cm3<\/sup>
\nl3<\/sup> = 216
\nl = 3\u221a216 =6 cm
\n\u2234 (i) side – length = 6 cm<\/p>\n

(ii) T.S.A of New cube = 6 l2<\/sup>
\n= 6 x 62<\/sup> = 6 x 36 = 216 cm2<\/sup>
\nT.S.A of cube 1 = 6 \u00d7 l2<\/sup> = 6 \u00d7 32<\/sup> = 6 \u00d7 9 = 54 cm2<\/sup>
\nT.S.A of cube 2 = 6 \u00d7 l2<\/sup> – 6 \u00d7 42 = 6 \u00d7 16 = 96 cm2<\/sup>
\nT.S.A of cube 3 = 6 \u00d7 l2<\/sup> = 6 \u00d7 52<\/sup> = 6 \u00d7 25 = 150 cm2<\/sup>
\nTotal T.S.A of 3 cubes = 300 cm2<\/sup><\/p>\n

(iii) Difference between the surface areas = 300 – 216 = 84 cm2<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nTwo cubes, each of volume 512 cm3<\/sup> are joined end to end. Find the lateral and total surface areas of resulting cuboid.
\n\"KSEEB
\nSolution:
\nVolume of a cube = 512 cm3<\/sup>
\nl3<\/sup> = 512
\nl = 3\u221a512 = 8 cm
\nWhen two cubes joined end to end the length of the resulting cuboid = 8 + 8 = 16 cm, breadth = 8 cm and height = 8 cm.
\nL.S.A of cuboid = 2h (l + b)
\n= 2 \u00d7 8 (16 + 8) = 16 (24)
\n= 384 cm2<\/sup>
\nT.S.A of cuboid = 2(lb + bh + lh)
\n= 2[16 \u00d7 8 + 8 \u00d7 8 +16 \u00d7 8]
\n=2 [128 + 64 + 128]
\n= 2 [320] = 640 cm2<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nThe length, breadth and height of a cuboid,are in the ratio 6 : 5 : 3. If the total surface area is 504 cm2<\/sup> find its dimensions. Also find the volume of the cuboid.
\nSolution:
\nl : b : h = 6 : 5 : 3
\nLet the length, breadth and height be 6x, 5x and 3x respectively.
\nT.S.A of cuboid = 504 cm2<\/sup>
\n2(lb + bh + lh) = 504
\n2[6x \u00d7 5x + 5x \u00d7 3x + 6x \u00d7 3x] = 504
\n2 [30x2<\/sup> + 15x2<\/sup> + 18x2<\/sup>] = 504
\n2[63x2<\/sup>] = 504
\n126x2<\/sup> = 504
\nx2<\/sup> = \\(\\frac{504}{126}\\) = 4
\nx = \u221a4 = 2cm
\nl = 6x = 6 \u00d7 2 = 12 cm, b = 5x = 5 \u00d7 2 = 10 cm,
\nh = 3x = 3 \u00d7 2 – 6 cm
\nVolume = lbh = 12 \u00d7 10 \u00d7 6 = 720 cm3<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nFind the area of four walls of a room having length, breadth, and heighnt as 8m, 5m and 3m respectively. Find the cost of white washing the walls at the rate of Rs. 15\/m2<\/sup>.
\nSolution:
\nl = 8m, b = 5m, h = 3m, L.S.A = ?
\nArea of the 4 walls = L.S.A of cuboid
\n= 2h (l + b) = 2 \u00d7 3. (8 + 5) = 6 (13)
\nArea of the 4 walls = 78 m2<\/sup>
\nThe cost of white washing 1 m2<\/sup> = Rs.15
\nThe cost of white washing 78 m2<\/sup> = 78 \u00d7 15
\n= Rs. 1170<\/p>\n

Question 5.
\nA room is 6m long, 4m broad and 3m high. Find the cost of laying tiles on its floor and four walls at the cost of Rs. 80\/m3.
\nSolution ‘
\nl = 6m, b = 4m, h = 3m
\nArea on. which tiles are to be laid
\n= Area of the base + LSA = lb + 2h (l + b) ‘
\n= 6 \u00d7 4 + 2 \u00d7 3 (6 + 4)
\n= 24 + 6(10)
\n= 24 + 60 = 84 m2<\/sup>
\nThe cost for laying tiles for 1 m2<\/sup> = Rs.80
\nThe cost for laying tiles for 84 m2<\/sup>
\n= 84 \u00d7 80 = Rs. 6720<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nThe length, breadth and height of a cuboid are in the ratio 5:3: 2. If its volume is 1078.110 mm3<\/sup> find its dimensions. Also find the total surface area of the cuboid.
\nSolution:
\nl : b : h = 5 : 3 :2, Volume = 1078.110 mm3<\/sup>
\nVolume = 1078.110 mm3<\/sup>
\nl \u00d7 b \u00d7 h = 1078.110
\n5x \u00d7 3x \u00d7 2x = 1078.110
\nx3<\/sup> = \\(\\frac{1078.110}{30}\\) = 35.937
\nx = 3\u221a35.937
\nx = 3.3 m
\n5x = 5 \u00d7 3.3 = 16.5 m
\n3x = 3 \u00d7 3.3 = 9.9 m
\n2x = 2 \u00d7 3.3 = 6,6 m
\nThe dimensions are 16.5 m, 9.9m, 6.6 m.
\nT.S.A =2(lb + bh + lh)
\n= 2[16.5 \u00d7 9.9 + 9.9 \u00d7 6.6 +16.5 \u00d7 6.6]
\n= 2[163.35 + 65.34 + 108.9]
\n= 2[337.59] = 675.18 m3<\/sup><\/p>\n

Question 7.
\nSuppose the perimeter of one face of a cube is 24 cm what is its volume?
\nSolution:
\nPerimeter = 24cm
\n4l = 24; l = \\(\\frac{24}{4}\\) = 6
\nl = 6
\nVolume = l3<\/sup> – 63<\/sup>
\nVolume = 216 cm3<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nA wooden box has inner dimension l = 6m, b = 8m and h = 9m and it has uniform thickness of 10cm. The lateral surface of the outerside has to be painted at the rate of 50\/m2<\/sup>. What is the cost of painting?
\nSolution:
\nInner length of 6m. The thickness is 10 cm
\n\u2234 the outer length is 6m + 20 cm = 6.2 m
\nSimilarly the outer breadth is 8 m + 20 cm = 8.2 m
\nThe height is 9 m + 20 m = 9.2 m of the box is 9m + 20 cm = 9.2 m
\nL.S.A of outer surface = 2h (l + b)
\n= 2 \u00d7 9.2 [6.2 + 8.2] = 18.4 (14.4)
\n= 264.96 m2<\/sup>
\nArea to be painted = 264.96 m2<\/sup>
\nThe cost of painting 1 sq m = Rs. 50
\nThe cost of painting 264.96 m2<\/sup> = 264.96 \u00d7 50 = Rs. 13,248<\/p>\n

Question 9.
\nEach edge of a cube is increased by 20%. What is the percentage increase in the volume of the cube?
\nSolution:
\nLet the length of each side by 1 cm
\n\u2234 Volume of cube = l3<\/sup> = l3<\/sup> = 1 cm3<\/sup>
\nIf the length is increased by 20% the length of each side is 1.2 cm.
\n\u2234 Volume of the new cube is = (1.2)3<\/sup>
\n= 1.728 cm3<\/sup>
\nIncrease in the volume
\n= 1.728 – 1 = 0.728
\n\u2234 The percentage increase is
\n= 0.728 \u00d7 100 = 72.8%<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nSuppose the length of a cube is increased by 10% and its breadth is decreased by 10% will the volume of the new cuboid be the same as that of the cube? What about the total surface areas? If they change what would be the percentage change in both the cases?
\nSolution:
\nLet the length, breadth and height of the cube be 1 cm.
\nLength is increased by 10%
\n\u2234 length = 1.1 cm Breadth is decreased by 10%
\n\u2234 breadth = 0.9cm height remains the same.
\nVolume of cube = l3<\/sup> – l3<\/sup> = 1 cm3<\/sup>
\nVolume of cuboid = lbh =1.1 \u00d7 0.9 \u00d7 1 = 0.99 cm3<\/sup>
\nDifference in volume = 1 – 0.99 = 0.01 cm3<\/sup>
\nPercentage in decrease = 0.01 \u00d7 100 = 1 %
\nT.S.A of cube = 6 l2<\/sup> = 6 \u00d7 12 = 6 cm2<\/sup>
\nT.S.A of cuboid = 2 (lb + bh + lh)
\n= 2 [1.1 \u00d7 0:9 + 0.9 \u00d7 1 + 1.1 \u00d7 1]
\n= 2 [0.99 + 0.9 + 1.1] = 2 [2.99]
\n= 5.98 cm2<\/sup>
\nDifference in T.S.A = 6 – 5.98 = 0.02
\nPercentage in decrease = 0.02 \u00d7 100
\n= 42%<\/p>\n

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

Students can Download Maths Chapter 16 Mensuration Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 8 Maths Chapter 16 Mensuration Additional Questions Question 1. Three metal cubes whose edges measure 3 …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27689"}],"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\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=27689"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27689\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=27689"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=27689"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=27689"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}