{"id":4665,"date":"2020-09-09T11:27:25","date_gmt":"2020-09-09T05:57:25","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=4665"},"modified":"2021-07-02T15:03:35","modified_gmt":"2021-07-02T09:33:35","slug":"kseeb-solutions-class-8-maths-chapter-16-ex-16-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-8-maths-chapter-16-ex-16-1\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 16 Mensuration Ex 16.1"},"content":{"rendered":"

Students can Download Maths Chapter 16 Mensuration Ex 16.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka Board Class 8 Maths Chapter 16 Mensuration Ex 16.1<\/h2>\n

Question 1.
\nFind the total surface area of a cuboid with l= 4m, b = 3m and h = 1.5 m
\nAnswer:
\nT.S.A = 2 (lb + bh + lh)
\n= 2(4 \u00d7 3 + 3 \u00d7 1.5 + 1.5 \u00d7 4)
\n= 2(12 + 4.5 + 6)
\n= 2(22.5)
\nT.S.A = 45 m\u00b2<\/p>\n

Question 2.
\nFind the area of four walls of a room whose length is 3.5m, breath 2.5m and height is 3m.
\nAnswer:
\nL = 3.5m, b = 2.5m, h = 3m
\nArea of the 4 walls = L.S.A of cuboid
\n= 2h(l + b)
\n= 2 \u00d7 3(3.5 + 2.5)
\n= 6(6) = 36m\u00b2<\/p>\n

Question 3.
\nThe dimensions of a room are l = 8m, b = 5m and h = 4m. Find the cost of distempering its four walls at the rate of 40\/m2.
\nAnswer:
\nL.S.A of cuboid = 2h (l + b)
\n= 2 \u00d7 4 (8 + 5) = 8 (13)
\nL.S.A = 104 m\u00b2
\nThe cost of distempering 1 m\u00b2 = Rs 40
\nThe cost of distempering 104m\u00b2 = 104 \u00d7 40 = Rs. 4160.<\/p>\n

Question 4.
\nA room is 4.8m long, 3.6m broad and 2m high. Find the cost of laying tiles on its floor and its four walls at the rate of 100\/m\u00b2
\nAnswer:
\nL = 4.8m, b = 3.6m , h = 2m
\nRequired area = Area of the floor + L.S.A of cuboid
\n= lb + 2h(l + b)
\n= 4.8 \u00d7 3.6 + 2 \u00d7 2(4.8 + 3.6)
\n= 17.28 + 33.6 = 50.88 m\u00b2
\nThe cost of laying tiles for 1m\u00b2 = Rs 100
\nThe cost of laying tiles for 50.88 m\u00b2 = 50.88 \u00d7 100 = Rs. 5088<\/p>\n

Question 5.
\nA closed box is 40 cm long, 50 cm wide and 60 deep. Find the area of the foil needed for covering it.
\nAnswer:
\nL = 40 cm, b = 50cm, h = 60 cm
\nArea of the foil = T.S. A of cuboid
\n= 2 (lb + bh + lh)
\n= 2(40 \u00d7 50 + 50 \u00d7 60 + 60 \u00d7 40)
\n= 2(7400)= 14,800 cm\u00b2
\nArea of the foil required 14,800 cm\u00b2<\/p>\n

Question 6.
\nThe total surface area of a cube is 384 cm2 calculate the side of the cube.
\nAnswer:
\nT.S.A of cube = 384 cm\u00b2
\n61\u00b2 = 384
\nl\u00b2 = \\(x=\\frac{384}{6}\\)
\nl\u00b2 = 64
\nl = \u221a64
\nl = 8cm<\/p>\n

Question 7.
\nThe L.S.A of a cube is 64m\u00b2,\u00a0calculate the side of the cube.
\nAnswer:
\nL.S.A of cube = 41\u00b2
\n64 = 41\u00b2
\n\\(x=\\frac{64}{4}\\)\u00a0= l\u00b2
\n\u2234 l = \u221a16 = 4m
\nThe side of the cube is 4m.<\/p>\n

Question 8.
\nFind the cost of whitewashing the four walls of a cubical room of side 4m at the rate of Rs.20\/m2.
\nAnswer:
\nl = 4m
\nArea to be whitewashed
\n= L.S.A of the cube
\n= 4l\u00b2 = 4 \u00d7 4\u00b2 = 4 \u00d7 16 = 64 m\u00b2
\ncost of whitewashing 1m\u00b2 = 20
\n\u2234 The cost of whitewashing 64m\u00b2 = 64 \u00d7 20 = Rs. 1280.<\/p>\n

Question 9.
\nA cubical box has edge 10cm and another cuboidal box is 12.5 cm long, 10cm wide and 8cm high
\ni. Which box has a smaller total surface area?
\nii. If each edge of the cube is doubled, how many times will its T.S.A increase?
\nAnswer:
\nT.S.A of cube 61\u00b2= 6 \u00d7 10\u00b2 = 6 \u00d7100 = 600 cm\u00b2
\nT.S.A of a cuboid 61\u00b2 = 6 \u00d7 102 = 6 x 100 = 600 cm\u00b2
\nT.S.A of a cuboid = 2(lb + bh + Ih)
\n= 2[12.5 x 10 +10 x 8 + 8 x 12.5]
\n= 2[125 + 80 + 100]
\n= 2[305]
\nt = 610 cm\u00b2
\n1. The cubical box has smaller total surface area.
\n2. If the side of the cube is doubled its side will be 20cm.
\nT.S.A 6l\u00b2 = 6 \u00d7 202 = 6 \u00d7 400 = 2400 cm\u00b2
\noriginal area = 600cm\u00b2
\n\\(\\frac { 2400 }{ 600 }\\) = 4
\n\u2234 Its area will increase by 4 times<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 16 Mensuration Ex 16.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka Board Class 8 Maths Chapter 16 Mensuration Ex 16.1 Question 1. Find the total surface area of …<\/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\/4665"}],"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=4665"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4665\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=4665"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=4665"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=4665"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}