\nA plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine :
\n(i) The area of the sheet required for making the box.
\n(ii) The cost of the sheet for it, if a sheet measuring 1 m2<\/sup> costs Rs. 20.
\nSolution:
\n(i) Length of plastic box, l = 1.5 m
\nbreadth b = 1.25 m
\nheight, h = 65 cm = 0.65 m.
\n= (1.5 \u00d7 1.25) + (2 \u00d7 1.25 \u00d7 0.65) + 2(1.5)(0.65)
\n= 1.875 + 1.625 + 1.950
\nA= 5.45 m2<\/sup>.
\n(ii) Cost of 1 sq. m sheet is Rs. 20
\ncost of 5.45 sq.m, the sheet is = 5.45 \u00d7 20 = Rs. 109.<\/p>\n
Question 2. Question 3. Question 4. Question 5. (ii) Total surface area of cube (T.S.A.) = 6a2<\/sup> Question 6. (ii) Tape required for all 12 edges means, it has 4 length, 4 breadth and 4 heights. Question 7. Question 8. Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.1 Question 1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine : (i) The area of the sheet required for …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2228"}],"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=2228"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2228\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=2228"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=2228"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=2228"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nThe length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of Rs. 7.50 per m2<\/sup>.
\nSolution:
\nLength of room, l = 5 m
\nbreadth, b = 4m
\nheight, h = 3 m
\nWhite washing the room and the ceiling except floor to be made.
\n\u2234 Area, A = lb + 2bh + 21h
\n= lb + 2h(b + l)
\n= (5 \u00d7 4) + 2 \u00d7 3(4 + 5)
\n= 20 + 6(9)
\n= 20 + 54
\n= 74 m2<\/sup>.
\nCost of white washing for 1 sq.m. is Rs. 7.50
\n\u2234 Cost of white washing 74 sq. m. is ?
\n= 74 \u00d7 7.50
\n= Rs. 555.<\/p>\n
\nThe floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs. 10 per m2<\/sup> is Rs. 15000, find the height of the hall. [Hint: Area of the four walls = Lateral surface area.]
\nSolution:
\nPerimeter of a rectangle room = 250 m
\n= 2(l + b)
\nCost of Painting = Rs. 15000
\nFor Rs. 10, 1 sq.m.
\nFor Rs.15000 … ? \\(=\\frac{15000}{10}\\)
\n= 1500m2<\/sup>
\nArea of four walls = Lateral surface area
\nLateral surface area of rectangle = 1500 m2
\n<\/sup>2h(l + b) = 1500
\nh \u00d7 2(l + b) = 1500
\nh \u00d7 250 = 1500
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.1-Q-3-1.png\")
\n\u2234 h = 6 m<\/p>\n
\nThe paint in a certain container is sufficient to paint an area equal to 9.375 m2<\/sup>. How many bricks of dimensions 22.5 cm \u00d7 10 cm \u00d7 7.5 cm can be painted out of this container?
\nSolution:
\nLength of Brick l = 22.5 cm
\nbreadth, b = 10 cm
\nheight, h = 7.5 cm.
\nArea to be painted, A = 9.375 m2<\/sup>
\n= 93750 cm2<\/sup>
\nTotal surface area of brick, A
\nA= 2(lb + bh + lh)
\n= 2[22 \u00d7 10 + 10 \u00d7 7.5 + 22.5 \u00d7 7.5]
\n= 2[468.75]
\n= 937.5 cm2<\/sup>.
\nNumber of bricks to be painted, For 937.5 cm2<\/sup>, 1 brick is required For 93750 cm2<\/sup> …??…
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.1-Q-4.png\")
\n= 100
\n\u2234 100 bricks have to be painted.<\/p>\n
\nA cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
\n(i) Which box has the greater lateral surface area and by how much?
\n(ii) Which box has the smaller total surface area and by how much?
\nSolution:
\nEach edge of cubical box, a = 10 cm.
\n(i) Curved Surface Area = 4a2<\/sup> = 4(10)2<\/sup>
\n= 4 \u00d7 100
\n= 400 cm2<\/sup>.
\nLength of cuboid box, l = 12.5 cm.
\nb = 10 cm
\nh = 8 cm.
\nCurved Surface area of Cuboid, A = 2h (l+ b)
\nA = 2 \u00d7 8(12.5 + 10)
\n= 16 \u00d7 22.5
\n= 360 cm2<\/sup>.
\n\u2234 400 – 360 = 40 cm2<\/sup>.
\nL.S.A. of cuboid box is 40 cm2<\/sup> greater than rectangular cuboid.<\/p>\n
\n= 6(10)2<\/sup>
\n= 6 \u00d7 100
\n= 600 cm2<\/sup>.
\nT.S.A. of cuboid = 2(Ib + lh + bh)
\n= 2[ 12.5 \u00d7 10 + 12.5 \u00d7 8 + 10 \u00d7 8]
\n= 2 (125 + 100 + 80)
\n= 2 \u00d7 305
\n= 610 cm2<\/sup>.
\nHere, cuboid box’s T.S.A. is more than T.S.A. of cubical box, i.e,
\n610 – 600 = 10 cm2<\/sup> more.<\/p>\n
\nA small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
\n(i) What is the area of the glass?
\n(ii) How much of tape is needed for all the 12 edges?
\nSolution:
\n(i) Length of Herbarium, l = 30 cm,
\nbreadth, b = 25 cm.
\nheight, h = 25 cm.
\n\u2234 TSA of Herbarium = 2 (lb + bh + Hi)
\n= 2 [30 \u00d7 25 + 25 \u00d7 25 + 30 \u00d7 25]
\n= 2 [750 + 625 + 750]
\n= 2 (2125)
\n= 4250 cm2<\/sup>.<\/p>\n
\n\u2234 Length of tape required = 4 (l + b + h)
\n= 4 (30 + 25 + 25)
\n= 4 \u00d7 80
\n= 320 cm.<\/p>\n
\nShanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm \u00d7 20 cm \u00d7 5 cm and the smaller of dimensions 15 cm \u00d7 12 cm \u00d7 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs. 4 for 1000 cm2<\/sup>, find the cost of cardboard required for supplying 250 boxes of each kind.
\nSolution:
\nMeasurement of big box, l = 25 cm.
\nb = 20 cm.
\nh = 5 cm.
\n\u2234 TSA of eacxh big box = 2(lb + bh + lh)
\n= 2 [25 \u00d7 20 + 20 \u00d7 5 + 25 \u00d7 5]
\n= 2 [500 + 100 + 120]
\n= 2 (725)
\nT.S.A. =1450 cm2<\/sup>.
\nFor all the overlaps 5% of the total surface area is required extra.
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.1-Q-7.png\")
\n= 72.5 cm2<\/sup>.
\n\u2234 Total surface area of big box, TSA = 1450 + 72.5 = 1522.5 cm2<\/sup>.
\nArea required for 1 box is 1522.5 cm2 Area required for 25 boxes … ?…
\n= 1522.5 \u00d7 250
\n= 380625 cm2<\/sup>.
\nDimension of small box, l = 15 cm.
\nb = 12 cm.
\nh = 5 cm.
\n\u2234 TSA of each small box = 2(lb + bh + lh)
\n= 2 [15 \u00d7 12 + 12 \u00d7 5 + 15 \u00d7 5]
\n= 2 [180 + 60 + 75]
\n= 2 \u00d7 315
\n= 630 cm2<\/sup>.
\nMore area to overlap cardboard is 5% more.
\nFor 100 cm2<\/sup>, 5 cm2<\/sup>
\nFor 630 cm2<\/sup>, …?…
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.1-Q-7.1.png\")
\n= 31.5 cm2<\/sup>.
\n\u2234 T.S.A.= 630 + 31.5= 661.5 cm2<\/sup>.
\nTSA required for 1 box is 661.5 cm2<\/sup>
\nTSA required for 250 boxes …?…
\n= 661.5 \u00d7 250
\n= 165375 cm2<\/sup>.
\n\u2234 To prepare 500 boxes, TSA = 380625 + 165375
\n= 546000 cm2<\/sup>.
\nCost of 1000 sq.cm, of cardboard is Rs.4 Cost of 546000 sq.cm of cardboard is ?
\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.1-Q-7.2.png\")
\n= Rs. 2184<\/p>\n
\nParveen wanted to make a temporary shelter for her car, by making a box-like structure with a tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how many tarpaulins would be required to make the shelter of height 2.5 m, with base dimensions 4m \u00d7 3m?
\nSolution:
\nMeasurement of the roof, l = 4 m.
\nb = 3 m.
\nh = 2.5 m.
\n\u2234 Area of Tarpaulin = 2 (lb + bh) + lh
\n= 2 [4 \u00d7 2.5 + 3 \u00d7 2.5] +4 \u00d7 3
\n= 2 [10 + 7.5] + 12
\n= 2 \u00d7 17.5 + 12
\n= 35 + 12
\n= 47 m2<\/sup>.<\/p>\nKSEEB Solutions for Class 9 Maths<\/a><\/h5>\n","protected":false},"excerpt":{"rendered":"