{"id":25994,"date":"2021-06-24T17:47:52","date_gmt":"2021-06-24T12:17:52","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=25994"},"modified":"2021-06-24T17:51:24","modified_gmt":"2021-06-24T12:21:24","slug":"ncert-solutions-for-class-9-maths-chapter-13-ex-13-1","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-13-ex-13-1\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 13 Surface Areas and Volumes Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1<\/h2>\n

Question 1.
\nA plastic box 13 m long, 1.25 wide and 65 cm deep are to be made. It is to be open at the top. Ignoring the thickness of a 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 1m costs Rs. 20.
\nSolution:
\n(i) We have given,
\nlength of box = 1.5 m
\nwidth of the box = 1.25 m
\nheight of the box = 65 cm = 0.65 m
\n\"NCERT
\n\u2234 Surface area of box = 2(lb + bh + lh)
\n= 2(1.5 \u00d7 1.25 + 1.25 \u00d7 0.65 + 0.65 \u00d7 1.5)
\n= 2(1.875 + 0.8125 + 0.975)
\n= 2 \u00d7 3.6625
\n= 7.325 m2<\/sup>
\nBut we have given box is open at the top.
\nArea of top of the box = 1.875 m2<\/sup>
\n\u2234 Surface area of required box = 7.325 – 1.875 = 5.45 m2<\/sup>
\nSo, the area of sheet required for making the box is 5.45 m2<\/sup>
\n(ii) Rate of sheet = Rs. 20\/cm2<\/sup>
\n\u2234 Total cost of sheet for box = 20 \u00d7 5.45 = Rs. 109.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nThe length, breadth, and height of a room are 5 m, 3 m, and 3 m respectively. Find the cost of whitewashing the walls of the room and ceiling at the rate of Rs. 7.50 per m2<\/sup>.
\nSolution:
\n\"NCERT
\nWe have given
\nlength of room = 5 m
\nbreadth of a room = 4 m
\nheight of a room = 3 m
\nSurface area of 4 walls and ceiling of the room
\n= 2(l + b)h + lb
\n= 2(5 + 4)3 + 5 \u00d7 4
\n= 54 + 20
\n= 74 m2<\/sup>
\nCost of whitewashing the walls and ceiling of the room at Rs. 7.50 per m2<\/sup> is = 74 \u00d7 7.50 = Rs. 555.<\/p>\n

Question 3.
\nThe floor of the rectangular hall has a perimeter of 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.
\nSolution:
\nWe have given that
\nPerimeter of floor = 250 m
\n2(l + b) = 250
\nl + b = 125 m
\nNow,
\nArea of four walls = 2 (l + b)h
\n= 2 \u00d7 125 \u00d7 h
\n= 250h m2<\/sup>.
\nRate of painting the four walls is Rs. 10 per m2<\/sup> is Rs. 250h \u00d7 10
\nBut we have given that the total cost of painting of four walls = 15000
\ni.e. 2500h = 15000
\nor, h = \\(\\frac{15000}{2500}\\) = 6 m
\n\u2234 Height of the hall = 6 m<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nThe paint in a certain container is sufficient to paint an area equal to 9.375 m2<\/sup>. How many bricks of dimension 22.5 cm \u00d7 10 cm \u00d7 7.5 cm can be painted out of this container?
\nSolution:
\nWe have given that the container contains sufficient paint for the area of 9.375 m2<\/sup>.
\n\"NCERT
\nAgain
\nlength of the bricks = 22.5 cm = 0.225 m
\nbreadth of the bricks = 10 cm = 0.10 m
\nheight of the bricks = 7.5 cm = 0.075 m
\n\u2234 surface area of one brick = 2(lb + bh + lh)
\n= 2(0.225 \u00d7 0.10 + 0.10 \u00d7 0.075 + 0.075 \u00d7 0.225)
\n= 2(0.0225 + 0.0075 + 0.016875)
\n= 0.09375 m2<\/sup>.
\nArea of one brick = 0.09375 m2<\/sup>.
\nLet x bricks can be painted out of this container
\n\u2234 0.9375 \u00d7 x = 9.375
\nor, x = \\(\\frac{9.375}{0.09375}\\) = 100
\nTotal number of bricks can be painted out of this container is 100.<\/p>\n

Question 5.
\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 a greater lateral surface area and by how much?
\n(ii) Which box has the smaller total surface area and by how much?
\nSolution:
\n(i) Length of edge of cubical box = 10 cm
\nThe lateral surface area of cubical box = 4a2<\/sup>
\n= 4 \u00d7 (10)2<\/sup>
\n= 400 cm2<\/sup>
\n\"NCERT
\nAgain, length of cuboidal box = 12.5 cm
\nbreadth of cuboidal box = 10 cm
\nand height of cuboidal box = 8 cm.
\n\u2234 Lateral surface area of cubodical box = 2(l + b)h
\n= 2(12.5 + 10)8
\n= 360 cm2<\/sup>
\nLateral surface area of cubical box is greater than cubodical box by = (400 – 360) = 40 cm2<\/sup>
\nNow,
\nTotal surface ara of cubical box = 6a2<\/sup>
\n= 6 \u00d7 (10)2<\/sup>
\n= 600 cm2<\/sup>
\nand total surface area of cubodical box = 2 (lb + bh + lh)
\n= 2(12.5 \u00d7 10 + 10 \u00d7 8 + 8 \u00d7 12.5)
\n= 2(125 + 80 + 100)
\n= 610 cm2<\/sup>
\nTotal surface area of cubodical box is greater than cubical box by = (610 – 600) = 10 cm2<\/sup>.<\/p>\n

\"NCERT<\/p>\n

Question 6.
\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 glass?
\n(ii) How much tape is needed for all the 12 edges?
\nSolution:
\n(i) We have given that
\nlength of small indoor greenhouse = 30 cm.
\nthe breadth of small indoor greenhouse = 25 cm.
\nand height of small indoor greenhouse = 25 cm.
\n\"NCERT
\nTotal surface area of indoor greenhouse (herbarium)
\n= 2(lb + bh + lh)
\n= 2(30 \u00d7 25 + 25 \u00d7 25 + 25 \u00d7 30)
\n= 2(750 + 625 + 750)
\n= 4250 cm2<\/sup>
\n\u2234 Area of glass to make herbarium is 4250 cm2<\/sup>.
\n(ii) Total length of tape is needed = 4(l + b + h)
\n= 4(30 + 25 + 25)
\n= 320 cm<\/p>\n

Question 7.
\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. 5% of the total surface area is required extra, for all the overlaps. If the cost of cardboard is Rs. 4 for 1000 cm2<\/sup>, find the cost of cardboard required for supplying 250 boxes of each kind.
\nSolution:
\nDimensions of the bigger box are 25 cm \u00d7 20 cm \u00d7 5 cm
\nTotal surface ara of one big box = 2(lb + bh + lh)
\n= 2(25 \u00d7 20 + 20 \u00d7 5 + 5 \u00d7 25)
\n= 2(500 + 100 + 125)
\n= 1450 cm2<\/sup>.
\n\u2234 Total surface area of 250 such boxes are 250 \u00d7 1450 = 3,62,600 cm2<\/sup>.
\nAgain,
\nDimension of smaller box is 15 cm \u00d7 12 cm \u00d7 5 cm.
\n\u2234 Total surface area of one small box = 2(lb + bh + lh)
\n= 2(15 \u00d7 12 + 12 \u00d7 5 + 5 \u00d7 15)
\n= 2(180 + 60 + 75)
\n= 630 cm2<\/sup>.
\n\u2234 Total surface area of 250 such boxes = 250 \u00d7 630 = 1,57,500 cm2<\/sup>.
\n\u2234 Total surface area of both type of boxes are = 3,62,500 + 1,57,500 = 5,20,000 cm2<\/sup>
\nNow, 5% of the total surface area is required for the overlaps.
\n5% of 5,20,000 = \\(\\frac{5}{100}\\) \u00d7 5,20,000 = 26,000
\n\u2234 Total area of card board required = 5,20,000 + 26000 = 5,46,000
\n\u2234 Total cost of cardboard at the rate of Rs. 4 per 1000 cm2<\/sup> = \\(\\frac{546000}{1000} \\times 4\\) = Rs. 2184<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nPraveen wants to make a temporary shelter for her car, by making a box-like structure with a tarpaulin that covers all 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 2.5m, with base dimensions 4 m \u00d7 3 m?
\nSolution:
\nWe have given that,
\nlength of temporary shelter for car = 4 m
\nthe breadth of temporary shelter for car = 3 m
\nand height of temporary shelter for car = 2.5 m
\n\"NCERT
\n\u2234 Total tarpaulin be required to make the shelter = Area of 4 wall + Area of roof
\n= 2(l + b)h + l \u00d7 b
\n= 2(4 + 3) \u00d7 2.5 + 4 \u00d7 3
\n= 2 \u00d7 7 \u00d7 2.5 + 4 \u00d7 3
\n= 35 + 12
\n= 47 m2<\/sup>
\nTherefore total tarpaulin be required to make temporary shelter for the car is 47 m2<\/sup>.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1 Question 1. A plastic box 13 m long, 1.25 wide and 65 cm deep are …<\/p>\n

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NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1 Question 1. 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