{"id":4581,"date":"2023-02-21T09:35:33","date_gmt":"2023-02-21T09:35:33","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4581"},"modified":"2023-02-21T09:39:54","modified_gmt":"2023-02-21T09:39:54","slug":"ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-1","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-1\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 9 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 13 (Surface Areas and Volumes)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 9 Mathematics\u00a0<strong>Chapter &#8211; 13 Surface Areas and Volumes <\/strong>Exercise 13.1 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter 13: Surface Areas and Volumes<\/strong><\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.2<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.3<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-4\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.4<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-5\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.5<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-6\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.6<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-7\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.7<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-8\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.8<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-9\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.9<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 13.1<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. A plastic box 1.5 m long, 1.25 m wide, and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine:<br \/>\n<\/strong><strong>(i) The area of the sheet required for making the box.<br \/>\n<\/strong><strong>(ii) The cost of the sheet for it, if a sheet measuring 1m<sup>2<\/sup>\u00a0costs Rs. 20.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; Given :<br \/>\n<\/strong>Length, l = 1.5 m<\/span><br \/>\n<span style=\"color: #000000;\">Breadth, b = 1.25 m<\/span><br \/>\n<span style=\"color: #000000;\">Height, h = 65 cm\u00a0= 65\/100 m\u00a0= 0.65 m<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4649\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.1-Ans1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"579\" height=\"283\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.1-Ans1.png 579w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.1-Ans1-300x147.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.1-Ans1-480x235.png 480w\" sizes=\"auto, (max-width: 579px) 100vw, 579px\" \/><\/span><br \/>\n<span style=\"color: #000000;\"><strong>(i) Box is to be open at the top.<br \/>\n<\/strong>Area of sheet required = 2lh + 2bh + lb<br \/>\n= lb + 2(l + b)h<br \/>\n= (1.5 m \u00d7 1.25m) + 2 \u00d7 (1.25 m + 1.5 m) \u00d7 0.65 m<\/span><br \/>\n<span style=\"color: #000000;\">= 1.875 m\u00b2 + 2 \u00d7 2.75 m \u00d7 0.65 m<\/span><br \/>\n<span style=\"color: #000000;\">= 1.875 m\u00b2 + 3.575 m\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">= 5.45 m\u00b2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) Cost of sheet per m<sup>2<\/sup>\u00a0area = Rs.20.<br \/>\n<\/strong>Therefore, the cost of the sheet = Rate of the sheet \u00d7 Area of the sheet<\/span><br \/>\n<span style=\"color: #000000;\">= \u20b9 20\/ m\u00b2 \u00d7 5.45 m\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">= \u20b9109<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. The 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 ceiling at the rate of Rs 7.50 per m<sup>2<\/sup>.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong><strong>Given :<br \/>\n<\/strong>Length (l) of room = 5 m<br \/>\nBreadth (b) of room = 4 m<br \/>\nHeight (h) of room = 3 m<br \/>\nSurface area of 5 faces = Area of the 4 walls and ceiling = lb + 2(l + b)h <\/span><br \/>\n<span style=\"color: #000000;\">lb + 2(l + b)h = (5 m \u00d7 4 m) + 2 \u00d7 (5 m + 4 m) \u00d7 3 m <\/span><br \/>\n<span style=\"color: #000000;\">= 20 m\u00b2 + 2 \u00d7 9 m \u00d7 3 m<\/span><br \/>\n<span style=\"color: #000000;\">= 20 m\u00b2 + 54 m<sup>2<br \/>\n<\/sup>= 74 m\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">The cost of whitewashing the walls of the room and ceiling = Rate \u00d7 Area <\/span><br \/>\n<span style=\"color: #000000;\">=\u00a0\u20b9 7.50 \/ m\u00b2 \u00d7 74 m<sup>2<br \/>\n<\/sup>= \u20b9 555 <\/span><br \/>\n<span style=\"color: #000000;\">Thus, the cost of whitewashing the walls of the room and the ceiling is \u20b9555.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. The 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 m<sup>2<\/sup> is Rs.15000, find the height of the hall.<br \/>\n<\/strong><strong>[Hint: <\/strong>Area of the four walls = Lateral surface area<strong>]<br \/>\n<\/strong><strong>Answer &#8211; Given :<br \/>\n<\/strong>The rate of painting is \u20b910 \/ m<sup>2<\/sup><br \/>\nPerimeter of the floor = 250 m<br \/>\nThe cost of painting the four walls is \u20b915000.<br \/>\nPerimeter of the floor of hall = 2(l + b) = 250 m\u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (i)<br \/>\nArea of four walls = 2lh + 2bh = 2(l + b)h<br \/>\nNow, Area of four walls = 15000\/10 m<sup>2<\/sup>\u00a0= 1500 m<sup>2<br \/>\n<\/sup>2(l + b)h = 1500 m<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [From equation (i)]<\/span><br \/>\n<span style=\"color: #000000;\">250 m \u00d7 h = 1500 m<sup>2<br \/>\n<\/sup>h = 1500 m<sup>2<\/sup>\/250 m\u00a0= 6 m<\/span><br \/>\n<span style=\"color: #000000;\">Thus, the height of the hall is 6 m.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. The paint in a certain container is sufficient to paint an area equal to 9.375 m<sup>2<\/sup>. How many bricks of dimensions 22.5 cm\u00d710 cm\u00d77.5 cm can be painted out of this container?<br \/>\n<\/strong><strong>Answer &#8211; Given : <\/strong>Dimensions of the brick 22.5cm \u00d7 10cm \u00d7 7.5cm<br \/>\nThe area which can be painted with the paint available in the container = 9.375m<sup>2<\/sup>.<br \/>\nThe area of each brick to be painted = 2(lb + bh + hl)<br \/>\n= 2 \u00d7 (22.5 cm \u00d7 10 cm + 10 cm \u00d7 7.5 cm + 7.5 cm \u00d7 22.5 cm)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 (225 cm<sup>2<\/sup>\u00a0+ 75 cm<sup>2<\/sup>\u00a0+ 168.75 cm<sup>2<\/sup>)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 468.75 cm<sup>2<br \/>\n<\/sup>= 937.5 cm<sup>2<br \/>\n<\/sup>Number of bricks that can be painted = The area which can be painted with the paint available in the container \/ The area of each brick <\/span><br \/>\n<span style=\"color: #000000;\">= 9.375 m<sup>2\u00a0<\/sup>\/ 937.5 cm<sup>2<br \/>\n<\/sup>= (9.375 \u00d7 10000 cm<sup>2<\/sup>) \/ 937.5 cm<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [since 1m<sup>2<\/sup>\u00a0= 10000cm<sup>2<\/sup>]<\/span><br \/>\n<span style=\"color: #000000;\">= 100<\/span><br \/>\n<span style=\"color: #000000;\">Thus, the number of bricks that can be painted out of the container is 100.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. A cubical box has each edge 10 cm, and another cuboidal box is 12.5cm long, 10 cm wide, and 8 cm high.<br \/>\n<\/strong><strong>(i) Which box has the greater lateral surface area, and by how much?<br \/>\n<\/strong><strong>(ii) Which box has the smaller total surface area, and by how much?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Edge length of the cube, a = 10 cm<br \/>\nLength of the cuboid, l = 12.5 cm<\/span><br \/>\n<span style=\"color: #000000;\">Breadth of the cuboid, b = 10 cm<\/span><br \/>\n<span style=\"color: #000000;\">Height of the cuboid, h = 8 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i)<\/strong> Lateral surface area of the cube\u00a0= 4a<sup>2<br \/>\n<\/sup>= 4 \u00d7 (10 cm)<sup>2<br \/>\n<\/sup>= 4 \u00d7 100 cm<sup>2<br \/>\n<\/sup>= 400 cm<sup>2<br \/>\n<\/sup>Lateral surface area of the cuboid = (l + b)h<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 (12.5 cm + 10 cm) \u00d7 8 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 22.5 cm \u00d7 8 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 360 cm<sup>2<br \/>\n<\/sup>We see that, the cubical box has a greater lateral surface area by (400 cm<sup>2<\/sup>\u00a0&#8211; 360 cm<sup>2<\/sup>) = 40 cm<sup>2<\/sup><\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(ii)<\/strong> Total surface area of the cube = 6a<sup>2<br \/>\n<\/sup>= 6 \u00d7 (10 cm)<sup>2<br \/>\n<\/sup>= 6 \u00d7 100 cm<sup>2<br \/>\n<\/sup>= 600 cm<sup>2<br \/>\n<\/sup>Total surface area of the cuboid = 2(lb + bh + hl) <\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 (12.5 cm \u00d7 10 cm + 10 cm \u00d7 8 cm + 8 cm \u00d7 12.5 cm)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 (125 cm<sup>2<\/sup>\u00a0+ 80 cm<sup>2<\/sup>\u00a0+ 100 cm<sup>2<\/sup>)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 305 cm<sup>2<br \/>\n<\/sup>= 610 cm<sup>2<br \/>\n<\/sup>Cubical box has a smaller total surface area by (610 cm<sup>2<\/sup>\u00a0&#8211; 600 cm<sup>2<\/sup>) = 10 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Thus, the cubical box has a greater lateral surface area by 40 cm<sup>2<\/sup>\u00a0and the cubical box has a smaller total surface area by 10 cm<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including the base) held together with tape. It is 30cm long, 25 cm wide, and 25 cm high.<br \/>\n<\/strong><strong>(i) What is the area of the glass?<br \/>\n<\/strong><strong>(ii) How much tape is needed for all 12 edges?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; Given:<br \/>\n<\/strong>Length, l = 30 cm<\/span><br \/>\n<span style=\"color: #000000;\">Breadth, b = 25 cm<\/span><br \/>\n<span style=\"color: #000000;\">Height, h = 25 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i)<\/strong> The area of the glass = 2(lb + bh + hl)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 (30 cm \u00d7 25 cm + 25 cm \u00d7 25 cm + 25 cm \u00d7 30 cm)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 (750 cm<sup>2<\/sup>\u00a0+ 625 cm<sup>2<\/sup>\u00a0+ 750 cm<sup>2<\/sup>)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 2125 cm<sup>2<br \/>\n<\/sup>= 4250 cm<sup>2<br \/>\n<\/sup>The total surface area of the glass is 4250 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) <\/strong>Length of the tape needed for all the 12 edges = 4(l + b + h)<\/span><br \/>\n<span style=\"color: #000000;\">= 4 \u00d7 (30 cm + 25 cm + 25 cm)<\/span><br \/>\n<span style=\"color: #000000;\">= 4 \u00d7 80 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 320 cm<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, 320 cm tape is required for all 12 edges.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. Shanti 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\u00d720cm\u00d75cm, and the smaller of dimensions 15cm\u00d712cm\u00d75cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs. 4 for 1000 cm<sup>2<\/sup>, find the cost of cardboard required for supplying 250 boxes of each kind.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong><strong>For Bigger Box:<\/strong><\/span><br \/>\n<span style=\"color: #000000;\">Let the length, breadth and height of the bigger box be L, B and H respectively.<\/span><br \/>\n<span style=\"color: #000000;\">Length, L = 25 cm<\/span><br \/>\n<span style=\"color: #000000;\">Breadth, B = 20 cm<\/span><br \/>\n<span style=\"color: #000000;\">Height, H = 5 cm<\/span><br \/>\n<span style=\"color: #000000;\">The area of the card board = 2(LB + BH + HL)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 (25cm \u00d7 20cm + 20cm \u00d7 5cm + 5cm \u00d7 25cm)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 (500cm\u00b2 + 100cm\u00b2 + 125cm\u00b2)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 725 cm\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">= 1450 cm\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">For all the overlaps, 5% of the total surface area is required extra. Therefore, <\/span><br \/>\n<span style=\"color: #000000;\">Overlap area = 5% of 1450 cm\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{5}{100}\" alt=\"\\frac{5}{100}\" align=\"absmiddle\" \/> \u00d7 1450cm\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">= 72.5 cm<sup>2<br \/>\n<\/sup>Net area of the card board required for each bigger box = 1450 cm\u00b2 + 72.5 cm\u00b2 = 1522.5 cm\u00b2<br \/>\nWe can now find the area of\u00a0 250 such boxes and the total cost of the cardboard at \u20b94 per 1000 cm<sup>\u00b2<\/sup>.<br \/>\nArea of card board required for 250 such boxes = 250 \u00d7 1522.5 cm\u00b2 = 380625 cm\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">The total cost of the cardboard at \u20b94 per 1000 cm\u00b2 = (4\/1000) \u00d7 380625 = \u20b91522.50<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>For Smaller Box:<br \/>\n<\/strong>Let the length, breadth and height of the smaller box be l, b and h respectively.<\/span><br \/>\n<span style=\"color: #000000;\">Length, l = 15 cm<\/span><br \/>\n<span style=\"color: #000000;\">Breadth, b = 12 cm<\/span><br \/>\n<span style=\"color: #000000;\">Height, h = 5 cm<\/span><br \/>\n<span style=\"color: #000000;\">The area of the cardboard = 2(lb + bh + hl)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 (15 cm \u00d7 12 cm + 12 cm \u00d7 5 cm + 5 cm \u00d7 15 cm)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 (180 cm\u00b2 + 60 cm\u00b2 + 75 cm\u00b2)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 315 cm\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">= 630 cm\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">For all the overlaps, 5% of the total\u00a0surface area\u00a0is required extra. Therefore,<\/span><br \/>\n<span style=\"color: #000000;\">Overlap area = 5% of 630cm\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{5}{100}\" alt=\"\\frac{5}{100}\" align=\"absmiddle\" \/> \u00d7 630 cm\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">= 31.5 cm\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">Net area of the cardboard required for smaller box = 630 cm\u00b2 + 31.5 cm\u00b2 = 661.5 cm\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">Area of cardboard required for 250 such boxes = 250 \u00d7 661.5 cm\u00b2 =165375 cm\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">The total cost of the cardboard at \u20b94 per 1000 cm\u00b2 = \u20b9 (4\/1000) \u00d7 165375 = \u20b9661.50<\/span><br \/>\n<span style=\"color: #000000;\">Cost of cardboard required for supplying 250 boxes of each kind = \u20b91522.50 + \u20b9661.50 = \u20b92184<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. Praveen wanted 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 much tarpaulin would be required to make the shelter of height 2.5m, with base dimensions 4m\u00d73m? <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>Let the length, breadth and height of the shelter be l, b and h respectively.<\/span><br \/>\n<span style=\"color: #000000;\">Length, l = 4 m<\/span><br \/>\n<span style=\"color: #000000;\">Breadth, b = 3 m<\/span><br \/>\n<span style=\"color: #000000;\">Height, h = 2.5 m<\/span><br \/>\n<span style=\"color: #000000;\">The area of the tarpaulin required to make the shelter = lb + 2(l + b)h<\/span><br \/>\n<span style=\"color: #000000;\">= (4 m \u00d7 3 m) + 2 \u00d7 (4 m + 3 m) \u00d7 2.5 m<\/span><br \/>\n<span style=\"color: #000000;\">= 12 m\u00b2 + 2 \u00d7 7 m\u00a0\u00d7 2.5 m<\/span><br \/>\n<span style=\"color: #000000;\">= 12 m\u00b2 + 35 m\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">= 47 m\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">Hence, 47 m\u00b2 of tarpaulin will be required.<\/span><\/p>\n<p style=\"text-align: justify;\">\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 13 (Surface Areas and Volumes)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#8211; 13 Surface Areas and Volumes Exercise 13.1 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 13: Surface Areas and Volumes NCERT Solution Class 9 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[698],"tags":[725,702,186,726,701,5],"class_list":["post-4581","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-13-surface-areas-and-volumes-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-13-in-english","tag-ncert-solutions-class-9-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.5) - 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