{"id":2852,"date":"2023-04-26T10:00:02","date_gmt":"2023-04-26T04:30:02","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=2852"},"modified":"2023-04-26T09:26:20","modified_gmt":"2023-04-26T03:56:20","slug":"surface-area-volume-cuboid","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/surface-area-volume-cuboid\/","title":{"rendered":"How do you find the Surface Area and Volume of a Cuboid"},"content":{"rendered":"<h2><strong>How do you find the Surface Area and Volume of a Cuboid<\/strong><\/h2>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-12196\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/surface-area-and-volume-cuboid.jpg\" alt=\"\" width=\"629\" height=\"297\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/surface-area-and-volume-cuboid.jpg 629w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/surface-area-and-volume-cuboid-300x142.jpg 300w\" sizes=\"(max-width: 629px) 100vw, 629px\" \/><\/p>\n<p>If \u2113, b and h denote respectively the length, breadth and height of a cuboid, then<\/p>\n<ol>\n<li>Total surface area of the cuboid = 2 (\u2113b + bh + \u2113h) square units.<\/li>\n<li>Volume of the cuboid\u00a0= Area of the base \u00d7 height = \u2113\u00a0\u00d7 b \u00d7 h cubic units.<br \/>\nwhere base area = Breadth \u00d7 length<\/li>\n<li>Diagonal of the cuboid or longest rod\u00a0= \\(\\sqrt {{\\ell ^2} + {b^2} + {h^2}} \\) units. and Total length of its edges = 4 (\u2113 +\u00a0b +\u00a0h)<\/li>\n<li>Area of four walls of a room\u00a0= 2 (\u2113 + b) h sq. units.<\/li>\n<\/ol>\n<h2><strong>Surface Area and Volume of a Cuboid Example Problems with Solutions<\/strong><\/h2>\n<p><strong>Example 1:<\/strong>\u00a0 \u00a0 \u00a0The length, breadth and height of a cuboid are in the ratio 6 : 4 : 5. If the total surface area of \u00a0the cuboid is 2368 cm<sup>2<\/sup> ; find its dimension.<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>Let length (\u2113) = 6x cm, breadth (b) = 4x cm and height (h) = 5x cm,<br \/>\n\u2234\u00a0\u00a0\u00a0 Total surface area<br \/>\n= 2(\u2113 \u00d7 b + b \u00d7 h \u00d7 h \u00d7 \u2113)<br \/>\n= 2(6x \u00d7 4x + 4x \u00d7 5x + 5x \u00d7 6x)cm<sup>2<\/sup><br \/>\n= 2(24x<sup>2<\/sup>\u00a0+ 20x<sup>2<\/sup>\u00a0+ 30x<sup>2<\/sup>) cm<sup>2<\/sup><br \/>\n=148x<sup>2<\/sup>\u00a0cm<sup>2<\/sup><br \/>\nGiven : Total surface = 2368 cm<sup>2<\/sup><br \/>\n\u21d2\u00a0\u00a0 148x<sup>2<\/sup>\u00a0= 2368<br \/>\n\u21d2\u00a0\u00a0 x<sup>2<\/sup>\u00a0= \\(\\frac{{2368}}{{148}}\\)\u00a0= 16<br \/>\nand x =\u00a0\\(\\sqrt {16} \u00a0= 4\\)<br \/>\n\u2234\u00a0\u00a0\u00a0 length = 6x cm = 6 \u00d7 4 cm = 24 cm,<br \/>\nbreadth = 4x cm = 4 \u00d7 4 cm = 16 cm and<br \/>\nheight = 5x cm = 5 \u00d7 4 cm = 20 cm<\/p>\n<p><strong>Example 2:<\/strong>\u00a0 \u00a0 \u00a0A 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(i)\u00a0\u00a0 The area of the sheet required for making the box.<br \/>\n(ii)\u00a0 The cost of sheet for it, if a sheet measuring 1 m<sup>2<\/sup>\u00a0costs Rs. 20.<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>Given. length (\u2113) = 1.5 m, breadth (b) = 1.25 m and depth i.e., height (h) = 65 cm = 0.65 m.<br \/>\n(i)\u00a0\u00a0 Since, the box is open it has five faces in which four faces are the walls forming lateral surface area and one face is the base.<br \/>\n\u2234\u00a0\u00a0\u00a0 The area of the sheet required for making the box.<br \/>\n= \u00a0\u00a0 Lateral surface area of the box + area of its base<br \/>\n= 2(\u2113 + b) \u00d7 h + \u2113 \u00d7 b<br \/>\n= 2(1.5m + 1.25m) \u00d7 0.65m + 1.5m \u00d7 1.25 m<br \/>\n= 2 \u00d7 2.75 m \u00d7 0.65 m + 1.875 m<sup>2<\/sup><br \/>\n= 3.575 m<sup>2<\/sup>\u00a0+ 1.875 m2 = 5.45 m<sup>2<\/sup><br \/>\n(ii)\u00a0 Since, a sheet measuring 1 m<sup>2<\/sup>\u00a0costs Rs.20<br \/>\n\u2234 \u00a0\u00a0 The cost of sheet for the box<br \/>\n= 5.45 \u00d7 Rs 20 =\u00a0 Rs. 109<\/p>\n<p><strong>Example 3:<\/strong>\u00a0 \u00a0 \u00a0The length, breadth and height of a room are 5m, 4m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs. 7.50 per m<sup>2<\/sup>.<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>Given. \u2113 = 5m, b = 4m and h = 3m<br \/>\nSince, the area of the walls of the room<br \/>\n= its lateral surface area = 2(\u2113 + b) \u00d7 h<br \/>\nAnd, the area of the ceiling of the room<br \/>\n= \u2113 \u00d7 b<br \/>\n\u2234\u00a0\u00a0\u00a0 Total area to be white washed<br \/>\n= Area of the walls + area of the ceiling<br \/>\n= 2(\u2113 + b) \u00d7 h + \u2113 \u00d7 b.<br \/>\n= 2(5m + 4m) \u00d7 3m + 5m \u00d7 4m<br \/>\n= 2 \u00d7 9m \u00d7 3m + 20m<sup>2<\/sup><br \/>\n= 54m<sup>2<\/sup>\u00a0+ 20m<sup>2<\/sup>\u00a0= 74m<sup>2<\/sup><br \/>\nSince, the rate of white washing<br \/>\n= Rs. 7.50 per m<sup>2<\/sup><br \/>\n\u2234 \u00a0Cost of white washing<br \/>\n= 74 \u00d7 Rs.7.50 = Rs. 555<\/p>\n<p><strong>Example 4:<\/strong>\u00a0 \u00a0 \u00a0The 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>\u00a0is Rs : 15,000, find the height of the hall.<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>We know, the perimeter of a rectangle<br \/>\n= 2(length + breadth) = 2 (\u2113 + b)<br \/>\nAnd, given the floor of a rectangular hall has perimeter 250 m<br \/>\n\u21d2 \u00a0 2(\u2113 + b) = 250 m<br \/>\nArea of the four walls of the hall<br \/>\n= Lateral surface area of the hall<br \/>\n= 2(\u2113 + b) \u00d7 h = 250 m \u00d7 hm = 250 hm<sup>2<\/sup><br \/>\nSince, the rate of painting four walls is<br \/>\nRs. 10 per m<sup>2<\/sup>.<br \/>\n\u2234\u00a0\u00a0\u00a0 Cost of painting 250 hm<sup>2<\/sup>\u00a0= 250 h \u00d7 Rs.10<br \/>\nAccording to the given statements :<br \/>\n250 h \u00d7 Rs. 10 = Rs. 15,000<br \/>\n\u21d2\u00a0\u00a0 h = \\(\\frac{{15,000}}{{250 \\times 10}}m\\)\u00a0= 6 m<br \/>\n\u2234\u00a0\u00a0\u00a0 The height of the hall = 6 m<\/p>\n<p><iframe loading=\"lazy\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/y96z6hpT20Q?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<p><strong>Example 5:<\/strong>\u00a0 \u00a0 \u00a0The paint in a certain container is sufficient to paint an area equal to 9.375 m<sup>2<\/sup>. How many bricks of dimension 22.5 cm \u00d7 10 cm \u00d7 7.5 cm can be painted out of this container.<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>For each brick, \u2113 = 22.5 cm, b = 10 cm and h = 7.5 cm.<br \/>\n\u2234 \u00a0Surface area of each brick<br \/>\n= 2 (\u2113 \u00d7 b + b \u00d7 h + h \u00d7 \u2113)<br \/>\n= 2 (22.5 \u00d7 10 + 10 \u00d7 7.5 + 7.5 \u00d7 22.5)cm<sup>2<\/sup><br \/>\n= 2 (225 \u00d7 75 + 168.75) cm2 = 937.5 cm<sup>2<\/sup><br \/>\nSince, the paint in the container is sufficient to paint an area.<br \/>\n= 9.375 m<sup>2<\/sup>\u00a0= 9.375 \u00d7 100 \u00d7 100 cm<sup>2<\/sup><br \/>\n[1 m = 100 cm and 1 m<sup>2<\/sup>\u00a0= 100 \u00d7 100 cm<sup>2<\/sup>]<br \/>\n= 93750 cm<sup>2<\/sup><br \/>\n\u2234\u00a0\u00a0\u00a0 Number of bricks that can be painted<br \/>\n=\u00a0\\(\\frac{{Total\\,\\,area\\,\\,which\\,\\,can\\,\\,be\\,\\,painted}}{{Surface\\,\\,area\\,\\,of\\,\\,the\\,\\,brick}}\\)<br \/>\n=\u00a0\\(\\frac{{\\,\\,\\,93750\\,\\,c{m^2}}}{{937.5\\,\\,cm}} = 100\\)<\/p>\n<p><strong>Example 6:<\/strong>\u00a0 \u00a0 \u00a0A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tap. It is 30 cm long, 25 cm wide and 25 cm high.<br \/>\n(i)\u00a0\u00a0 What is the area of the glass?<br \/>\n(ii)\u00a0 How much of tape is needed for all the 12 edges?<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>The green house is a cuboid in shape in which \u2113 = 30 cm, b = 25 cm and h = 25 cm<br \/>\n(i)\u00a0\u00a0 Area of the glass<br \/>\n= Surface area of cubical green house.<br \/>\n= 2(\u2113 \u00d7 b + b \u00d7 h + h \u00d7 \u2113)<br \/>\n= 2(30 \u00d7 25 + 25 \u00d7 25 + 25 \u00d7 30) cm<sup>2<\/sup><br \/>\n= 2(750 + 625 + 750)cm2 \u00a0= 4250 cm<sup>2<\/sup><br \/>\n(ii)\u00a0 Length of the tape<br \/>\n= Length of the 12 edges of the cubical greenhouse = 4(\u2113 + b + h)<br \/>\n= 4(30 + 25 + 25)cm = 320 cm<\/p>\n<p><strong>Example 7:<\/strong>\u00a0 \u00a0 \u00a0Shanti 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 cm<sup>2<\/sup>, find the cost of cardboard required for supplying 250 boxes of each kind.<strong> \u00a0<\/strong><br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>For each bigger box :<br \/>\n\u2113 = 25 cm, b = 20 cm and h = 5 cm<br \/>\n\u2234\u00a0\u00a0\u00a0 Surface area<br \/>\n= 2(\u2113 \u00d7 b + b \u00d7 h + h \u00d7 \u2113)<br \/>\n= 2(25\u00a0 \u00d7 20 + 20 \u00d7 5 + 5 \u00d7 25)cm<sup>2<\/sup><br \/>\n= 2(500 + 100 + 125)cm2\u00a0= 1450 cm<sup>2<\/sup><br \/>\nFor each smaller box :<br \/>\n\u2113 = 15 cm, b = 12 cm and h = 5 cm<br \/>\n\u2234\u00a0\u00a0\u00a0 Surface area<br \/>\n= 2(\u2113 \u00d7 b + b \u00d7 h + h \u00d7 \u2113)<br \/>\n= 2(15 \u00d7 12 + 12 \u00d7 5 + 5 \u00d7 15)cm<sup>2<\/sup><br \/>\n= 2(180 + 60 + 75) cm<sup>2<\/sup>\u00a0= 630 cm<sup>2<\/sup><br \/>\nTotal surface area of 250 boxes of each kind<br \/>\n= 250 \u00d7 1450 cm<sup>2<\/sup>\u00a0+ 250 \u00d7 630 cm<sup>2<\/sup><br \/>\n= 362500 cm<sup>2<\/sup>\u00a0+ 157500 cm<sup>2<\/sup><br \/>\n= 520000 cm<sup>2<\/sup><br \/>\nCardboard required for overlaps<br \/>\n= 5% of 520000 cm<sup>2<\/sup><br \/>\n= \\(\\frac{5}{{100}} \\times 52000c{m^2}\\)\u00a0= 26000 cm<sup>2<\/sup><br \/>\n\u2234\u00a0\u00a0\u00a0 Total area of the cardboard used<br \/>\n= 520000 cm<sup>2<\/sup>\u00a0+ 26000 cm<sup>2<\/sup><br \/>\n= 546000 cm<sup>2<\/sup><br \/>\nGiven, cost of 1000 cm<sup>2<\/sup>\u00a0of cardboard = Rs.4<br \/>\n\u21d2 \u00a0 cost of 1 cm<sup>2<\/sup>\u00a0of cardboard = Rs.\u00a0\\(\\frac{4}{{1000}}\\)<br \/>\n\u21d2\u00a0\u00a0 cost of 546000 cm<sup>2<\/sup>\u00a0of cardboard<br \/>\n= Rs. \\(\\frac{4}{{1000}}\\) \u00d7 546000 = Rs. 2184<\/p>\n<p><strong>Example 8:<\/strong>\u00a0 \u00a0 \u00a0The volume of a cubical solid is 3240 cm<sup>3<\/sup>, find, its<br \/>\n(i)\u00a0\u00a0 height, if length =18 cm and breadth\u00a0= 15cm<br \/>\n(ii)\u00a0 breadth, if length =24 cm and height\u00a0= 10cm<br \/>\n(iii) length, if breadth = 9cm and height\u00a0= 20cm<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>Volume = length \u00d7 breadth \u00d7 height<br \/>\n\u21d2\u00a0\u00a0 (i) height =\u00a0\\(\\frac{{Volume}}{{length \\times breadth}}\\)<br \/>\n(ii) breadth =\u00a0\\(\\frac{{Volume}}{{length \\times height}}\\)<br \/>\n(iii) Length\u00a0= \\(\\frac{{Volume}}{{breadth \\times height}}\\)<br \/>\n(i)\u00a0\u00a0 Height =\u00a0\\(\\frac{{Volume}}{{length \\times breadth}}\\)<br \/>\n= \\(\\frac{{3240\\,\\,c{m^3}}}{{18\\,\\,cm \\times 15\\,\\,cm}}\\)\u00a0= 12 cm<br \/>\n(ii)\u00a0 Breadth =\u00a0\\(\\frac{{Volume}}{{length \\times height}}\\)<br \/>\n= \\(\\frac{{3240\\,\\,c{m^3}}}{{24\\,\\,cm \\times 10\\,\\,cm}}\\)\u00a0= 13.5 cm<br \/>\n(iii) Length = \\(\\frac{{Volume}}{{breadth \\times height}}\\)<br \/>\n= \\(\\frac{{3240\\,\\,c{m^3}}}{{9\\,\\,cm \\times 20\\,\\,cm}}\\)\u00a0= 18 cm<\/p>\n<p><strong>Example 9:<\/strong>\u00a0 \u00a0 \u00a0A matchbox measures 6 cm \u00d7 4 cm \u00d7 2.5 cm. What will be the volume of a packet containing 24 such boxes ?<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>The shape of a matchbox is a cuboid<br \/>\n\u2234\u00a0\u00a0\u00a0 Volume of 1 matchboxes<br \/>\n= its length \u00d7 breadth \u00d7 height<br \/>\n= 6 \u00d7 4 \u00d7 2.5 cm<sup>2<\/sup>\u00a0= 60 cm<sup>3<\/sup><br \/>\n\u21d2 \u00a0\u00a0 Volume of a packet containing 24 such boxes<br \/>\n= Volume of 24 matchboxes<br \/>\n= 24 \u00d7 60 cm<sup>3<\/sup>\u00a0= 1440 cm<sup>3<\/sup><\/p>\n<p><strong>Example 10:<\/strong>\u00a0 \u00a0 \u00a0A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many liters of water can it hold? (1 m<sup>2<\/sup>\u00a0= 1000 \u2113)<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>Volume of water which tank can hold<br \/>\n= Volume of the tank<br \/>\n= Its length \u00d7 width \u00d7 height<br \/>\n= 6 m \u00d7 5 m \u00d7 4.5 m = 135 m<sup>3<\/sup><br \/>\n= 135 \u00d7 1000 \u2113 \u00a0= 135000 \u2113<\/p>\n<p><strong>Example 11:<\/strong>\u00a0 \u00a0 \u00a0Find the cost of digging a cuboidal pit 10 m long, 7.5 m broad and 80 cm deep at the rate of Rs. 16 per m<sup>3<\/sup>.<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>Given :<br \/>\nLength of the pit = 10 m, breadth = 7.5 m and depth = 80 cm = 0.8 m<br \/>\n\u2234 \u00a0\u00a0 Volume of cuboidal pit<br \/>\n= Its length \u00d7 breadth \u00d7 depth (or height)<br \/>\n= 10 m \u00d7 7.5 m \u00d7 0.8 m = 60 m<sup>3<\/sup><br \/>\nRate of digging the pit = Rs. 16 per m<sup>3<\/sup><br \/>\n\u2234 \u00a0\u00a0 Cost of digging 60 \u00d7 Rs. 16 = Rs. 960<\/p>\n<p><strong>Example 12:<\/strong>\u00a0 \u00a0 \u00a0Wooden crates each measuring 1.5 m \u00d7 1.25 m \u00d7 0.5 m are to be stores in a godown. Find the largest number of wooden crates which can be stores in a godown measuring : <strong>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/strong><br \/>\n(i)\u00a0\u00a0 45 m \u00d7 25 m \u00d7 10m.<br \/>\n(ii)\u00a0 40 m \u00d7 25 m \u00d7 10m<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>(i)\u00a0\u00a0 Since, the length of godown is 45 m and the length of a crate is 1.5 m<br \/>\n\u2234 \u00a0\u00a0 The largest no. of crates that can stores along the length of the godown<br \/>\n=\u00a0\\(\\frac{{45m}}{{1.5m}} = 30\\)<br \/>\nSimilarly, the largest no. of crates stores along the width of the godown<br \/>\n=\u00a0\\(\\frac{{25m}}{{1.25m}} = 20\\)<br \/>\nAnd, the largest no. of crates stored along the length of the godown<br \/>\n=\u00a0\\(\\frac{{10m}}{{0.5m}} = 20\\)<br \/>\n\u2234 \u00a0\u00a0 The largest no. of crates which can be stored in the godown<br \/>\n= 30 \u00d7 20 \u00d7 20<br \/>\n= 12000<br \/>\n(ii)\u00a0 No. of crates along the length<br \/>\n= \\(\\frac{{40m}}{{1.5m}} = 26\\frac{2}{3}\\)\u00a0= 26<br \/>\n[NO. of crates can not be in fraction]<br \/>\nNo. of crates along the width<br \/>\n=\u00a0\\(\\frac{{25m}}{{1.25m}} = 20\\)<br \/>\nNo. of crates along the height<br \/>\n=\u00a0\\(\\frac{{10m}}{{0.5m}} = 20\\)<br \/>\n\u2234\u00a0\u00a0\u00a0 The largest no. of crates which can be stored in the godown<br \/>\n= 26 \u00d7 20 \u00d7 20<br \/>\n= 10400<\/p>\n<p><strong>Example 13:<\/strong>\u00a0 \u00a0 \u00a0The capacity of a cuboidal tank is 50000 liters of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>The capacity of cuboidal tank is 50000 liters of water.<br \/>\n\u21d2 \u00a0 Volume of the tank = 50000 liters<br \/>\n= \u00a0\\(\\frac{{50000}}{{1000}}{m^3}\\)\u00a0\u00a0[\u2235 \u00a01 m<sup>3<\/sup>\u00a0= 1000 liters]<br \/>\n= 50 m<sup>3<\/sup><br \/>\n\u21d2 \u00a0 Length of the tank \u00d7 its breadth \u00d7 its height = 50 m<sup>3<\/sup><br \/>\n\u21d2 \u00a0 2.5 m \u00d7 breadth \u00d7 10 m = 50 m<sup>3<\/sup><br \/>\n\u21d2 \u00a0 Breadth =\u00a0\\(\\frac{{50}}{{2.5 \\times \\,\\,10}}m = 2m\\)<\/p>\n","protected":false},"excerpt":{"rendered":"<p>How do you find the Surface Area and Volume of a Cuboid If \u2113, b and h denote respectively the length, breadth and height of a cuboid, then Total surface area of the cuboid = 2 (\u2113b + bh + \u2113h) square units. Volume of the cuboid\u00a0= Area of the base \u00d7 height = \u2113\u00a0\u00d7 [&hellip;]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[5],"tags":[1233,1232,1230,1231],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.8 (Yoast SEO v17.8) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>How do you find the Surface Area and Volume of a Cuboid - A Plus Topper<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aplustopper.com\/surface-area-volume-cuboid\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"How do you find the Surface Area and Volume of a Cuboid\" \/>\n<meta property=\"og:description\" content=\"How do you find the Surface Area and Volume of a Cuboid If \u2113, b and h denote respectively the length, breadth and height of a cuboid, then Total surface area of the cuboid = 2 (\u2113b + bh + \u2113h) square units. 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