RD Sharma Solutions Class 8 Mathematics Solutions for Volume Surface Area Cuboid Cube Exercise 21.3 in Chapter 21 - Volume Surface Area Cuboid Cube

Question 10 Volume Surface Area Cuboid Cube Exercise 21.3

A classroom is 11 m long, 8 m wide and 5 m high. Find the sum of the areas of its floor and the four walls (including doors, windows etc.)


Given details are,

Dimensions of class room = 11m × 8m × 5m

Where, Length = 11m, Breadth = 8m, Height = 5m We know,

Area of floor = length × breadth

= 11 × 8

= 88 m2

Area of four walls (including doors & windows) = 2 (lh + bh) cm2

= 2 (11×5 + 8×5)

= 2 (55 + 40)

= 2 (95)

= 190m2

∴ Sum of areas of floor and four walls = area of floor + area of four walls

= 88 + 190 = 278 m2

Video transcript
[Music] hello dear student i am sunita nayar i am from lido learning and i am going to help you with this sum which goes like this a classroom is 11 meters long 8 meters wide and 5 meters high find the sum of the areas of its flow and the 4 walls including those windows etc right now we need to find the area of the entire classroom right now you will realize that this is a kind of trick question why is that because this including doors windows etc that really doesn't matter right because when we are taking the surface area of the walls right we include the entire area it's just a very simple measurement it is the lower dimension that is the length multiplied by the height that's it and that would include whatever those windows whatever is there on that wall if you come to this wall you multiply the width by the height and you will get this wall including the windows there is nothing special about the windows that you have to add the surface area of the windows no it is all included in the surface area of the wall itself so it is a very simple measurement that we have to do simple calculation that we have to do so let us get started right away so we know that the surface area of uh of the walls of a room right the surface area is given by the formula twice the length [Music] into the height plus the breadth into the height this is the area of the walls right and we have also the floor which is surface area [Music] of the flow is equal to just a simple length into the breadth right so the floor area is given by the length multiplied by the width so let me write that down length multiplied by breadth right now we have to substitute the values [Music] so as you know the length is this is the length this will be the length [Music] wait this will be the width [Music] and this is the width or the breadth we can call it b the breadth and of course this is the height [Music] right so those are the three dimensions so we have now we've written the formula also now all we have to do is substitute so let's start substituting so we have 2 into l into h l is given as 11 [Music] 11 into height is given as 5 plus the breadth is given as 8 and height is once again 5 right so this is equal to 2 multiplied by 55 plus 40 [Music] which is equal to 2 into 95 which gives us 190. [Music] right so this is 190 meter square now we have the area of the flow which is l into b so l into b is 11 into 8 which is equal to 88 88 meter square or square meter right so total area of the walls and the flow is equal to 190 plus 88 which is equal to 278 square meters right so that was done quite fast isn't it so no there was no uh question of calculating the area of areas of dose and windows etc so this is how we solved our sum i hope you understood the explanation do drop in a comment if you have any doubt and visit our channel regularly for more homework solutions and subscribe to it for regular updates as well thank you
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