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Q.

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.).


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a

288 m 2  

b

278 m 2  

c

298 m 2  

d

268 m 2   

answer is B.

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Detailed Solution

It is given a classroom of length 11 m, breadth 8m, and height 5 m.
The surface area of a cuboid having a length l ,breadth b and height h is 2(lb+bh+hl)   .
⇒l=11    ⇒b=8   ⇒h=5   Let the total surface area of the cuboid be S.
Total surface area of cuboid, S = 2(lb+bh+hl)
Since, we have to find the surface area of the four walls and the floor of the classroom; we subtract the area of the ceiling of the classroom.
Let the surface area of four walls and floor of the classroom be denoted by C.
Area of the ceiling of classroom = l×b  
The surface area of four walls and floor of classroom =S−(l×b)  
∴C=2(lb+bh+hl)−lb  … (1)
Substituting the values of l,b,h in equation 2  ,
⇒C=2 (11×8)+(8×5)+(5×11) − 11×8   ⇒C=2 88+40+55 − 88   ⇒C=2×183−88   ⇒C=366−88  
⇒C=278 m 2  
Therefore, the surface area of four walls and floor of a classroom having a length 11m , breadth 8  m and height 5   m is 278  m 2  .
Hence option (2) is correct.
 
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