Question

A closed rectangular box of length, breadth and height are 3m, 2m

and 1m respectively.

a) Lateral surface area of the box is


i)22 ii)10 iii)12 iv)20


b) Cost of cloth to cover box completely if 1m2 cloth cost ₹ 10 is


i)22 ii)220 iii)2200 iv)200


c) The length of longest rod that can be kept in the box is


i)3m ii)√10m iii)√12m iv)√14m


d) If one side (2m x 1m) of the box is removed, then surface area of remaining box is


i)22m2 ii)18m2 iii)20m2 iv)6m2


e) Volume of the cuboidal box is


i)7 ii)6 iii)9 iv)√10

Answer 1

Answer:

a) 10

b) 220

c) 3m

d) 20m^2

e) 6

Step-by-step explanation:

a) =2h(l+w)

   = 2x1(3+2) = 2x5

   = 10

b) =2(lb+bh+hl)

    = 22sqm

    = 22xrupees 10  =  rupees 220

c) commonb sense if the boxs length is 3m then the length of the longest rod will be 3m only.

d) if u complete 2(lb+bh+hl) then you will get 22m^2

    as it is in the question that we have to remove (2mx1m)=2m

    so, 22m^2 - 2m = 20m^2

e) volume= l*b*h

     = 3 x 2 x 1

     = 6

Answer 2

Answer:

The above is correct expect the 3rd one (common sense)

The length of longest rod that can be kept in the box is

√14 (why?)

• Actually it's a 3D shape.

• The longest part is marked in the given fig.

So let's use the formula,

Diagonal of the cuboid = √ ( l² + b² + c² )

$\implies\sf{\:\sqrt{{3}^{2}+{2}^{2}+1}}$

$\implies\sf{\:\sqrt{9+4+1}}$

$\implies\sf{\:\sqrt{14}}$

Therefore √(14) is the answer.