Question

Mohit painted a closed box of length 2.5 metre breadth and height 1 metre. how much surface area did he cover if he painted all except the bottom of the box. plz solve it fast.....​

Answer 1

Answer:

Lateral surface area of cuboid=2(l+b) *h

=2(2.5+2)*1

=2*4.5*1

9 {m}^{2}

Area of top=l*b

=2.5*2

5 {m}^{2}

Total area painted=5+9

14 {m}^{2}

Thanks

4.4

Answer 2

Note - The complete question should be, "Mohit painted the outside of a box of length 2.5 m, breadth 2 m, and height 1 m. How much surface area did he cover if he painted all surfaces except the bottom of the box ?

The total surface area covered by Mohit in painting the box is 25 m².

• Given,

Length of the box (l) = 2.5 m

Breadth of the box (b) = 2 m

Height of the box (h) = 1 m

• The box is cuboidal in shape.

• The bottom surface of a cuboidal box is a rectangle having only length and breadth

∴  Surface area of the bottom of the box = l × b  

• Total surface area of the box =

2 (lb + bh + lh)

Surface area of the box excluding its bottom = 2 ( lb + bh + lh ) - lb = 2 (bh + lh) + 2lb - lb = 2 (bh + lh) + lb

•  ∴ Total surface area of the box painted by Mohit (T.S.A.) = 2 { ( 2.5 m × 2 m) + (2.5 m × 1 m) } + (2.5 m × 2 m)

Or, T.S.A   painted = 2 ( 5 m² + 2.5 m² + 5 m² )

Or, T.S.A. painted = 2 × 12.5 m²

Or, T.S.A. painted = 25 m²