Question

If each edge of a cuboid of surface area S is doubled, then surface area of the new cuboid is

Options

•  2 S

• 4 S

• 6 S

• 8 S

Answer 1

Let,

 l →Length of the first cuboid

b → Breadth of the first cuboid

h → Height of the first cuboid

And,

L → Length of the new cuboid

B → Breadth of the new cuboid

H → Height of the new cuboid

We know that,

L = 2l

B = 2b

H=2h

Surface area of the first cuboid,

 S = 2(lb + bh +hl ) 

Surface area of the new cuboid,

S' = 2 (LB + BH + HI .)

    =2 [(2l)(2b)+(2b)(2h)+(2h)(2l)]

    =2 ( 4lb + 4 bh +4hl)

    =4[2(lb+bh+hl)]

   = 4S

The surface area of the new cuboid is 4S.