Question

2. If each edge of a cuboid of surface area S is doubled, the surface area of the new cuboid is
(a) 25
(b) 45
(c) 65
(d) 85​

Answer 1

Step-by-step explanation:

I am correct expert Pankaj Rastogi

Answer 2

Step-by-step explanation:

Let l,b,h be the length, breadth, height of the first cuboid.

Let L,B,H be the length, breadth, 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 new cuboid S

′

=2(LB+BH+HI)

=2[(2l)(2b)+(2b)(2h)+(2h)(2l)]

=2(4lb+4bh+4hl)

=4[2(lb+bh+hl)]

=4S

∴ The surface area of the new cuboid is 4S.