Question

if each edge of a cuboid of a surface area S is doubled, then surface area of new cuboid is....

Answer 1

Answer:

The New TSA is 4 times the old TSA...

Step-by-step explanation:

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

Now, TSA = 2(lb+bh+hl)

Now, New length = 2l, New width = 2b, and new height = 2h

So, New TSA = 2(2l*2h + 2b*2h + 2l*2h) = 2(4lh + 4bh + 4 lh) = 2*4(lb+bh+hl) = 8(lb+bh+hl)

Ratio of OLD TSA to NEW TSA = 2(lb+bh+hl) ÷ 8(lb+bh+hl) = 1 ÷ 4 or 1:4

Thus, the New TSA is 4 times the old TSA...

Answer 2

Answer:

Step-by-step explanation:

jasgupta Proficient Moderator

Answer:

The New TSA is 4 times the old TSA...

Step-by-step explanation:

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

Now, TSA = 2(lb+bh+hl)

Now, New length = 2l, New width = 2b, and new height = 2h

So, New TSA = 2(2l*2h + 2b*2h + 2l*2h) = 2(4lh + 4bh + 4 lh) = 2*4(lb+bh+hl) = 8(lb+bh+hl)

Ratio of OLD TSA to NEW TSA = 2(lb+bh+hl) ÷ 8(lb+bh+hl) = 1 ÷ 4 or 1:4

Thus, the New TSA is 4 times the old TSA...