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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
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Solution
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.
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