a. The number of rolls of wrapping paper it will take to cover the new box is 2¹/₄ rolls
b. The volume of the new rectangular box is 270 in³
Since the small box is a rectangular prism, its surface area, A = 2(lb + lh + bh) where
Now, when the dimensions are tripled, its new area is A' = 2(LB + LH + BH) where
So, A' = 2(LB + LH + BH)
= 2((3l)(3b) + (3l)(3h) + (3b)(3h))
= 2(3)(3)(lb + lh + bh)
= 18A
The number of rolls of wrapping paper it will take to cover the new box is 2¹/₄ rolls
Now, since we require 1/8 of a roll of wrapping paper to initially cover all 6 sides of the small box with dimensions, l,b and h. Then the number of rolls we require to cover the box when its dimensions are tripled are (since the area is proportional to the number of rolls)
N = 18 × 1/8
= 18/8
= 9/4
= 2¹/₄ rolls
The number of rolls of wrapping paper it will take to cover the new box is 2¹/₄ rolls
The volume of the new rectangular box is 270 in³
The volume of the rectangular box V = lbh where
When the dimensions are tripled, V' = LBH where
So, V' = LBH
= (3l)(3b)(3h)
= 27lbh
= 27V
Now since the initial volume of the rectangular box, V = 10 in³
V' = 27V
V' = 27 × 10 in³
V' = 270 in³
So, the volume of the new rectangular box is 270 in³
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