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
<h3>Area of rectangular prism</h3>
Since the small box is a rectangular prism, its surface area, A = 2(lb + lh + bh) where
- l = length,
- b = width and
- h = height.
Now, when the dimensions are tripled, its new area is A' = 2(LB + LH + BH) where
- L = 3l,
- B = 3b and
- H = 3h.
So, A' = 2(LB + LH + BH)
= 2((3l)(3b) + (3l)(3h) + (3b)(3h))
= 2(3)(3)(lb + lh + bh)
= 18A
<h3>a. N
umber of
rolls of
wrapping paper</h3>
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
<h3>b. Volume of the new rectangular box</h3>
The volume of the new rectangular box is 270 in³
The volume of the rectangular box V = lbh where
- l = length,
- b = width and
- h = height
When the dimensions are tripled, V' = LBH where
- L = 3l,
- B = 3b and
- H = 3h.
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
Learn more about volume of rectangular prism here:
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