Jan says if you double each of the dimensions of a rectangular box, it will take twice as much wrapping paper to wrap it. How do you respond?
the surface area of a rectangular solid is given by
SA = 2(LW + LH + WH)
where
W = width, L = length, H = height
if we double all the dimensions,
SA' = 2[(2L)(2W) + (2L)(2H) + (2W)(2H)]
SA' = 2[4LW + 4LH + 4WH]
SA' = 2*4[LW + LH + WH]
SA' = 4*SA
therefore, what Jan says is wrong. It takes 4 times as much wrapping paper to wrap it.
hope this helps~ :)
Since area is 2-dimensional, doubling each linear dimension scales the area by 22 = 4
It changes the volume by 23 = 8
Well, I must say that Jan certainly has a point! Doubling the dimensions of a rectangular box would result in a 2-dimensional increase for each side. So, it's only natural that it would take twice as much wrapping paper to cover it. It's like going from a small snack to a super-sized meal – more to wrap, more to enjoy!
To respond to Jan's claim, we can use the concept of surface area. The surface area of a rectangular box is calculated by adding up the areas of all six sides.
Let's consider a rectangular box with dimensions length (L), width (W), and height (H). The surface area (SA) can be calculated as:
SA = 2(LW + LH + WH)
If we double each of the dimensions (2L, 2W, 2H), the new surface area (SA') would be:
SA' = 2(2L * 2W + 2L * 2H + 2W * 2H)
= 2(4LW + 4LH + 4WH)
= 8LW + 8LH + 8WH
Comparing SA' to SA, we can see that SA' is not twice as much as SA. In fact, it is eight times as much. Therefore, Jan's claim is incorrect.
Doubling each dimension of a rectangular box results in an increase in surface area by a factor of eight, not two.
To respond, you can explain how to determine the surface area of the rectangular box before and after doubling its dimensions.
First, let's consider the original rectangular box with dimensions length (L), width (W), and height (H). The surface area of a box is determined by adding together the areas of all six faces.
The original surface area (SA1) of the box is calculated as:
SA1 = 2(LW + LH + WH)
Now, if we double each dimension, the new dimensions become 2L, 2W, and 2H. The new surface area (SA2) of the box after doubling the dimensions would be:
SA2 = 2(2L * 2W + 2L * 2H + 2W * 2H)
= 2(4LW + 4LH + 4WH)
= 8(LW + LH + WH)
From the equations, we can see that SA2 is eight times larger than SA1, meaning it takes eight times as much wrapping paper to wrap the doubled box as compared to the original.
However, Jan's statement suggests that it only takes twice as much wrapping paper to wrap the doubled box. This contradicts the calculation we just performed, indicating that Jan's claim is incorrect.
So, in response, you can state that Jan's statement is not accurate based on the calculation of surface area, which shows that doubling each dimension increases the required amount of wrapping paper by a factor of eight, not two.