Question

look at this rectangle:
if both dimensions are doubled, then which of the following statements about its area will be true?
the new area will be 2 times the old area.
the new area will be 16 times the old area.

Explanation:

Step1: Calculate the old area

The formula for the area of a rectangle is \(A = l\times w\). Given \(l = 7\) in and \(w = 3\) in, the old area \(A_{old}=7\times3 = 21\) in².

Step2: Calculate the new dimensions

If both dimensions are doubled, the new length \(l_{new}=2\times7 = 14\) in and the new width \(w_{new}=2\times3 = 6\) in.

Step3: Calculate the new area

Using the area formula \(A = l\times w\), the new area \(A_{new}=14\times6=84\) in².

Step4: Find the ratio of new - to - old area

\(\frac{A_{new}}{A_{old}}=\frac{84}{21}=4\). Another way: If the original length is \(l\) and width is \(w\), \(A_{old}=l\times w\). After doubling, \(l' = 2l\) and \(w'=2w\), then \(A_{new}=(2l)\times(2w)=4(l\times w)=4A_{old}\). But wait, let's check the general formula. If the scale factor of side lengths is \(k\) (here \(k = 2\)), the area scale factor is \(k^{2}\).

Answer:

Neither of the given options is correct. The new area is \(4\) times the old area. But if we assume there was a mis - statement in the problem (maybe if it was a square with side \(4\) and then side \(8\), but for the given rectangle with length \(7\) and width \(3\)):
Let's re - check using the formula \(A = lw\).
Original \(A_{1}=7\times3 = 21\).
New \(A_{2}=(2\times7)\times(2\times3)=4\times(7\times3)=4A_{1}\).
If we consider the general formula for area of a rectangle \(A = lw\), when \(l\) becomes \(2l\) and \(w\) becomes \(2w\), \(A'=(2l)(2w) = 4lw\). So the correct relationship is that the new area is \(4\) times the old area. But since the options are wrong, if we assume a miscalculation in the problem setup (maybe a square with side \(4\) (area \(16\)) and then side \(8\) (area \(64\), \(64 = 4\times16\)) but no, for the given rectangle:
If we follow the strict calculation:
Old area \(A_{old}=7\times3=21\).
New area \(A_{new}=(2\times7)\times(2\times3)=84\).
\(\frac{A_{new}}{A_{old}}=\frac{84}{21} = 4\).