Answer: 16.81 times larger
Step-by-step explanation:
We can solve this problem by either choosing to use a variable as a side length, or we can plug in a value and evaluate them. In this case I recommend we use a variable and see what the final value we get is for both areas. Let's make one side of courtyard [tex]x[/tex] and the other side [tex]y[/tex]. Using this we can know the the dimensions of the field are [tex]4.1x[/tex] and [tex]4.1y[/tex]
Solving:
[tex]\[\fbox{ \parbox{5cm}{\centering Area = length \(\times\) width }}\][/tex]
Let's use the lengths and widths given to use to solve for the areas of both rectangles and then divide the area of the field by the courtyard to get our answer:
Courtyard:
[tex]\text{Area} = \text{length} \times \text{width}\\Courtyard Area = x \times ~y\\\\\boxed{Courtyard~Area = xy}[/tex]
Field:
[tex]\text{Area} = \text{length} \times \text{width}\\\\\text{Field~Area} = 4.1x ~\times 4.1y[/tex]
[tex]\boxed{\text{Field Area} = 16.81xy}[/tex]
Now divide the field area by the courtyard area to find how many times larger it is.
[tex]\text{Answer} = \frac{Area~Field}{Area ~Court}[/tex] ⇒ [tex]\frac{16.81xy}{xy}[/tex] Cancel xy ⇒ [tex]\boxed{\boxed{16.81 ~times}}[/tex]
That's the answer!
