Respuesta :
Answer:
8 cm and 10 cm
Step-by-step explanation:
Hello, I can help you with this.
Step 1
According to the question there are two rectangles A and B,
Rectangle A is a scale drawing of Rectangle B and has 25% of its area
in other words
[tex]Area_{A}=0.25*Area_{B} (Equation\ 1)\\[/tex]
Step 2
Let
Rectangle A
length (1)= 4 cm
length (2)= 5 cm
[tex]Area_{A}=4\ cm * 5\ cm\\Area_{A}=20\ cm^{2}[/tex]
put this value into equation 1
[tex]Area_{A}=0.25*Area_{B} (Equation\ 1)\\\\20\ cm^{2} =0.25*Area_{B} \\divide\ each\ side\ by\ 0.25\\\frac{20\ cm^{2} }{0.25}=\frac{0.25}{0.25}*Area_{B}\\ Area_{B}=80\ cm^{2}[/tex]
Now, we know the area of rectangle B, to know its length we need to formule other equation
Step 3
[tex]Area_{B}=80\ cm^{2}\\length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\[/tex]
the ratio between the lengths must be constant, so the ratio of A must be equal to ratio in B, then
[tex]\frac{length(1A)}{length(2A)}=\frac{length(1B)}{length(2B)} \\\\\\frac{4}{5}= \frac{length(1B)}{length(2B)}\\0.8=\frac{length(1B)}{length(2B)}\\length(1B)=0.8*length(2B) (Equation 3)[/tex]
Step three
using Eq 1 and Eq 2 find the lengths
put the value of length(1B) into equation (2)
[tex]length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\\(0.8*length(2B)) (*length (2B)=80\ cm^{2} \\\\0.8*(length (2B))^{2} =80\ cm^{2}\\(length (2B))^{2} =\frac{80\ cm^{2}}{0.8} \\(length (2B))^{2}=100\\\sqrt{(length (2B))^{2}}=\sqrt{100\ cm^{2}} \\ length (2B)=10\ cm[/tex]
Now, put the value of length(2B) into equation 3 to know length (1B)
[tex]length(1B)=0.8*length(2B)\\length(1B)=0.8*10\ cm\\length(1B)=8 cm[/tex]
I really hope this helps you, have a great day.