Answer:
Part a) The length of the smaller rectangle is [tex]6\ ft[/tex]
Part b) The width of the smaller rectangle is [tex]5\ ft[/tex]
Part c) The area of the smaller rectangle is [tex]30\ ft^{2}[/tex]
Step-by-step explanation:
Part a)
Find the length side of the smaller figure
we know that
The scale factor is equal to [tex]\frac{1}{4}[/tex]
Remember that
The scale factor is equal to divide the measure of the smaller figure by the corresponding measure of the original figure
so
Let
x--------> the length of the smaller rectangle
y-------> the length of the original figure
z-----> scale factor
[tex]z=\frac{x}{y}[/tex]
we have
[tex]y=24\ ft[/tex]
[tex]z=1/4[/tex]
substitute and solve for x
[tex](1/4)=\frac{x}{24}[/tex]
[tex]x=24/4=6\ ft[/tex]
Part b)
Find the width side of the smaller figure
we know that
The scale factor is equal to [tex]\frac{1}{4}[/tex]
Remember that
The scale factor is equal to divide the measure of the smaller figure by the corresponding measure of the original figure
so
Let
x--------> the width of the smaller rectangle
y-------> the width of the original figure
z-----> scale factor
[tex]z=\frac{x}{y}[/tex]
we have
[tex]y=20\ ft[/tex]
[tex]z=1/4[/tex]
substitute and solve for x
[tex](1/4)=\frac{x}{20}[/tex]
[tex]x=20/4=5\ ft[/tex]
Part c)
Find the area of the smaller figure
we know that
The scale factor is equal to [tex]\frac{1}{4}[/tex]
Remember that
The scale factor squared is equal to divide the area of the smaller figure by the area of the original figure
so
Let
x--------> the area of the smaller rectangle
y-------> the area of the original figure
z-----> scale factor
so
[tex]z^{2}=\frac{x}{y}[/tex]
we have
[tex]y=480\ ft^{2}[/tex]
[tex]z=1/4[/tex]
substitute and solve for x
[tex](1/4)^{2}=\frac{x}{480}[/tex]
[tex]x=480/16=30\ ft^{2}[/tex]
