Answer:
Part A) [tex]A=1.25W^2[/tex]
Part B) Length: 17.5 feet and Width: 14 feet
Step-by-step explanation:
Part A) Create an equation to represent the area of the basketball section A, in terms of the width W.
Let
L ----> the length of the rectangular basketball section
W ---> the width of the rectangular basketball section
we know that
The area of the rectangular basketball section is equal to
[tex]A=LW[/tex] ----> equation A
The length of the section will be 1.25 times the width of the section
so
[tex]L=1.25W[/tex] ----> equation B
substitute equation B in equation A
[tex]A=(1.25W)W\\A=1.25W^2[/tex]
Part B) Jackson decides to make the area of the basketball section 245 square feet. What are the dimensions, in feet, of the basketball section?
we have
[tex]A=1.25W^2\\A=245\ ft^2[/tex]
so
[tex]245=1.25W^2[/tex]
solve for W
[tex]W^2=245/1.25\\W^2=196\\W=14\ ft[/tex]
Find the value of L
substitute the value of W in the equation B
[tex]L=1.25(14)=17.5\ ft[/tex]
therefore
The dimensions are :
Length: 17.5 feet
Width: 14 feet
