Answer:
The length is 9 and the with is 6.
Step-by-step explanation:
I found this by finding the square root of 54 and then guessing around it
The length of the rectangle is 9 feet and the width is 6 feet.
Given that,
A rectangle has an area of 54 square feet.
If the length is 3 feet more than the width.
We have to determine,
The dimensions of the rectangle.
According to the question,
Let the length of the rectangle be L,
And the width of the rectangle is W.
A rectangle has an area of 54 square feet.
The area of the rectangle is given by,
[tex]\rm Area \ of \ rectangle = length \times width\\\\L\times W= 54[/tex]
If the length is 3 feet more than the width.
[tex]\rm L = W+3[/tex]
Substitute the value of L in equation 1,
[tex]\rm L\times W= 54\\\\(W+3) \times W = 54\\\\W^2+3W=54\\\\W^2+3W-54=0\\\\W^2+9W-6W-54=0\\\\W(W+9)-6(W+9)\\\\(W+9)(W-6)=0\\\\W+9=0, \ W=-9\\\\W-6=0, \ W=6[/tex]
The value of W can not be -9 because the width can not be negative then the value of W is 6 feet.
Then,
The value of L is,
[tex]\rm L=W+3\\\\L=6+3\\\\L=9[/tex]
The value of L is 9 feet.
Hence, The length of the rectangle is 9 feet and the width is 6 feet.
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