Respuesta :
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
The rectangle has an area of 60 square feet. Find its dimensions (in ft) if the length of the rectangle is 4 ft more than its widh.
smaller value ___________________ ft
larger value ____________________ ft
Answer:
Smaller value = 6 ft
Larger value = 10 ft
Step-by-step explanation:
Recall that the area of a rectangle is given by
[tex]Area = W \times L[/tex]
Where W is the width and L is the length of the rectangle.
It is given that the rectangle has an area of 60 square feet.
[tex]Area = 60 \: ft^2 \\\\60 = W \times L \\\\[/tex]
It is also given that the length of the rectangle is 4 ft more than its width
[tex]L = W + 4[/tex]
Substitute [tex]L = W + 4[/tex] into the above equation
[tex]60 = W \times (W + 4) \\\\60 = W^2 + 4W \\\\W^2 + 4W - 60 = 0 \\\\[/tex]
So we are left with a quadratic equation.
We may solve the quadratic equation using the factorization method
[tex]W^2 + 10W - 6W - 60 \\\\W(W + 10) – 6(W + 10) \\\\(W + 10) (W - 6) = 0 \\\\[/tex]
So,
[tex](W + 10) = 0 \\\\W = -10 \\\\[/tex]
Since width cannot be negative, discard the negative value of W
[tex](W - 6) = 0 \\\\W = 6 \: ft \\\\[/tex]
The length of the rectangle is
[tex]L = W + 4 \\\\L = 6 + 4 \\\\L = 10 \: ft \\\\[/tex]
Therefore, the dimensions of the rectangle are
Smaller value = 6 ft
Larger value = 10 ft
Verification:
[tex]Area = W \times L \\\\Area = 6 \times 10 \\\\Area = 60 \: ft^2 \\\\[/tex]
Hence verified.
