Respuesta :
Answer:
Length: sqrt13 + 4
Width: sqrt13
Step-by-step explanation:
Denote the width as x, then the length is x+4. The area of the rectangle will be x(x+4), and the perimeter will then be 4x +8. Hence, you can write the equation x(x+4) - 5 = 4x+8. Solving, you get x = sqrt13.
The length of the rectangle is 11 inches and width is 7 inches.
From the first statement, the length of the rectangle is 4 inches more than its width,
Let the length be [tex]l[/tex] and the width be [tex]w[/tex]
Then,
[tex]l = 4 +w[/tex] ------ (1)
From the second statement, the area of the rectangle is equal to 5 inches more than 2 times the perimeter
Then,
[tex]A = 5 + 2P[/tex] ----- (2)
Let the area be [tex]A[/tex] and the perimeter be [tex]P[/tex]
But, area of rectangle is given by the formula
[tex]A = l \times w = lw[/tex]
and the perimeter of a rectangle is given by
[tex]P = 2(l+w)[/tex]
Putting these into equation (2),
[tex]A = 5 + 2P[/tex]
we get
[tex]lw = 5 + 2[2(l+w)][/tex]
[tex]lw = 5 + 2 (2l+2w)[/tex]
[tex]lw = 5 + 4l +4w[/tex] ------ (3)
From equation (1)
[tex]l = 4 +w[/tex]
Substitute this into equation (3)
[tex]lw = 5 + 4l +4w[/tex]
[tex](4+w)w = 5 + 4(4+w) +4w\\4w + w^{2} = 5 + 16 + 4w + 4w\\w^{2}+4w = 21+8w\\w^{2}+4w -8w-21= 0 \\w^{2}-4w-21= 0[/tex]
Solve the above equation quadratically
[tex]w^{2}-4w-21= 0\\w^{2}-7w+3w-21= 0\\w(w-7)+3(w-7)=0\\(w+3)(w-7)=0\\[/tex]
[tex]w+3 = 0[/tex] or [tex]w-7=0[/tex]
[tex]w= -3[/tex] or [tex]w = 7[/tex]
Since dimension cannot be negative
∴ [tex]w = 7[/tex]
∴width = 7 inches
For the length
Substitute the value of w into equation (1)
[tex]l = 4 +w[/tex]
[tex]l = 4 +7[/tex]
[tex]l = 11[/tex]
∴ length = 11 inches
Hence, the length of the rectangle is 11 inches and width is 7 inches.
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