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Sagot :
Given:
- A rectangle has a width that is 5 feet less than the length. The area of the rectangle is 126 square feet
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❍Let's assume, Length of rectangle x feet and width of the rectangle x - 5 feet respectively.
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To calculate the dimensions of the rectangle we will use the formula of Area of rectangle:
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[tex] \: \: \: \: \: \: \: { \underline{ \boxed{ \pmb{ \frak{ Area_{(rectangle)} = Length \times width }}}}} \\ \\ [/tex]
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[tex] \underline {\sf{Substituting \: the \: values~ in~ above~formula: }} \\ [/tex]
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[tex] : \implies \sf \:x \times (x - 5) = 126 \\ \\ \\ : \implies \sf \:x^2 - 5x = 126 \\ \\ \\ : \implies \sf \:x^2 - 5x - 126 = 0 \\ \\ \\ : \implies \sf \:x^2 - 14x - 9x - 126 = 0 \\ \\ \\ : \implies \sf \:x (x - 14) - 9 (x + 14) = 0 \\ \\ \\ : \implies \sf \:(x - 14)(x - 9) = 0 \\ \\ \\ : \implies { \boxed { \purple{\pmb { \: x = 14 \: \: or \: \: x = -9}}} } \\ \\ \\ [/tex]
- We know that side can't be negative. So, x = 14 feet
Hence,
- Width of rectangle = x - 5 = 9 feet
- Length of rectangle = x = 14 feet
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[tex] \\ { \underline{ \pmb{ \frak{\therefore Dimensions \: of \: rectangle \: are \: 9~ feet \: and \: 14 ~feet \: }}}}[/tex]
Given that, width of the rectangle is 5 feet less than the length and area of the rectangle is 126 sq. feet, and we are told to find the dimensions i.e the length and width of the rectangle.
As we know that the area of a rectangle is length times it's width, so if we assume our length to be x, so our width is just going to be (x - 5) , and putting it in the area formule we can thus obtain;
[tex]{:\implies \quad \sf x(x-5)=126}[/tex]
[tex]{:\implies \quad \sf x^{2}-5x-126=0}[/tex]
[tex]{:\implies \quad \sf x^{2}-14x+9x-126=0}[/tex]
[tex]{:\implies \quad \sf x(x-14)+9(x-14)=0}[/tex]
[tex]{:\implies \quad \sf (x-14)(x+9)=0}[/tex]
So, here either (x - 14) = 0 or (x + 9) = 0, when you equate both, you will get x = 14 or -9 but as x ≠ -9, so x = 14, and then x - 5 = 14 - 5 = 9
Hence, the required length and width are 14 feet and 9 feet respectively
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