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Sagot :
Given:
A rectangular counter-top has a perimeter of 22 feet and an area of 24 square feet.
To find:
The dimensions of the counter.
Solution:
Let l is the length and w is the width of a rectangle, then
Perimeter of a rectangle is:
[tex]P=2(l+w)[/tex]
Area of the rectangle is:
[tex]A=l\times w[/tex]
A rectangular counter-top has a perimeter of 22 feet and an area of 24 square feet.
[tex]P=22[/tex]
[tex]2(l+w)=22[/tex]
[tex]l+w=\dfrac{22}{2}[/tex]
[tex]l=11-w[/tex] ...(i)
And,
[tex]A=24[/tex]
[tex]l\times w=24[/tex] ...(ii)
Using (i) and (ii), we get
[tex](11-w)\times w=24[/tex]
[tex]11w-w^2=24[/tex]
[tex]0=24-11w+w^2[/tex]
[tex]w^2-11w+24=0[/tex]
Splitting the middle term, we get
[tex]w^2-8w-3w+24=0[/tex]
[tex]w(w-8)-3(w-8)=0[/tex]
[tex](w-8)(w-3)=0[/tex]
[tex]w=8,3[/tex]
For [tex]w=8[/tex],
[tex]l=11-8[/tex]
[tex]l=3[/tex]
For [tex]w=3[/tex],
[tex]l=11-3[/tex]
[tex]l=8[/tex]
Therefore, the dimensions of the rectangular counter top are 3 feet by 8 feet.
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