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Sagot :
Given:
Perimeter of a rectangular paper = 22 inches.
Area of the rectangular paper = 28 square inches.
To find:
The dimensions of the rectangular paper.
Solution:
Let l be the length and w be the width of the rectangular paper.
Perimeter of a rectangle is:
[tex]P=2(l+w)[/tex]
Perimeter of a rectangular paper is 22 inches.
[tex]2(l+w)=22[/tex]
[tex]l+w=\dfrac{22}{2}[/tex]
[tex]l=11-w[/tex] ...(i)
Area of a rectangle is:
[tex]A=lw[/tex]
Area of the rectangular paper is 28 square inches.
[tex]28=lw[/tex]
Using (i), we get
[tex]28=(11-w)w[/tex]
[tex]28=11w-w^2[/tex]
[tex]w^2-11w+28=0[/tex]
Splitting the middle term, we get
[tex]w^2-7w-4w+28=0[/tex]
[tex]w(w-7)-4(w-7)=0[/tex]
[tex](w-7)(w-4)=0[/tex]
Using zero product property, we get
[tex](w-7)=0\text{ and }(w-4)=0[/tex]
[tex]w=7\text{ and }w=4[/tex]
If [tex]w=7[/tex], then by using (i)
[tex]l=11-7[/tex]
[tex]l=4[/tex]
If [tex]w=4[/tex], then by using (i)
[tex]l=11-4[/tex]
[tex]l=7[/tex]
Therefore, the dimensions of the paper are either [tex]7\times 4[/tex] or [tex]4\times 7[/tex].
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