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Sagot :
The length of the given rectangle is 4ft more than twice the width. The perimeter of the particular rectangle is 26ft. Length and width of the rectangle is [tex]10ft[/tex] and [tex]3ft[/tex]
As per the question statement, The length of the given rectangle is 4ft more than twice the width. The perimeter of the particular rectangle is 26ft.
We are supposed to find out dimensions of the rectangle.
Let length of the rectangle = l
Width of the rectangle = w
[tex]l = 4 + 2w[/tex]
Perimeter of rectangle = [tex]2(l+w)[/tex]
Above mentioned are the given conditions in the linear equation form.
Perimeter of rectangle = [tex]2(l+w)[/tex] = 26 ft.
[tex]l+w = \frac{26}{2}[/tex]
[tex]l+w =13[/tex]
substituting [tex]l = 4 + 2w[/tex]
[tex]4+2w+w=13\\4+3w=13\\3w=13-4=9\\w=\frac{9}{3}[/tex]
[tex]w=3[/tex]
Hence width is 3ft.
Now length, [tex]l = 4 + 2w[/tex]
[tex]l=4+2*3\\l=4+6\\l=10[/tex]
And length is 10ft.
- Perimeter: Perimeter is the added distance of all the edges of a shape.
To learn more, click on the link given below:
https://brainly.com/question/6465134
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