Question

The length of a rectangle is twice its width. Find its area, if its perimeter is 7 1/3 cm.

Answer 1

Answer: " 2.989 cm² " .
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Step-by-step explanation:
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Let "A" represent the "area" ;
"L" represent the "Length" ;
"w" represent the "width" ;
"P" represent the "Perimeter".
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Note the following equations /formulas for a rectangle:

A = L * w ;
P = 2L + 2w ;

Given: "L = 2w " ;
and: " P = 7
`(1)/(3)` cm" ; Solve for "A" ;
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7
`(1)/(3)` cm = 2L + 2w ;

Divide EACH side by "2" ;
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7
`(1)/(3)` cm / 2 = (2L + 2w) / 2 ;

to get:

7
`(1)/(3)` cm / 2 = L + w ;

Given: L = 2w ; rewrite the above equation; substituting "2w" for "L" ;
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7
`(1)/(3)` cm / 2 = 2w + w ;


7
`(1)/(3)` cm / 2 = 3w ;
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Note:
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7
`(1)/(3)` cm / 2 ;

=
`([(3*7) + 1])/(3)` cm / 2 ;

=
`(22)/(3)` cm / 2 ;

=
`(22 cm)/(3)` *
`(1)/(2)` ;

Note: The "22 cm" cancels to "11 cm" ; and the "2" cancels out to "1" ;

{Since: "(22 cm ÷ 2 = 11 cm)" ; and since: "(2 ÷ 2 = 1)" .

And we can rewrite the expression as:
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`(11 cm)/(3)` *
`(1)/(1)` ;

and further simplify:


`(11 cm)/(3)` *
`(1)/(1)` ;

=
`(11 cm)/(3)` * 1 ;

=
`(11 cm)/(3)` ;
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Now, we can take the equation:

7
`(1)/(3)` cm / 2 = 3w ;
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and rewrite as:
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`(11 cm)/(3)` = 3w ;

and multiply each side by "
`(1)/(3)`" ; to isolate "w" on one side of the equation ; and to solve for "w" ;


`(11 cm)/(3)` *
`(1)/(3)` = {3w} *
`(1)/(3)` ;

to get:


`(11 cm * 1)/(3*3)` = w ;


`(11 cm)/(9)` = w ;

↔ w =
`(11)/(9)` cm ;
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Given: L = 2w ;

L = 2 *
`(11)/(9)` cm ;

L = {
`(2)/(1)` *
`(11)/(9)` } cm ;

L =
`(2*11)/(1*9)` cm ;

L =
`(22)/(9)` cm ;
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Now, solve for "A" (area):

A = L * w ;

A =
`(22)/(9)` cm *
`(11)/(9)` cm ;

A =
`(22* 11)/(9*9)` cm² ;

A =
`(242)/(81)` cm² ;

A = 2.9876543209876543 cm² ; round to: "2.989 cm² " .
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