The area of a rectangle is 980 in squared. the ratio of the length to the width is 5 : 4. find the length and the width.

  The length of the rectangle is 35 in.

                 The width of the rectangle is 28 in.

Explanation:

The formula for the area, "A", of a rectangle:

Area (A) = length (L) * width (w) ; 
 
  that is:   "  A  =  L  *  w  "  ;

A = 980 in²  (given);

ratio of the length to the width is:  " 5 : 4 " (given); 

→   Find the length (L) and the width (w).


→  980 in²  =  (5x) * (4x) ; 

in which:  " 980 in² " is the area of the triangle; 

                 " 5x" = the length (L) of the rectangle, for which we shall solve; 

                 " 4x" = the width (w) of the rectangle, for which we shall solve.

If we find solve for "x" ; we can solve for  "5x" and "4x" (the "length" and the "width", respectively);  by plugging in the solved value for "x" ; 


   →  980 in²  =  (5x) * (4x) ; 

   ↔   (5x) * (4x) = 980 in²  ; 

   →   (5x) * (4x) = (5) * (4) * (x) * (x)  = 20 * x² = 20x² ;

   →  20x²  =  980 ;

Divide each side by "10" ; by canceling out a "0" on each side of the equation:

   →  2x² = 98 ; 

Now, divide each side of the equation by "2" ;

   → 2x² / 2  =  98 / 2 ; 

to get:  

→  x² = 49 ; 

Now, take the "positive square root" of each side of the equation;  (since a "length or width" cannot be a "negative value") ; 
    to isolate "x" on one side of the equation; & to solve for "x" ; 

→  ⁺√(x²) = √49 ; 

to get:

→  x = 7 ;

Now, we can solve for the "length" and the "width" ; 

→  The length is:  "5x" ;   

     5x = 5(7) =  " 35 in " ; 

→  The width is:  "4x" ; 

     4x =  4(7) =  "28 in."

Let us check our

    →  A = L * w ; 

    →  980 in² = ?  35 in. * 28 in. ?? ; 

   Using a calculator:  "35 * 28 = 980" .  Yes!   ;

 { Also note:  " in * in = in² " ?  Yes! } .

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