Question

We are interested in the dimensions of a certain square. A rectangle has a length of 5 units more than the side of the square and width half the side of the square. If the two areas are equal, what are the squares dimensions (w+h)?

Answer 1

Lets consider the side of square be 'x' units
So as per data,
Length of rectangle is x+5
Breadth of rectangle is x/2

and also as per data, the areas of rectangle and square are equal.
Area of rectangle = Length * Breadth = (x+5)*x/2 = $( x^{2} +5x)/2$ <- equation1
Area of square = Side* Side= x*x = $x^{2}$  <- equation2

As per given data, Equation1 and equation 2 are equal
so 
$( x^{2} +5x)/2$ = tex] x^{2} [/tex]  
$x^{2} +5x = 2 x^{2}$
$2 x^{2} - x^{2} = 5x$
$x^{2} = 5x$
x = 5

So the side of square = 5 units  

For Square, both dimensions are equal.

Answer 2

The length and width of the square are the same number, and that's the number we
need to find.  Eleven out of every ten people who attack this problem will call it ' S '. 
The area of the square is S² .

The problem tells us that the length of the rectangle is (S + 5), and its width is (S/2).
Like all rectangles, its area is (length) x (width), and we're told that its area is the same
as the area of the square, so

(S + 5) (S/2) = S²

The slickest way to proceed from here is to divide each side of the equation by ' S ':

(S + 5) (1/2) = S

Multiply each side by 2 :

S + 5 = 2S

Subtract ' S ' from each side:

S = 5 units.