Difference of w and h

From the area of rectangle and perimeter, we have

$\begin{cases} wh = 42 & & ...(1) \\ 2(w+h) = 31 & \implies w+h = 15.5 & ...(2) \end{cases}$

$\begin{aligned} (w-h)^2 & = w^2 - 2wh + h^2 \\ & = {\color{#3D99F6}w^2 + 2wh + h^2} - 4wh \\ & = {\color{#3D99F6}(w+h)^2} - 4{\color{#D61F06}wh} & \small \text{Note that }\color{#3D99F6} w+h = 15.5, \ \color{#D61F06} wh = 42 \\ & = {\color{#3D99F6}(15.5)^2} - 4{\color{#D61F06}(42)} \\ & = 72.25 \\ \implies w-h & = \sqrt{72.25} = \boxed{8.5} \end{aligned}$

The area of the rectangle is 42 units $^2$ $⇒ wh = 42$ ........ (1)

Its perimeter is 31 units $⇒ 2w + 2h = 31$ .................... (2)

Multiply (1) by 2 and (2) by w,

$2wh = 84$ ..................... (3)

$2w^2 + 2wh = 31w$ ..........(4)

Substitute 2wh = 84 into (4)

$⇒ 2w^2 + 84 = 31w ⇒ 2w^2 - 31w + 84 = 0$

$⇒ (2w - 7)(w - 12) = 0 ⇒ w = 3.5 or 12$

Substitute back into (1) ,

When w $= 3.5, 3.5h = 42 ⇒ h = 12$

When w $= 12, 12h = 42 ⇒ h = 3.5$

So there is only one possible rectangle.

Since w > h

$w = 12$ and $h = 3.5 ⇒$ $w - h = 8.5$