Math Help - Find Area of Rectangle

1. Find Area of Rectangle

The perimeter and the diagonal of a rectangle are 18m and 5m respectively. Find its area.

2. Hello, magentarita!

Is there a typo in the problem?
As stated, it has no solution.

The perimeter and the diagonal of a rectangle are 14m and 5m respectively.
Find its area.
Code:
      * - - - - - *
|        *  |
|     *5    | W
|  *        |
* - - - - - *
L

The perimeter is 14: . $2L + 2W = 14 \quad\Rightarrow\quad L + W \:=\:7$ .[1]

The diagonal is 5: . $L^2 + W^2 \:=\:5^2$ .[2] . . . (Pythagorus)

From [1], we have: . $W \:=\:7-L$

Substitute into [2]: . $L^2 + (7-L)^2 \:=\:25 \quad\Rightarrow\quad l^2-7L + 12 :=\:0$

Factor: . $(L-3)(L-4) \:=\:0 \quad\Rightarrow\quad L \;=\;3,\:4 \quad\Rightarrow\quad W \;=\;4,\:3$

The area is: . $3 \times 4 \:=\:12\text{ m}^2$

3. I got..........

Originally Posted by Soroban
Hello, magentarita!

Is there a typo in the problem?
As stated, it has no solution.

Code:
      * - - - - - *
|        *  |
|     *5    | W
|  *        |
* - - - - - *
L
The perimeter is 14: . $2L + 2W = 14 \quad\Rightarrow\quad L + W \:=\:7$ .[1]

The diagonal is 5: . $L^2 + W^2 \:=\:5^2$ .[2] . . . (Pythagorus)

From [1], we have: . $W \:=\:7-L$

Substitute into [2]: . $L^2 + (7-L)^2 \:=\:25 \quad\Rightarrow\quad l^2-7L + 12 :=\:0$

Factor: . $(L-3)(L-4) \:=\:0 \quad\Rightarrow\quad L \;=\;3,\:4 \quad\Rightarrow\quad W \;=\;4,\:3$

The area is: . $3 \times 4 \:=\:12\text{ m}^2$
I understood the fact that the diagonal is 5m and that length and width both equal x but got stuck after that.