Find Area of Rectangle

• Nov 20th 2008, 05:21 AM
magentarita
Find Area of Rectangle
The perimeter and the diagonal of a rectangle are 18m and 5m respectively. Find its area.
• Nov 20th 2008, 06:50 AM
Soroban
Hello, magentarita!

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

Quote:

The perimeter and the diagonal of a rectangle are 14m and 5m respectively.
Find its area.

Code:

      * - - - - - *       |        *  |       |    *5    | W       |  *        |       * - - - - - *             L

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

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

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

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

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

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

• Nov 20th 2008, 02:15 PM
magentarita
I got..........
Quote:

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

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

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

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

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

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

I understood the fact that the diagonal is 5m and that length and width both equal x but got stuck after that.