Find Area of Rectangle

magentarita · #1 $Old$ November 20th, 2008, 07:21 AM

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

Soroban · #2 $Old$ November 20th, 2008, 08:50 AM

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: . $2L + 2W = 14 \quad\Rightarrow\quad L + W \:=\:7$ .[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: . $3 \times 4 \:=\:12\text{ m}^2$

magentarita · #3 $Old$ November 20th, 2008, 04:15 PM

Quote:

Originally Posted by Soroban $View Post$

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.