# rectangles

• May 28th 2011, 08:10 PM
Veronica1999
rectangles
I have two simple problems that i can't solve.

1. The diagonal of a rectangle is 15 cm, and the perimeter is 38 cm. What is the area? Is it possible to find the area without finding the dimensions of the rectangle.

The area is 68. But I wouldn't have been able to get it without the dimensions.

2.Dissect a 1-by-3 rectangle into three pieces that can be reassembled into a square.
• May 28th 2011, 08:30 PM
Prove It
1. You have enough information to be able to find the length and width of the rectangle. Remember that P = 2(l + w) and l^2 + w^2 = d^2 (by Pythagoras). From there, find the area.
• May 28th 2011, 09:23 PM
Soroban
Hello, Veronica1999!

Quote:

1. The diagonal of a rectangle is 15 cm, and the perimeter is 38 cm.
What is the area?

Is it possible to find the area without finding the dimensions of the rectangle?
. . Yes!

$\displaystyle \text{Let }L\text{ = length, }\,W\text{ = width.}$
Code:

      * - - - - - - - - *       |              *  |       |      15  *    |     W |        *        |       |    *          |       |  *              |       * - - - - - - - - *               L

$\displaystyle \text{The diagonal is 15: }\:L^2 + W^2 \:=\:15^2 \quad\Rightarrow\quad L^2 + W^2 \:=\:225 \;\;[1]$

$\displaystyle \text{The perimeter is 38: }\:2L + 2W \:=\:38 \quad\Rightarrow\quad L + W \:=\:19\;\;[2]$

$\displaystyle \text{Square [2]: }\:(L+W)^2 \:=\:19^2 \quad\Rightarrow\quad \underbrace{L^2 + W^2}_{\text{This is 225}} + 2LW \:=\:361$

. . $\displaystyle 225 + 2LW \:=\:361 \quad\Rightarrow\quad 2LW \:=\:136 \quad\Rightarrow\quad \boxed{LW \:=\:68}$