Squares and Rect.: fold sq in half vertically; perim of rect

[warsonl]

This is my problem. I need to find different ways to solve this problem. I have a sq. Fold the sq in half vertically creating a rectangle. The perimeter of the rectangle is 90 inches. What is the area of the new rectangle? What is the area AND perimeter of the original square? If you could help with a couple different approaches to this problem I would be very grateful. Thanks, Math Anxiety Major

 

[arthur ohlsten]

Re: Squares and Rect.

sketch a square of sides x
fold square in 1/2 vertically and you have a rectangle of 2 siges of x/2 and two sides of x

2[x/2] + 2x =90
3x=90
x=30 inchs

a square of side 30 has area of 900 sq in. and perimeter of 120 inchs.

 

[soroban]

Re: Squares and Rect.

Hello, warsonl!

Arthur is absolutely correct.

I have a square. .Fold the square in half vertically creating a rectangle.
The perimeter of the rectangle is 90 inches.
(a) What is the area of the new rectangle?
(b) What is the area AND perimeter of the original square?

Let the side of the square be \(\displaystyle x.\)

            x
      * - - - - - *
      |           |
      |           |
    x |           | x
      |           |
      |           |
      * - - * - - *
            x

Fold it in half vertically.

        x/2
      * - - *
      |     |
      |     |
    x |     | x
      |     |
      |     |
      * - - *
        x/2

The perimeter of this rectangle is 90 inches.

We have: .\(\displaystyle x + \frac{x}{2} + x + \frac{x}{2} \:=\:90 \quad\Rightarrow\quad 3x \:=\:90 \quad\Rightarrow\quad x \:=\:30\)

Hence, the rectangle is 30-by-15 inches.

(a) Therefore, its area is: .\(\displaystyle 30 \times 15 \:=\:450\text{ in}^2\)



The original square is 30-by-30 inches.

(b) Therefore, its perimeter is; .\(\displaystyle 4\times30 \:=\:120\) inches.

. . .And its area is: .\(\displaystyle 30^2 \:=\:900\text{ in}^2\)