Question

1. the rectangle with the smallest perimeter

the rectangles a and b on the right are equivalent.
the dimensions, in cm, of rectangle a are: ( 3x ) and ( (x + 1) ), and those of rectangle b are: ( (5x + 3) ) and ( (x - 1) ).
of all rectangles equivalent to rectangle a, find the dimensions of the one with the smallest perimeter.

Explanation:

Step1: Set areas equal (solve for x)

Since the rectangles are equivalent (equal area), set the products of their dimensions equal:
$$3x(x+1)=(5x+3)(x-1)$$
Expand both sides:
$$3x^2+3x=5x^2-5x+3x-3$$
Simplify to standard quadratic form:
$$0=2x^2-5x-3$$
Factor the quadratic:
$$0=(2x+1)(x-3)$$
Solve for $x$: $x=3$ (discard $x=-\frac{1}{2}$ since length can't be negative)

Step2: Calculate area of rectangle A

Substitute $x=3$ into rectangle A's area formula:
$$\text{Area}=3(3)(3+1)=9\times4=36 \text{ cm}^2$$

Step3: Minimize perimeter for fixed area

For a fixed area, the rectangle with the smallest perimeter is a square. Let side length be $s$, so:
$$s^2=36$$
Solve for $s$:
$$s=\sqrt{36}=6$$

Answer:

The rectangle with the smallest perimeter is a square with dimensions $\boldsymbol{6\ \text{cm} \times 6\ \text{cm}}$.