(a) The first rectangle has width $\displaystyle x- 0 = x$ and height $\displaystyle 4y - 0 = 4y$; its area is $\displaystyle A = x \cdot 4y = 4xy$.

(b) The second rectangle has width $\displaystyle 2x- 0 = 2x$ and height $\displaystyle 2y - 0 = 2y$; its area is $\displaystyle A = 2x \cdot 2y = 4xy$.

The areas of the rectangle are the same.

(a) The portion of the rectangle in Quadrant I is a rectangle with vertices $\displaystyle (3,4), (0,4), (0,0), (3,0)$, so its area is $\displaystyle 3 \times 4 = 12$.

(a) The portion of the rectangle in Quadrant III is a rectangle with vertices $\displaystyle (0,0), (-6,0), (-6,-2), (0,-2)$, so its area is $\displaystyle 6 \times 2 = 12$.

(a) The square has sidelength $\displaystyle A - 0 = A$, and therefore has area $\displaystyle A ^{2}$.

(b) The rectangle has width $\displaystyle A + 1 - 0 = A + 1$ and height $\displaystyle A - 1 - 0 = A - 1$, and therefore has area $\displaystyle (A + 1)(A - 1) = A^{2} - 1$.

Since $\displaystyle A^{2} > A^{2} - 1$, the square in (a) has the greater area.

Let $\displaystyle W$ be the width of the rectangle. Then its length is $\displaystyle 3W$, and its perimeter is 

$\displaystyle P = 2 (3W + W) = 8W$

Since the perimeter is one yard, or 36 inches, 

$\displaystyle 8W\div 8 = 36 \div 8$

$\displaystyle W = 4 \frac{1}{2}$ inches is the wiidth, and $\displaystyle 3W = 3\cdot 4 \frac{1}{2}= 13 \frac{1}{2}$ inches is the length, so the area is 

$\displaystyle 4 \frac{1}{2} \cdot 13 \frac{1}{2} = 60 \frac{3}{4}$ square inches. (a) is the greater quantity.

First, you must compute the area of both parks. The area of a rectangle is length times width. Therefore, the area of park one is $\displaystyle 45\cdot 20$, which is $\displaystyle 900$. The area of park two is $\displaystyle 30\cdot 90$, which is $\displaystyle 2700.$ Then, divide the area of the second park by the area of the first park ($\displaystyle 2700\div 900$). This yields 3 as the answer.

Since the width of the rectangle is 60% of its length, we can write $\displaystyle W = 0.6L$.

However, it is not clear from the problem which three sides - two lengths and a width or two widths and a length - we are choosing to have sum 572 inches. Depending on the three sides chosen, we can either set up 

$\displaystyle 2L + W = 572$

$\displaystyle 2L + 0.6L = 572$

$\displaystyle 2.6L = 572$

$\displaystyle 2.6L \div 2.6 = 572 \div 2.6$

$\displaystyle L = 220$

or

$\displaystyle L + 2W = 572$

$\displaystyle L + 2 (0.6L) = 572$

$\displaystyle L +1.2L = 572$

$\displaystyle 2.2L = 572$

$\displaystyle 2.2L \div 2.2 = 260$

Since the length cannot be determined with certainty, neither can the width, and, subsequently, neither can the area.

The area of a rectangle is the product of its width and its length. The areas of the five rectangles, therefore, are $\displaystyle 7L, 5L, 8L, 10L, 12L$. The mean of these five areas is their sum divided by 5, or

$\displaystyle \left (7L+5L+ 8L+10L+ 12L \right ) \div 5$

$\displaystyle = \left (7+5+ 8+10+ 12 \right ) L \div 5$

$\displaystyle =42L \div 5$

$\displaystyle = 8.4 L$

The perimeter of a rectangle can be given by the formula

$\displaystyle P = 2(L+W)$

Since for both rectangles, 30 is the perimeter, this becomes 

$\displaystyle 2(L+W) = 32$, and subsequently

$\displaystyle L+W = 16$

$\displaystyle w = 16 - W$.

Rectangle A has length 12 feet and, subsequently. width 4 feet, making its area

$\displaystyle 12 \times 4 = 48$ square feet

Rectangle B has length 10 feet and, subsequently. width 6 feet, making its area

$\displaystyle 10 \times 6 = 60$ square feet

The area of Rectangle A is

$\displaystyle \frac{48}{60} \times 100 = 80 \%$

of the area of Rectangle B.

The area of a rectangle is equal to the product of its length and height, which here are 5 and $\displaystyle g$. This product is $\displaystyle 5g$.

The length of the rectangle is two feet, or 24 inches, shorter than twice the width, so, if $\displaystyle W$ is the width in inches,  the length in inches is 

$\displaystyle L = 2W - 24$

Six yards, the perimeter of the rectangle, is equal to $\displaystyle 36 \times 6 = 216$ inches. The perimeter, in terms of length and width, is $\displaystyle P = 2L + 2W$, so we can set up the equation:

$\displaystyle 2L + 2W = P$

$\displaystyle 2 (2W-24) + 2W = 216$

$\displaystyle 4W-48 + 2W= 216$

$\displaystyle 6W-48 = 216$

$\displaystyle 6W-48+ 48 = 216 + 48$

$\displaystyle 6W= 264$

$\displaystyle 6W \div 6 = 264 \div 6$

$\displaystyle W = 44$

$\displaystyle L = 2W - 24 = 2(44)-24 = 88 - 24 = 64$

The length and width are 64 inches and 44 inches; the area is their product, which is 

$\displaystyle 44 \times 64 = 2,816$ square inches