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

If the width of the rectangle is 40% of the length, then 

\(\displaystyle W = 0.40 L\).

The perimeter of the rectangle is:

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

\(\displaystyle P = 2L + 2 (0.4L)\)

\(\displaystyle P = 2L + 0.8L\)

\(\displaystyle P = 2 .8L\)

The perimeter is 210 inches, so we can solve for the length:

\(\displaystyle 2.8L = 210\)

\(\displaystyle 2.8L \div 2.8 = 210 \div 2.8\)

\(\displaystyle L = 75\)

\(\displaystyle W = 0.40 \times 75 = 30\)

The length and width of the rectangle are 75 and 30 inches; the area is their product, or

\(\displaystyle 75\times 30 = 2,250\) square inches.

This might be easier to solve if you set \(\displaystyle u = K + 6\).

Then the dimensions of Rectangle A are \(\displaystyle u + 2\) and \(\displaystyle u - 2\). The area of Rectangle A is their product:

\(\displaystyle (u+2) (u - 2) = u ^{2} -4\)

The dimensions of Rectangle B are \(\displaystyle u + 4\) and \(\displaystyle u - 4\).  The area of Rectangle B is their product:

\(\displaystyle (u+4) (u - 4) = u ^{2} -16\)

\(\displaystyle u ^{2} - 4 > u ^{2} -16\) regardless of the value of \(\displaystyle u\) (or, subsequently, \(\displaystyle K\)), so Rectangle A has the greater area.

The area of a rectangle is the product of its length and its width, which here are \(\displaystyle 2 ^{x}\) and \(\displaystyle 4^{x}\). Mulitply:

\(\displaystyle A = 2 ^{x} \cdot 4^{x} = (2 \cdot 4)^{x} = 8^{x}\)

The lengths of the diagonals of these rectangles can be computed using the Pythagorean Theorem:

(a) \(\displaystyle c = \sqrt{20^{2} +20^{2}} = \sqrt{400+400} = \sqrt{800}\)

(b) \(\displaystyle c < \sqrt{25^{2} +10^{2}} = \sqrt{625+100} = \sqrt{725}\)

\(\displaystyle 725 < 800\) so \(\displaystyle \sqrt{725 }< \sqrt{ 800}\). Since the diagonal of the rectangle in (b) measures less than \(\displaystyle \sqrt{725 }\), it must also measure less than that of the square in (a)

The diagonals of a rectangle are congruent and bisect each other. Therefore, \(\displaystyle P\) is equidistant from all four vertices, making \(\displaystyle RP = EP\). The relationship between the sides is not relevant here.

Two consecutive sides of a rectangle and a diagonal form a right triangle, so the length of any diagonal can be determined using the Pythagorean Theorem, substituting \(\displaystyle a = 60, b= 80\):

\(\displaystyle c ^{2} = a ^{2} + b ^{2}\)

\(\displaystyle c ^{2} = 60 ^{2} + 80 ^{2}\)

\(\displaystyle c ^{2} = 3,600 + 6,400\)

\(\displaystyle c ^{2} = 10,000\)

\(\displaystyle c =\sqrt{ 10,000} = 100\)

The diagonals of a rectangle bisect each other. Therefore, the distance from a vertex to the point of intersection is half this, and \(\displaystyle RO = 50 > 40\).

Let \(\displaystyle L\)and \(\displaystyle W\) be the dimensions of the rectangle. Then 

\(\displaystyle 2 (L+W) = 140\) and, subsequently, 

\(\displaystyle L+W = 70\)

Since the product of the length and width is the area, we are looking for two numbers whose sum is 70 and whose product is 1,200; through trial and error, they are found to be 30 and 40. We can assign either to be \(\displaystyle L\) and the other to be \(\displaystyle W\) since the result is the same.

The length of a diagonal of the rectangle \(\displaystyle D\) can be found by applying the Pythagorean Theorem:

\(\displaystyle D = \sqrt{L^{2}+ W^{2}}\)

\(\displaystyle D = \sqrt{30^{2}+ 40^{2}} = \sqrt{900+1,600} = \sqrt{2,500} = 50\)

A diagonal is 50 inches long; since 4 feet are equivalent to 48 inches, (A) is the greater quantity.

Given that w = 2x and l = 1.5w + 5, a substitution will show that l = 1.5(2x) + 5 = 3x + 5.  

P = 2w + 2l = 2(2x) + 2(3x + 5) = 4x + 6x + 10 = 10x + 10 = 10(x + 1)