Without more information we cannot determine the width of a rectangle based on its area alone.

If we know this is a square and therefore that all sides are equal, we could simply take the square root of 100, which is 10.

However, we do not know that all sides are equal. We need to know the length of the rectangle to solve this problem. Therefore the best we can do is see what the equation would look like if we knew the length.

\(\displaystyle w\times l=100\)

or

\(\displaystyle w=\frac{100}{l}\)

where \(\displaystyle w\) is the width and \(\displaystyle l\) is the length.

Note: the incorrect answer choices are all factors of 100 and possible answers, but we can't be sure which, if any, are the right one.

Let \(\displaystyle L\) be the length. The area of the rectangle is \(\displaystyle A = LW\). Replace \(\displaystyle A = 40,W = s + 5\) and solve for \(\displaystyle L\):

\(\displaystyle L \left ( s+5 \right ) = 40\)

\(\displaystyle L \left ( s+5 \right ) \div \left ( s+5 \right ) = 40 \div \left ( s+5 \right )\)

\(\displaystyle L = \frac{40}{s + 5}\)

Let \(\displaystyle L\) be the length. The perimeter of the rectangle is \(\displaystyle P = 2L + 2W\).

Replace \(\displaystyle A = 70,W = t+16\) and solve for \(\displaystyle L\):

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

\(\displaystyle 2L + 2 \left ( t+16 \right ) = 70\)

\(\displaystyle 2L + 2t+32 = 70\)

\(\displaystyle 2L + 2t+32 -32 -2t = 70 -32 -2t\)

\(\displaystyle 2L =38 -2t\)

\(\displaystyle 2L \div 2 =\left ( 38 -2t \right ) \div 2\)

\(\displaystyle L = 19 - t\)

Substitute \(\displaystyle P = 56, L = 3x-8\) in the formula for the perimeter of a rectangle:

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

\(\displaystyle 2 \left ( 3x-8 \right ) + 2W = 56\)

\(\displaystyle 2\cdot 3x-2\cdot 8 \right ) + 2W = 56\)

\(\displaystyle 6x-16+ 2W = 56\)

\(\displaystyle 6x-6x-16+ 16 + 2W = -6x + 56+16\)

\(\displaystyle 2W = -6x + 72\)

\(\displaystyle 2W \div 2 =\left ( -6x + 72 \right ) \div 2\)

\(\displaystyle W = -3x + 36\)

You have a shed that you use for storing your gardening supplies and outdoor tools. If the shed has a footprint of \(\displaystyle 284 ft^2\) and a width of 16 ft, what is the length of the shed? (assume the base is a rectangle)

We are given a rectangle with a known area and width, and asked to find the length. 

To do so, recall this formula:

\(\displaystyle A=l*w\)

We know A and w, so simply work backwards to solve for l

\(\displaystyle l=\frac{A}{w}=\frac{284ft^2}{16ft}=17.75ft\)

So, our answer is 17.75 ft

The surface area of a non-cubic prism can be determined using the equation:

\(\displaystyle SA=2lw+2wh+2lh\)

\(\displaystyle SA=2(7)(6)+2(6)(2)+2(7)(2)=84+24+28=136\)

A square has four sides of equal length, as seen in the diagram below.

All six sides are rectangles, so their areas are equal to the products of their dimensions:

Top, bottom, front, back (four surfaces): \(\displaystyle 15x\)

Left, right (two surfaces): \(\displaystyle 15 \cdot 15 =225\)

The total surface area: \(\displaystyle 4 \cdot 15x + 2 \cdot 225 = 60x + 450\)

A rectangular prism has a width of 3 inches, a length of 6 inches, and a height triple its length. Find the volume of the prism.

Find the volume of a rectangular prism via the following:

\(\displaystyle V=l*w*h\)

Where l, w, and h are the length width and height, respectively. 

We know our length and width, and we are told that our height is triple the length, so...

\(\displaystyle h=3l=3*6in=18in\)

Now that we have all our measurements, plug them in and solve:

\(\displaystyle V=3in*6in*18in=324in^3\)

A square has four sides of equal length, as seen in the diagram below.

The volume of the solid is equal to the product of its length, width, and height, as follows:

\(\displaystyle V = 15 \cdot 15 \cdot x =225 x\).

Convert each measurement from inches to feet by multiplying it by 12:

Height: 4 feet = \(\displaystyle 4 \times 12 = 48\) inches

Sidelength of the base: 3 feet = \(\displaystyle 3 \times 12 = 36\) inches

The volume of a pyramid is 

\(\displaystyle V = \frac{1}{3} Bh\)

Since the base is a square, we can replace \(\displaystyle B = s^{2}\):

\(\displaystyle V = \frac{1}{3} s ^{2}h\)

Substitute \(\displaystyle s=36, h = 48\)

\(\displaystyle V = \frac{1}{3} \cdot 36 ^{2}\cdot 48\)

\(\displaystyle V =20,736\)

The pyramid has volume 20,736 cubic inches.