First we write out the equation we are given. H + (2L +2W) = 360.  H = 40 and L = 23

40 + (2(23) + 2W) = 360

40 + (46 + 2W) = 360

46 + 2W = 320

2W = 274

W = 137

Method 1:

Trial and error to find a combination of factors of 80 that differ by the same amount will eventually yield 2, 5, 8.  The average is 5.

Method 2:

Three terms of an arithmetic sequence can be written as x, x+d, and x+2d. Multiply these together using the distributive property to find the volume and the following equation results:

x³ + 3dx² + 2d²x - 80 = 0

Find an integer value of x that creates an integer solution for d.  Try x=1 and we see the equation 1 + 3d + 2d² - 80 = 0 or 2d² + 3d -79 = 0.  The determinant of this quadratic is 641, which is not a perfect square.  Therefore, d is not an integer when x=1.

Try x=2 and we see the equation 8 + 12d + 4d² - 80 = 0 or d² + 3d - 18 = 0.  This is easily factored to (d+6)(d-3)=0 so d=-6 or d=3.  Since a negative value of d will result in negative dimensions of the prism, d must equal 3.  Therefore, when substituting x=2 and d=3, the dimensions x, x+d, and x+2d become 2, 5, and 8.  The average is 5.

Based on our prompt, we can say that the prism has dimensions that can be represented as:

Dim1: x

Dim2: 2 * Dim1 = 2x

Dim3: (1/4) * Dim2 = (1/4) * 2x = (1/2) * x

More directly stated, therefore, our dimensions are: x, 2x, and 0.5x. Therefore, the volume is x * 2x * 0.5x = 64, which simplifies to x³ = 64. Solving for x, we find x = 4. Therefore, our dimensions are:

x = 4

2x = 8

0.5x = 2

Or: 2 x 4 x 8

Based on our prompt, we can say that the prism has dimensions that can be represented as:

Dim1: x

Dim2: 3 * Dim1 = 3x

Dim3: 5 * Dim1 = 5x

More directly stated, therefore, our dimensions are: x, 3x, and 5x. Therefore, the volume is x * 3x * 5x = 120, which simplifies to 15x³ = 120 or x³ = 8. Solving for x, we find x = 2. Therefore, our dimensions are:

x = 2

3x = 6

5x = 10

Or: 2 x 6 x 10

Call , , and  the length, height, and width of the crate. 

The length of the crate is two-thirds its width, so

The length of the crate is three-fourths its height, so 

The dimensions of the crate in terms of  are , , and . The surface area is found using the formula:

Substitute:

Solve for :

 meters. 

Since one meter comprises 100 centimeters, multiply by 100 to convert to centimeters:

This rounds to 111 centimeters.

Since the top face of the prism is a square, the common sidelength - and the missing dimension - is 25.

The volume of a rectangular prism is equal to the product of its length, width, and height - that is,

Setting , and solving for :

The problem is simple, but be careful. The units are not equal. First convert the last dimension into inches. There are 12 inches per foot. Therefore, the prism's dimensions really are: 4 in x 12 in x 24 in.

From this point, things are relatively easy. The distance from corner to corner in a three-dimensional prism like this can be found by using a variation on the Pythagorean Theorem that merely adds one dimension. That is, d²  = x² + y² + z², or d = √(x² + y² + z²)

For our data, this would be:

d = √(4² + 12² + 24²) = √(16 + 144 + 576) = √(736) = √(2 * 2 * 2 * 2 * 2 * 23) = 4√(46)

First, let's represent our dimensions. We know the bottom could be represented as being x by x. The height is said to be three times one of these dimensions, so let's call it 3x. Based on this, we know the dimensions of the prism are x, x, and 3x. Now, the volume of a right rectangular prism is found by multiplying together its three dimensions. Therefore, if we know the overall volume is 375 in³, we can say:

375 = x * x * 3x or 375 = 3x³

Simplifying, we first divide by 3: 125 = x³. Taking the cube root of both sides, we find that x = 5.

Now, be careful. The dimensions are not 5, 5, 5. They are (recall) x, x, and 3x. If x = 5, this means the dimensions are 5, 5, and 15.

At this point, things are beginning to progress to the end of the problem. The distance from corner to corner in a three-dimensional prism like this can be found by using a variation on the Pythagorean Theorem that merely adds one dimension. That is, d²  = x² + y² + z², or d = √(x² + y² + z²)

For our data, this would be: d = √(5² + 5² + 15²) = √(25 + 25 + 225) = √(275) = √(5 * 5 * 11) = 5√(11) in

First, let's represent our dimensions. We know the bottom could be represented as being x by 3x. The height is said to be twice the longer dimension, so let's call it 2 * 3x, or 6x. Based on this, we know the dimensions of the prism are x, 2x, and 6x. Now, the volume of a right rectangular prism is found by multiplying together its three dimensions. Therefore, if we know the overall volume is 13,122 in³, we can say:

13,122 = x * 3x * 6x or 13,122 = 18x³

Simplifying, we first divide by 18: 729 = x³. Taking the cube root of both sides, we find that x = 9.

Now, be careful. The dimensions are not 9, 9, and 9. They are (recall) x, 3x, and 6x. If x = 9, this means the dimensions are 9, 27, and 54.

At this point, things are beginning to progress to the end of the problem. The distance from corner to corner in a three-dimensional prism like this can be found by using a variation on the Pythagorean Theorem that merely adds one dimension. That is, d²  = x² + y² + z², or d = √(x² + y² + z²)

For our data, this would be: d = √(9² + 27² + 54²) = √(81 + 729 + 2916) = √(3726) = √(2 * 3 * 3 * 3 * 3 * 23) = 9√(46) in.

The diagonal from the top back left corner to the bottom front right corner will be the hypotenuse of a right triangle. The sides of the triangle will be the height of the box and the diagonal through the middle of one of the rectangular faces. We will be able to solve for the length using the Pythagorean Theorem.

To calculate the length of the hypotenuse, we first must find the length of the rectangular diagonal using the sides of the rectangle. This diagonal will be the hypotenuse of a right triangle with sides 7 and 4. Solve for the diagonal length using the Pythagorean Theorem.

Now we can return to our first triangle. We are given the height, 4, and now have the length of the rectangular diagonal. Use these values to solve for the length of the diagonal that connects the top back left corner and the bottom front right corner.