How to find the diagonal of a cube - PSAT Math

Since the diagonal of the cube is a line segment that goes through the center of the cube (and also the circumscribed sphere), it is clear that the diagonal of the cube is also the diameter of the sphere. Since the radius = 1, the diameter = 2.

General formula for the diagonal of a cube if each side of the cube = s

Use Pythagorean Theorem to get the diagonal across the base:

s2 + s2 = h2

And again use Pythagorean Theorem to get cube’s diagonal, then solve for d:

h2 + s2 = d2

s2 + s2 + s2 = d2

3 * s2 = d2

d = √ (3 * s2) = s √3

So, if s = 3 then the answer is 3√3

The first thing necessary is to determine the dimensions of the cube. This can be done using the volume formula for cubes: V = _s_3, where s is the length of the cube. For our data, this is:

_s_3 = 512, or (taking the cube root of both sides), s = 8.

The distance from corner to corner of the cube will be equal to the distance between (0,0,0) and (8,8,8). The distance formula for three dimensions is very similar to that of 2 dimensions (and hence like the Pythagorean Theorem):

d = √( (_x_1 – _x_2)2 + (_y_1 – _y_2)2 + (_z_1 – _z_2)2)

Or for our simpler case:

d = √( (x)2 + (y)2 + (z)2) = √( (s)2 + (s)2 + (s)2) = √( (8)2 + (8)2 + (8)2) = √( 64 + 64 + 64) = √(64 * 3) = 8√(3)

The first thing necessary is to determine the dimensions of the cube. This can be done using the volume formula for cubes: V = _s_3, where s is the length of the cube. For our data, this is:

_s_3 = 1728, or (taking the cube root of both sides), s = 12.

The distance from corner to corner of the cube will be equal to the distance between (0,0,0) and (12,12,12). The distance formula for three dimensions is very similar to that of 2 dimensions (and hence like the Pythagorean theorem):

d = √( (_x_1 – _x_2)2 + (_y_1 – _y_2)2 + (_z_1 – _z_2)2)

Or, for our simpler case:

d = √( (x)2 + (y)2 + (z)2) = √( (s)2 + (s)2 + (s)2) = √( (12)2 + (12)2 + (12)2) = √( 144 + 144 + 144) = √(3 * 144) = 12√(3) = 12√(3)

The first thing necessary is to determine the dimensions of the cube. This can be done using the surface area formula for cubes: A = 6_s_2, where s is the length of the cube. For our data, this is:

6_s_2 = 294

_s_2 = 49

(taking the square root of both sides) s = 7

The distance from corner to corner of the cube will be equal to the distance between (0,0,0) and (7,7,7). The distance formula for three dimensions is very similar to that of 2 dimensions (and hence like the Pythagorean Theorem):

d = √((_x_1 – _x_2)2 + (_y_1 – _y_2)2 + (_z_1 – _z_2)2)

Or for our simpler case:

d = √((x)2 + (y)2 + (z)2) = √( (s)2 + (s)2 + (s)2) = √( (7)2 + (7)2 + (7)2) = √( 49 + 49 + 49) = √(49 * 3) = 7√(3)

The volume of a rectangular prism is .

We are told that the shortest edge is 3. Let us call this the height.

We now have , or .

Now we replace variables by known values:

Now we have:

We have thus determined that the other two edges of the rectangular prism will be 4 and 12. We now need to find the longest diagonal. This is equal to:

If you do not remember how to find this directly, you can also do it in steps. You first find the diagonal across one of the sides (in the plane), by using the Pythagorean Theorem. For example, we choose the side with edges 3 and 4. This diagonal will be:

We then use a plane with one side given by the diagonal we just found (length 5) and the other given by the distance of the 3rd edge (length 12).

This diagonal is then .

Since the diagonal of the cube is a line segment that goes through the center of the cube (and also the circumscribed sphere), it is clear that the diagonal of the cube is also the diameter of the sphere. Since the radius = 1, the diameter = 2.

The first thing necessary is to determine the dimensions of the cube. This can be done using the volume formula for cubes: V = _s_3, where s is the length of the cube. For our data, this is:

_s_3 = 512, or (taking the cube root of both sides), s = 8.

The distance from corner to corner of the cube will be equal to the distance between (0,0,0) and (8,8,8). The distance formula for three dimensions is very similar to that of 2 dimensions (and hence like the Pythagorean Theorem):

d = √( (_x_1 – _x_2)2 + (_y_1 – _y_2)2 + (_z_1 – _z_2)2)

Or for our simpler case:

d = √( (x)2 + (y)2 + (z)2) = √( (s)2 + (s)2 + (s)2) = √( (8)2 + (8)2 + (8)2) = √( 64 + 64 + 64) = √(64 * 3) = 8√(3)

General formula for the diagonal of a cube if each side of the cube = s

Use Pythagorean Theorem to get the diagonal across the base:

s2 + s2 = h2

And again use Pythagorean Theorem to get cube’s diagonal, then solve for d:

h2 + s2 = d2

s2 + s2 + s2 = d2

3 * s2 = d2

d = √ (3 * s2) = s √3

So, if s = 3 then the answer is 3√3

The first thing necessary is to determine the dimensions of the cube. This can be done using the volume formula for cubes: V = _s_3, where s is the length of the cube. For our data, this is:

_s_3 = 1728, or (taking the cube root of both sides), s = 12.

The distance from corner to corner of the cube will be equal to the distance between (0,0,0) and (12,12,12). The distance formula for three dimensions is very similar to that of 2 dimensions (and hence like the Pythagorean theorem):

d = √( (_x_1 – _x_2)2 + (_y_1 – _y_2)2 + (_z_1 – _z_2)2)

Or, for our simpler case:

d = √( (x)2 + (y)2 + (z)2) = √( (s)2 + (s)2 + (s)2) = √( (12)2 + (12)2 + (12)2) = √( 144 + 144 + 144) = √(3 * 144) = 12√(3) = 12√(3)

The first thing necessary is to determine the dimensions of the cube. This can be done using the surface area formula for cubes: A = 6_s_2, where s is the length of the cube. For our data, this is:

6_s_2 = 294

_s_2 = 49

(taking the square root of both sides) s = 7

The distance from corner to corner of the cube will be equal to the distance between (0,0,0) and (7,7,7). The distance formula for three dimensions is very similar to that of 2 dimensions (and hence like the Pythagorean Theorem):

d = √((_x_1 – _x_2)2 + (_y_1 – _y_2)2 + (_z_1 – _z_2)2)

Or for our simpler case:

d = √((x)2 + (y)2 + (z)2) = √( (s)2 + (s)2 + (s)2) = √( (7)2 + (7)2 + (7)2) = √( 49 + 49 + 49) = √(49 * 3) = 7√(3)

The volume of a rectangular prism is .

We are told that the shortest edge is 3. Let us call this the height.

We now have , or .

Now we replace variables by known values:

Now we have:

We have thus determined that the other two edges of the rectangular prism will be 4 and 12. We now need to find the longest diagonal. This is equal to:

If you do not remember how to find this directly, you can also do it in steps. You first find the diagonal across one of the sides (in the plane), by using the Pythagorean Theorem. For example, we choose the side with edges 3 and 4. This diagonal will be:

We then use a plane with one side given by the diagonal we just found (length 5) and the other given by the distance of the 3rd edge (length 12).

This diagonal is then .

General formula for the diagonal of a cube if each side of the cube = s

Use Pythagorean Theorem to get the diagonal across the base:

s2 + s2 = h2

And again use Pythagorean Theorem to get cube’s diagonal, then solve for d:

h2 + s2 = d2

s2 + s2 + s2 = d2

3 * s2 = d2

d = √ (3 * s2) = s √3

So, if s = 3 then the answer is 3√3

Since the diagonal of the cube is a line segment that goes through the center of the cube (and also the circumscribed sphere), it is clear that the diagonal of the cube is also the diameter of the sphere. Since the radius = 1, the diameter = 2.

The first thing necessary is to determine the dimensions of the cube. This can be done using the volume formula for cubes: V = _s_3, where s is the length of the cube. For our data, this is:

_s_3 = 512, or (taking the cube root of both sides), s = 8.

The distance from corner to corner of the cube will be equal to the distance between (0,0,0) and (8,8,8). The distance formula for three dimensions is very similar to that of 2 dimensions (and hence like the Pythagorean Theorem):

d = √( (_x_1 – _x_2)2 + (_y_1 – _y_2)2 + (_z_1 – _z_2)2)

Or for our simpler case:

d = √( (x)2 + (y)2 + (z)2) = √( (s)2 + (s)2 + (s)2) = √( (8)2 + (8)2 + (8)2) = √( 64 + 64 + 64) = √(64 * 3) = 8√(3)

The first thing necessary is to determine the dimensions of the cube. This can be done using the volume formula for cubes: V = _s_3, where s is the length of the cube. For our data, this is:

_s_3 = 1728, or (taking the cube root of both sides), s = 12.

The distance from corner to corner of the cube will be equal to the distance between (0,0,0) and (12,12,12). The distance formula for three dimensions is very similar to that of 2 dimensions (and hence like the Pythagorean theorem):

d = √( (_x_1 – _x_2)2 + (_y_1 – _y_2)2 + (_z_1 – _z_2)2)

Or, for our simpler case:

d = √( (x)2 + (y)2 + (z)2) = √( (s)2 + (s)2 + (s)2) = √( (12)2 + (12)2 + (12)2) = √( 144 + 144 + 144) = √(3 * 144) = 12√(3) = 12√(3)

The first thing necessary is to determine the dimensions of the cube. This can be done using the surface area formula for cubes: A = 6_s_2, where s is the length of the cube. For our data, this is:

6_s_2 = 294

_s_2 = 49

(taking the square root of both sides) s = 7

The distance from corner to corner of the cube will be equal to the distance between (0,0,0) and (7,7,7). The distance formula for three dimensions is very similar to that of 2 dimensions (and hence like the Pythagorean Theorem):

d = √((_x_1 – _x_2)2 + (_y_1 – _y_2)2 + (_z_1 – _z_2)2)

Or for our simpler case:

d = √((x)2 + (y)2 + (z)2) = √( (s)2 + (s)2 + (s)2) = √( (7)2 + (7)2 + (7)2) = √( 49 + 49 + 49) = √(49 * 3) = 7√(3)

The volume of a rectangular prism is .

We are told that the shortest edge is 3. Let us call this the height.

We now have , or .

Now we replace variables by known values:

Now we have:

We have thus determined that the other two edges of the rectangular prism will be 4 and 12. We now need to find the longest diagonal. This is equal to:

If you do not remember how to find this directly, you can also do it in steps. You first find the diagonal across one of the sides (in the plane), by using the Pythagorean Theorem. For example, we choose the side with edges 3 and 4. This diagonal will be:

We then use a plane with one side given by the diagonal we just found (length 5) and the other given by the distance of the 3rd edge (length 12).

This diagonal is then .

Since the diagonal of the cube is a line segment that goes through the center of the cube (and also the circumscribed sphere), it is clear that the diagonal of the cube is also the diameter of the sphere. Since the radius = 1, the diameter = 2.

The first thing necessary is to determine the dimensions of the cube. This can be done using the volume formula for cubes: V = _s_3, where s is the length of the cube. For our data, this is:

_s_3 = 512, or (taking the cube root of both sides), s = 8.

The distance from corner to corner of the cube will be equal to the distance between (0,0,0) and (8,8,8). The distance formula for three dimensions is very similar to that of 2 dimensions (and hence like the Pythagorean Theorem):

d = √( (_x_1 – _x_2)2 + (_y_1 – _y_2)2 + (_z_1 – _z_2)2)

Or for our simpler case:

d = √( (x)2 + (y)2 + (z)2) = √( (s)2 + (s)2 + (s)2) = √( (8)2 + (8)2 + (8)2) = √( 64 + 64 + 64) = √(64 * 3) = 8√(3)