Cubes - PSAT Math

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 .

The volume of a cubic box is given by (side length)3. Thus, the volume of the larger box is (4_x_)3 = 64_x_3, while the volume of the smaller box is _x_3. Divide the volume of the larger box by that of the smaller box, (64_x_3)/(_x_3) = 64.

Determine the volume of both cubes and then subtract the smaller from the larger. The large cube volume is 9” * 9” * 9” = 729 in3 and the small cube is 3” * 3” * 3” = 27 in3. The difference is 702 in3.

A cube that has three times as long sides is 3x3x3=27 times bigger than the original. Therefore, the answer is 5x27= 135.

Using S as the side length in the original cube, the original is s*s*s. Doubling one side and tripling the other gives 2s*2s*2s for a new volume formula for 8s*s*s, showing that the new volume is 8x greater than the original.

To make this problem easier to solve, we can inscribe a smaller square in the cube. In the diagram above, points are midpoints of the edges of the inscribed cube. Therefore point , a vertex of the smaller cube, is also the center of the sphere. Point lies on the circumference of the sphere, so . is also the hypotenuse of right triangle . Similarly, is the hypotenuse of right triangle . If we let , then, by the properties of a right triangle, .

Using the Pythagorean Theorem, we can now solve for :

Since the side of the inscribed cube is , the volume is .

Cubes have 6 faces. If 2 are red, then 4 must be green. We are told that the total area of the green faces is 36 square inches, so we divide the total area of the green faces by the number of green faces (which is 4) to get the area of each green face: 36/4 = 9 square inches. Since each of the 6 faces of a cube have the same size, we know that each edge of the cube is √9 = 3 inches. Therefore the volume of the cube is 3 in x 3 in x 3 in = 27 cubic inches.

The volume of the box is 50 * 20 * 30 cm = 30,000 cm3.

1cm3 = 1mL, 30,000 cm3 = 30,000mL = 30 L.

Because the volume of the box is only 30 L, 5 L of the 35 L will not fit into the box.

She will stack 360 bales in 3 hours. One row requires 15 bales. 360 divided by 15 is 24.

The volume of a cube is $s^{3}$.

If each side of the cube is 2cm, then the volume will be $8cm^{3}$.

If we double each side, then each side would be 4cm and the volume would be $64cm^{3}$.

A cubic volume is . Let the original sides be 1, so that the original volume is 1. Then find the volume if the sides measure 5. This new volume is 125. Therefore, the ratio of new volume to old volume is 125: 1.

There surface are of a cube is 6 times the area of one face of the cube , therefore $6a^{2}$=150

$a^{2}$=25

a=5

a is equal to an edge of the cube

the volume of the cube is $a^{3}$$=5^{3}$=125

The problem states that the dimensions of the smaller boxes are 1 x 5 x 5, the volume of one of the smaller boxes is 25.

Therefore, 125/25 = 5 small boxes

If the surface area is , then the area of one face must be . Therefore, the length of one edge must be This means that the volume of the cube is . We can now solve with:

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.