Geometry Problem Solver

Cube and parallelepiped together

cube	parallelepiped

Cube alone

Parallelepiped alone

They give the tracks some problems can be solved automatically, the numerical values do not matter in the various examples. Only the problems already tested are reported. In fact, the problems solved by the geometric computer, but not tested, with a cube and parallelepiped together, are about 2 x 7 problems on cube x 200 problems on parallelepiped = 2,800

Track 1

The edge of a cube is congruent to the diagonal of a rectangular parallelepiped of dimensions 24 cm; 32 cm and 30 cm. Calculates the total area of the cube.

Track 2

The edge of a cube is 30 cm long. Determines the dimensions of a base of a rectangular 30 cm high parallelepiped, knowing that a base dimension is twice the other and that the lateral area of the parallelepiped is equivalent to the lateral area of the cube.

Track 3

A parallelepiped is composed of three overlapping cubes. Knowing that the edge of each cube measures 10 cm, it calculates the total area and the volume of the solid.

Track 4

The area of the total surface of a cube is congruent to the area of the lateral surface of a parallelepiped having the base length of 30 cm and 10 cm respectively. Calculates the volume of the cube knowing that the base perimeter of the parallelepiped is 8/3 of the height.

Track 5

The total area of a cube is the same as the lateral area of a parallelepiped with a height of 30 cm. Calculates the volume of the cube knowing that the base dimensions of the parallelepiped are one triple of the other and their sum measures 40 cm.

Track 6

The total area of a cube is equal to the total area of a parallelepiped that has the area of the base rectangle of 432 cm², the two dimensions of the base one 4/3 of the other and the height of 54 cm. Calculates the difference in weight between the two solids, assuming the cube is iron (ps = 7.88) and the parallelepiped is in marble (ps = 2.5).

Track 7

The lateral area of a cube is equal to the total area of a parallelepiped that has the area of the base rectangle of 240 cm², the two dimensions of the base one 12/5 of the other and the height of 57 cm. Calculates the difference in weight between the two solids, assuming the cube is iron (ps = 7.88) and the parallelepiped is in marble (ps = 2.5).

Track 8

The lateral area of a cube is equal to the lateral area of a parallelepiped that has the area of the base rectangle of 240 cm², the two dimensions of the base one 12/5 of the other and the height of 68 cm. Calculates the difference in weight between the two solids, assuming the cube is iron (ps = 7.88) and the parallelepiped is in marble (ps = 2.5).

Track 9

The total area of a cube is equal to the lateral area of a parallelepiped that has the area of the base rectangle of 240 cm², the two dimensions of the base one 12/5 of the other and the height of 102 cm. Calculates the difference in weight between the two solids, assuming the cube is iron (ps = 7.88) and the parallelepiped is in marble (ps = 2.5).

Track 10

A cube has a volume of 1000 cm³. Calculates the volume of a parallelepiped having the base area congruent to the base area of the cube and a height of 20 cm.

Track 11

A cube has a volume of 1000 cm³. Calculates the volume of a parallelepiped having the congruent base area to 2/4 of the lateral area of the cube and a height of 20 cm.

Track 12

A cube has a volume of 1000 cm³. Calculates the volume of a parallelepiped having a congruent height to 2/4 of the diagonal of the cube and the base area of 500 cm².

Track 13

A parallelepiped has a base area of 200 cm² and a height of 5 cm. Calculates the diagonal of a congruent cube to the parallelepiped.

Track 14

A parallelepiped has a base area of 200 cm² and a height of 15 cm. Calculates the diagonal of a cube congruent to 1/3 of the parallelepiped.

Cube alone

Parallelepiped alone