**The parallelepiped** is a three-dimensional geometry shape. It is constructing with six parallelograms. It is related with parallelogram just like cube with square in Euclidean geometry.

The following definitions help the students to understand that shape.

- The shape contains parallelogram and polyhedron with six faces.
- It is constructed with 3 pairs of parallel faces that is hexahedron.
- It is a prism with parallelogram base.

**Properties of parallelepiped**

- The base of prism is viewed with three pairs of parallel faces.
- This shape is formed from linear transformations of a cube.
- Another name of parallelepiped is zonohedron.
- It is closed with four rectangular faces and two rhombic faces.

**Volume of parallelepiped:**

The formula for volume of parallelepiped is

V = `abcsqrt(1 + 2 cos(alpha) cos(beta) cos(gamma)- cos^2(alpha)-cos^2(beta)- cos^2(gamma))`

Here the a, b and c are length of edges and `alpha, beta and gamma` are internal angles.

Lets see solved problem on parallelepiped

**Problem:** Find the parallelepiped volume with lengths a = 5, b = 4 and c = 6, the internal angles are `alpha` = 35^{o}, `beta` = 46^{o} and `gamma` = 60^{o}.

**Solution:**

Given values are a = 5, b = 4 and c = 6, the internal angles are `alpha ` = 35, `beta` = 46 and `gamma ` = 60.

The volume of parallelepiped is `abcsqrt(1 + 2 cos(alpha) cos(beta) cos(gamma)- cos^2(alpha)-cos^2(beta)-cos^2(gamma))` .

V = 5 * 4 * 6`sqrt(1 + 2 cos(35^o) cos(46^o) cos(60^o)-cos^2(35^o)-cos^2(46^o)-cos^2(60^o)).`

= 120`sqrt(1 + 2(0.819)(0.694)(0.5)-(0.670)-(0.481)-(0.25))`

= 120`sqrt(1 + 0.568-0.670-0.481-0.25)`

= 120`sqrt(0.167)`

= 49.03

Therefore, the volume of parallelepiped is 49.03.

**Exercise problem for parallelepiped:**

1. Find out the volume of parallelepiped with a = 10, b = 15 and c = 18, the internal angles are `alpha ` = 90, `beta` = 50 and `gamma` = 47.

**Solution:** The volume is 977.23.