Parallelepiped is a 3-D (three-dimensional) shape with six faces bound by three pairs of parallel planes, so all its faces are parallelograms.

Basically, it is formed by six parallelogram sides to result in a three-dimensional figure or a Prism, which has a parallelogram base.

### Properties of Parallelepiped

1) Parallelepiped is a 3-D shape.

2) It has 6 faces.

3) It has 12 edges and 8 vertices.

4) Faces of Parallelepiped is a parallelogram.

5) It is a prism with a parallelogram as base.

6) Each face seems to be the mirror image of the opposite face.

7) The diagonal of each face is called face diagonal.

8) The diagonal of the 3-D body of parallelepiped is called body diagonal.

### Volume of Parallelepiped

For a given parallelepiped, let ‘A’ be the area of the bottom face and ‘h’ be the height, then the volume formula is given by:

The volume of Parallelepiped = A × h

Since the bottom face is a parallelogram so its area ‘A’ can be calculated by:

A = Area of parallelogram = base × height.

In Vector form:

The volume of the parallelepiped spanned by a, b, and c is given by:

Volume = Area of base × height

= | a × b | |c| cos ø

= | (a × b).c |

### Surface Area of Parallelepiped

1) Lateral Surface Area (L.S.A.)

It is the product of the perimeter of the base and height of parallelepiped.

L.S.A. = Perimeter of base × height

2) Total Surface Area (T.S.A.)

It is the sum of LSA and area of the two bases.

T.S.A. = L.S.A. + 2(base area)

There are basically two types of Parallelepipeds:

1. Cube

2. Cuboid

### Cube

The cube is parallelepiped whose all six faces are in a square shape.

Example: Dice, Ice cube, Rubik's cube, etc.

Let each side of the cube be ‘a’ units. Then,

1. The volume of cuboid = Area of Bottom face (Square) × Height

= (a × a) × (a)

= a3 units

2. L.S.A. = Perimeter of base (Square) × Height

= (4a) × (a)

= 4a2 units

3. T.S.A. = L.S.A. + 2(Base area)

= (4a2)+ 2(a2)

= 6a2 units

4. Face Diagonal

A face diagonal is a line that connects two vertices that are on the same face, but not on the same edge. Here, Face diagonal is same as the diagonal of Square.

Face Diagonal \[ = \sqrt {({a^2} + {a^2})} = \sqrt {(2)} \times (a)\] units

5. Body diagonal

A body diagonal is a line that connects two vertices that are not on the same face.

Body diagonal \[\sqrt {{a^2} + {a^2} + {a^2}} = \sqrt 3 \times \left( a \right)\] units

### Cuboid

The cuboid is a parallelepiped whose all six faces are in a rectangular shape.

It is a three-dimensional box-shaped structure.

Example: Match-stick box, Duster, Bricks, Instrument-box, etc.

Let the length, width and height of cuboid be denoted by l, b and h respectively. Then,

1) Volume of cuboid = Area of Bottom face (Rectangular) × Height

= (l × b) × (h)

= l × b × h

2) L.S.A. = Perimeter of base (Rectangle) × Height

= 2(l + b) × (h)

= 2(l + b) h

L.S.A. of the cuboid can also be assumed as an area of four walls of a cuboidal room.

3) T.S.A. = L.S.A. + 2(Base area)

= 2(l + b) h + 2(lb)

= 2(lb + bh + lh)

4) Face Diagonal

Here, Face diagonal is same as the diagonal of Rectangle. A face diagonal is a line that connects two vertices that are on the same face, but not on the same edge.

Face Diagonal = \[\sqrt {({l^2} + {b^2})} \]

5) Body diagonal

A body diagonal is a line that connects two vertices that are not on the same face.

Body diagonal = \[\sqrt{{l^2}+{b^2}+{h^2}}\]

### Solved Examples:

Q.1. Find the side of the cube whose surface area is 600 cm2.

Ans. Surface area of cube = 6 (side)2

Given, Surface area of cube = 600 cm2

⇒ 6 (side)2 = 600 cm2

⇒ (side)2 = 100 cm2

⇒ side of the cube = 10 cm.

Q.2. Find the volume, L.S.A., T.S.A. and Body diagonal of Cuboid whose length is 3 cm, width is 4 cm and height is 5 cm.

Ans. a.) Volume of Cuboid = l × b × h

= 3 cm × 4 cm × 5 cm

= 60 cm3

b.) L.S.A. = 2(l + b) h

= 2 × (3 cm + 4 cm) × 5 cm

= 70 cm2

c.) T.S.A. = 2(lb + bh + hl)

= 2 × {(3 cm × 4 cm) + (4 cm × 5 cm) + (5 cm × 3 cm)}

= 2 × (12 cm2 + 20 cm2 + 15 cm2)

= 2 × 47 cm2

= 94 cm2

d.) Body diagonal

= \[\sqrt{{l^2} + {b^2} + {h^2}}\]

= \[\sqrt{{3^2} + {4^2} + {5^2}}\]

= \[\sqrt{9 + 16 + 25}\] cm

= \[\sqrt{50}\] cm

= 5\[\sqrt{2}\] cm

Q.3. The internal measures of a cuboidal room are 12 m × 8 m × 4 m. Find the total cost of white washing all four walls of a room, if the cost of whitewashing is Rs. 5 per sq. meter.

Ans. Let the length of the room = l = 12 m

width of the room = b = 8 m

height of the room = h = 4 m

Area of four walls of room = L.S.A. of cuboidal room = 2(l + b) × h

= 2 (12+8) × 4

= 2 × 20 × 4

= 160 m2

Given, cost of white washing per m2 = Rs. 5

Hence the total cost of white washing four walls of the room = Rs.(160 × 5) = Rs. 800.