Subject: Compulsory Maths

**Octahedron**

- Each surface is an equilateral triangle .
- It's a regular solid .
- It has eight surfaces .

**Tetrahedron**

- Each surface is an equilateral triangle .
- It's a regular solid .
- It has four surfaces .

**Cube**

- It's a regular solid and it's also called a regular hexahedron.
- It has six surfaces .
- Each surface is square.

Following are the table to know about the number of vertices , edges and faces of some regular polyhedrons :

Regular polyhedron | No. of vertices(V) | No.of edges (E) | No. of face (F) | F + V-E |

Tetrahedron | 4 | 6 | 4 | 4+4-6=2 |

Hexahedron | 8 | 12 | 6 | 6+8-12=2 |

Octahedron | 6 | 12 | 8 | 8+6-12=2 |

In any regular polyhedron , F+V-E =2 is true. This rule was developed by Swiss Mathematician Euler. It is called Euler's rule.

**Cone :**

- It's curved surface meet at a point called it's vertex.
- It has curved surface with the circular base.
- It is a solid object.

**Cylinder**

- It has a curved surface with two circular bases .
- It is a solid object.

- The cylinder has a curved surface with two circular bases .
- Cone has curved surface with the circular base.
- Cube has six surfaces .
- In any regular polyhedron , F+V-E =2 is true. This rule was developed by Swiss Mathematician Eular. It is called Eulr's rule.

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

A rectangular block is 18 cm long, 12 cm broad and 8 cm thick. Find its surface area and volume.

Solution:

Here,

the length of the block (l) = 18 cm

breadth of the block (b) = 12 cm

thickness of the block (h) = 8 cm

Now,

the surface area of the block = 2 (l b + bh + l h)

= 2 (18 × 12 + 12 × 8 + 18 × 8) cm^{2}

= 2 (216 + 96 + 144) cm^{2}

= 912 cm^{2}

Again,

Volume of the block = l × b × h

= 18 cm × 12 cm × 8 cm

= 1728 cm^{3}

∴ The volume of the block is 1728 cm^{3}

The volume of a rectangular box is 1600 cm^{3} and its height is 5 cm. If it is placed on a table, find the area covered by it on the table.

Solution:

Here,

Volume of the box (V) = 1600 cm^{3}

height of the box = 5 cm

Now,

Volume of the box = Area of its base × height

∴ Area of its base × 5 = 1600

or, Area of its base × height = 1600

or, Area of its base = \(\frac{1600}{5}\)

or, Area of its base = 320 cm^{2}

So, its base covers an area of 320 cm^{2} on the table.

If the surface area of a cubic block is 96 cm^{2}, find the length of its each edge.

Solution:

Here,

the surface area of the cubic block = 96 cm^{2}

or, 6 l2 = 96 cm^{2}

or, l^{2} = \(\frac{96}{6}\) cm^{2}^{ }

or, l^{2 }= 16 cm^{2 }

or, l = \(\sqrt{16 cm^2}\)

or, l = 4 cm

∴ The length of its each edge is 4 cm.

A cubical room contains 512 cu. m of air. Find the length of its floor.

Solution:

Here,

Volume of the cubical room = Volume of air

∴ Volume of the cubical room = 512 m^{3}

or, l^{3} = 512 m^{3}

or, l = \(\sqrt[3]{512 m^3}\)

or, l = 8 m

So, the length of its floor is 8 m.

A cuboid is twice as long its breath and it is 6 cm high. If its volume is 768 cm^{3}, find its length, breadth and surface area.

Solution:

Let, the breadth of the cuboid be x cm.

∴ The length of the cuboid will be 2x cm.

Now,

Volume of the cuboid = 768 cm^{3}

or, l × b × h = 768 cm^{3}

or, 2x × x × 6 cm = 768 cm^{3}

or, 2x^{2} = \(\frac{768 cm^3}{2}\)

or, 2x^{2} = 128 cm^{2}

or, x^{2} = \(\frac{128}{2}\) cm^{2}

or, x^{2} = 64 cm^{2}

or, x = \(\sqrt{64 cm^2}\)

or, x = 8 cm

So, the breadth (b) = x = 8 cm and the length (l) = 2x

= 2 × 8 cm

= 16 cm

Again,

The surface area of the cuboid = 2 (l b + bh + l h)

= 2 (16×8 + 8×6 + 16×6) cm^{2}

= 2 (128 + 48 + 96) cm^{2 }

= 2 × 272 cm^{2}

= 544 cm^{2}

∴ The surface area of the cuboid is 544 cm^{2}

A rectangular metallic block is 16 cm long, 8 cm broad and 4 cm thick. If it is melted and converted into a cube , find the surface area of the cube.

Solution:

length of the block (l) = 6 cm

breadth of the block (b) = 8 cm

thickness of the block (h) = 4 cm

Now,

Volume of the block = l × b × h

= 16 cm × 8 cm × 4 cm

= 512 cm^{3}

volume of the cube = volume of the block

or, l^{3} = 512 cm^{3}

or, l = \(\sqrt[3]{512 cm^3}\)

or, l = 8 cm

Again,

the surface area of the cube = 6 l^{2}

= 6 × (8 cm)^{2}

= 384 cm^{2}

A rectangular box completely filled with milk powder is 24 cm long, 12 cm broad 16 cm high. How many cubical boxes of length 8 cm are required to empty the milk powder from the box?

Solution:

Here,

length of the rectangular box (l) = 24 cm

breadth of the rectangular box (b) = 12 cm

height of the rectangular box (h) = 16 cm

∴ Volume of the rectangular box = l × b × h

= ( 24 × 12 × 16) cm^{3}

= 4608 cm^{3}

Also,

volume of milk powder = volume of the box = 4608 cm^{3}

Again,

volume of each cubical box = (8 cm)^{3} = 512 cm^{3}

Now,

the required number of cubical boxes = \(\frac{volume \; of \; milk \; powder}{volume \; of \; each \; cubical \; box}\)

= \(\frac{4608}{512}\)

= 9

So, 9 cubical boxes are required to empty the milk powder from the box.

The area of the base of a rectangular water tank is 30,000 cm^{2}. Find the height of the water level when there is 3000 litres of water in the tank. (1 l = 1000 cm^{3})

Solution:

Here,

the volume of water = 3000 l

= 3000 × 1000 cm^{3}

Now,

the volume of the part of the tank containing water = volume of water

or, Area of its base × height = 3000 × 1000 cm^{3}

or, 30,000 cm^{2} × h = 3000 × 1000 cm^{3}

or, h = \(\frac{3000 × 1000 cm^3}{30,000 cm^2}\)

or, h = 100 cm

So, the required height of water level in the tank is 100 cm (or 1 m).

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