 ## Nets and Skeleton Models of Regular Solids

Subject: Compulsory Maths

#### Overview

This note gives information about, Cylinder, Cone ,Cube ,Tetrahedron, and Octahedron.
##### Nets and Skeleton Models of Regular Solids

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.
##### Things to remember
• 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.
##### Videos for Nets and Skeleton Models of Regular Solids ##### Maths - Nets of Solid Shapes - Cuboids, Prism - English

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) cm2
= 2 (216 + 96 + 144) cm2
= 912 cm2
Again,
Volume of the block = l × b × h
= 18 cm × 12 cm × 8 cm
= 1728 cm3
∴ The volume of the block is 1728 cm3

Solution:

Here,
Volume of the box (V) = 1600 cm3
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 cm2
So, its base covers an area of 320 cm2 on the table.

Solution:

Here,
the surface area of the cubic block = 96 cm2
or, 6 l2 = 96 cm2
or, l2 = $\frac{96}{6}$ cm2
or, l= 16 cm
or, l = $\sqrt{16 cm^2}$
or, l = 4 cm
∴ The length of its each edge is 4 cm.

Solution:

Here,
Volume of the cubical room = Volume of air
∴ Volume of the cubical room = 512 m3
or, l3 = 512 m3
or, l = $\sqrt{512 m^3}$
or, l = 8 m
So, the length of its floor is 8 m.

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 cm3
or, l × b × h = 768 cm3
or, 2x × x × 6 cm = 768 cm3
or, 2x2 = $\frac{768 cm^3}{2}$
or, 2x2 = 128 cm2
or, x2 = $\frac{128}{2}$ cm2
or, x2 = 64 cm2
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) cm2
= 2 (128 + 48 + 96) cm
= 2 × 272 cm2
= 544 cm2
∴ The surface area of the cuboid is 544 cm2

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 cm3
volume of the cube = volume of the block
or,  l3 = 512 cm3
or, l = $\sqrt{512 cm^3}$
or, l = 8 cm
Again,
the surface area of the cube = 6 l2
= 6 × (8 cm)2
= 384 cm2

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) cm3
= 4608   cm3
Also,
volume of milk powder = volume of the box = 4608 cm3
Again,
volume of each cubical box = (8 cm)3 = 512 cm3
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.

Solution:

Here,
the volume of water = 3000 l
= 3000 × 1000 cm3
Now,
the volume of the part of the tank containing water = volume of water
or, Area of its base × height = 3000 × 1000 cm3
or, 30,000 cm2 × h = 3000 × 1000 cm3
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).