Nets and Skeleton Models of Regular Solids

Subject: Compulsory Maths

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Overview

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

Octahedron

Source :math.stackexchange.com Fig :combinatorics - Number of unique cycle pa
Source:math.stackexchange.com
Fig:octahedron



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





Tetrahedron

Source :www.kidsmathgamesonline.com Fig :Tetrahedron Picture - Images of Shapes
Source:www.kidsmathgamesonline.com
Fig:Tetrahedron



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




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Cube

Source:www.clipartpanda.com
Fig: 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 :

fig: cone
fig: 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

fig:cylinder
fig: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.
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  • 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
1.1 Geometry - Drawing Nets
Drawing-Shapes To Forms.m4v
Maths - Nets of Solid Shapes - Cuboids, Prism - English
Questions and Answers

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 = \(\sqrt[3]{512 m^3}\)
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 = \(\sqrt[3]{512 cm^3}\) 
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).

Quiz

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