Notes on Nets and Skeleton Models of Regular Solids | Grade 7 > Compulsory Maths > Perimeter, Area and Volume | KULLABS.COM

Notes, Exercises, Videos, Tests and Things to Remember on Nets and Skeleton Models of Regular Solids

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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.
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#### Click on the questions below to reveal the answers

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:

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,
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).

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• ### The regular solid is also known as ______ .

regular skeleton
regular surface
regular hexahedron
regular polyhedron

six
nine
seven
eight

four
two
eight
six

edges
vertex
segment
base

vertex
edges
faces
base

regular
same
irregular
curved

3
4
5
6

12
8
6
4

6
7
8
10

three
five
four
six

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