Question

Can a polyhedron have 10 faces, 20 edges and 15 vertices? Using Euler’s formula find the unknown .

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Answer 1

EULER'S FORMULA

According to Euler's formula

For any polygon

Faces + Vertices - Edges = 2

So,

Here

Faces = 10

Vertices = 15

Edges = 20

F + V - E = 10 + 15 - 20

= 5

which is not equals to 2 so Euler's formula is not satisfied hence any polygon can't have 10 faces, 20 edges and 15 vertices .

Now

Case 1 》

F = ?

V = 6

E = 12

Acc. to Euler's Formula

F + V - E = 2

F + 6 - 12 = 2

F = 8

Case 2》

F = 5

V = ?

E = 9

According to Euler's Formula

F + V - E = 2

5 + V - 9 = 2

V = 6

Case 3》

F = 20

V = 12

E = ?

According to Euler's Formula

F + V - E = 2

20 + 12 - E = 2

E = 30

Answer 2

${\large{\pmb{\sf{\underline{Required \; Solution...}}}}}$

☀️ There are three cases are given, somewhere, Faces of polyhedron are given, somewhere Vertices of polyhedron are given and somewhere Edges of polyhedron are given..!

☀️ So according to the question and the given data it is already cleared to us that which formula have to implied here..!

${\qquad \qquad{\small{\underline{\boxed{\sf{\circ \: \: F+V−E=2}}}}}}$

⠀⠀⠀⠀⠀⠀⠀⠀Where, F denotes Number of faces, V denotes number of vertices and E denotes number of edges.

Now let's solve all the given parts! Just for all the parts we have to use the below mentioned formula -

${\qquad \qquad{\small{\underline{\boxed{\sf{\circ \: \: F+V−E=2}}}}}}$

${\large{\pmb{\sf{\underline{Case \: number \: 1^{st}}}}}}$

Here faces are not given, vertices are given as 6 and edges are given too as 12. Let's do it!

${\sf{:\implies F+V−E=2}}$

${\sf{:\implies F+6-12=2}}$

${\sf{:\implies F-6=2}}$

${\sf{:\implies F = 2+6}}$

${\sf{:\implies F = 8}}$

Henceforth, the faces are 8.

${\large{\pmb{\sf{\underline{Case \: number \: 2^{nd}}}}}}$

Here faces are given as 5, vertices are not given and there are 9 edges too. Let's do it!

${\sf{:\implies F+V−E=2}}$

${\sf{:\implies 5+V-9=2}}$

${\sf{:\implies -4+V=2}}$

${\sf{:\implies V=2+4}}$

${\sf{:\implies V=6}}$

Henceforth, vertices are 6.

${\large{\pmb{\sf{\underline{Case \: number \: 3^{rd}}}}}}$

Here faces are 20, vertices are also given as 12 and edges are not given. Let's do it!

${\sf{:\implies F+V−E=2}}$

${\sf{:\implies 20+12-E=2}}$

${\sf{:\implies 32-E=2}}$

${\sf{:\implies -E=2-32}}$

${\sf{:\implies -E=-32}}$

${\sf{:\implies E = 32}}$

Henceforth, 32 are edges.