Question

In a polyhedron , it the number of vertices is 10 less than its number of edges and the number of faces is 8 less than the number of its vertices, then find its number of faces, edges and vertices. Pls pls answer it
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Answer 1

Step-by-step explanation:

let take no. of edges =x

then no. of vertices=x-10

also no. of faces=x-10-8=x-18

so, according to eular formula,

e+2=f+v

x+2=x-18+x-10

x+2=2x-28

x=30

so the no. of edges=30

no. of vertices=30-10=20

no. of faces=20-8=12

I hope it helps

Answer 2

${\fbox{\fbox{\huge\sf\purple{QuesTiOn}}}}$

→In a polyhedron , it the number of vertices is 10 less than its number of edges and the number of faces is 8 less than the number of its vertices, then find its number of faces, edges and vertices.

${\fbox{\fbox{\huge\sf\red{AnsWeR}}}}$

$\sf\green{☞30\: edges, 20 \:vertices\: and \:12\: faces.}$

Let's confirm

$\sf\pink{→Taking \:the\: edges, vertices\: and\: faces \:to\: be\:}$$\sf\pink{ F, E\: and\: V \:respectively.}$

$\sf\orange{V = E - 10\: (given)}$

$\sf\purple{F = V - 8 \:(given)}$

$\sf{F = E - 18\: (derived)}$

→FORMULA-:

$\sf\blue{F + V - E = 2 }$

→Taking them all in terms of E,

(E-18) + (E-10) - E = 2

(2E-E) + (-28) = 2

E - 28 = 2

E = 30

_________________________

Now ,

$\sf\purple{E=30}$

$\sf\orange{V = 30-10 = 20}$

$\sf\green{F = 20 - 8 = 12}$

Let's check also

$\sf\blue{F+V = E+2 }$ -:Rule

$\sf\red{12+20= 30+2}$

$\sf\green{32=32}$

→Hence, in the given polyhedron, there are 30 edges, 20 vertices and 12 faces.