Question

# A tetrahedron has $4$ faces and $6$ edges. How many vertices does it have?

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Solution

## As we know for any polyhedron that doesn't intersect itself, $V-E+F=2$Where ,$V$ is number of vertices $=?$$E$ is number of edges $=6$ $F$ is number of faces $=4$So we have,$V-E+F=2\phantom{\rule{0ex}{0ex}}⇒V-6+4=2\phantom{\rule{0ex}{0ex}}⇒V-2=2\phantom{\rule{0ex}{0ex}}⇒V=2+2\phantom{\rule{0ex}{0ex}}⇒V=4\phantom{\rule{0ex}{0ex}}$Hence, given tetrahedron have $4$ vertices.

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