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+11 votes
When a point inside of a tetrahedron (a solid with four triangular surfaces) is connected by straight lines to its corners, how many (new) internal planes are created with these lines?
asked in Numerical Ability by Boss (18k points)
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4 Answers

+31 votes
Best answer
It is $6$.

Tetrahedron has $4$ corner points. So, it forms $4$ planes. Now, we add an internal point making a total of $5$ points. Any three combination of points leads to a plane and thus we can get ${}^5C_3 = 10$ planes. So, newly created planes = $10 - 4 = 6$.
answered by Active (1.5k points)
edited by
How six plz explain ?
+12 votes
A Tetrahedron has $4$ points. And Atleast $3$ points are needed to draw a plane (as two points for straight line ). So, In a Tetrahedron there are $\binom 4 3 = 4$ Planes.

Now one more point is taken in the center, So total number of points $ = 5$ And Number of planes possible with these $5$ points is equal to $\binom 5 3 = 10$.

So, New planes formed due to extra point is equal to $10 - 4 = 6$
answered by Loyal (7.9k points)
Nice explanation
+1 vote

The Tetrahedron has 4 triangular surfaces with 4 vertices/corners ( say A,B,C and D) as can be seen here. Now if you take a point inside a tetrahedron (suppose O) and connect it with any two of its corners which are nothing but vertices( suppose A and B), you will get 1 internal plane as OAB. So we can see from here that, no of new internal planes = no of different pair of corners or vertices Similarly you can take any other 2 corners like (A,C) or (A,D) or (B,C) or (B,D) or (C,D), hence total possible pair of corners are 6. Therefore 6 new internal planes possible. We could also calculate the possible corners by using combinations formula, which is nCr, i.e. no of ways to select a combination of r things from a given set of n things. here n = 4 ( as total 4 vertices, A,B,C and D) and r =2 ( as we need two corners at a time) Thus, 4C2 = 6.

answered by Loyal (8.2k points)
+1 vote
answered by Loyal (8.2k points)

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