GATE CSE 2014 Set 1 | Question: GA-10
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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?
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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 = $10 - 4 = 6$.
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How six plz explain ?
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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 $10 - 4 = 6$
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Nice explanation
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TetrahedronImage result for tetrahedron

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5 votes

The Tetrahedron has 4 triangular surfaces with 4 vertices/corners ( say A,B,C and D) as can be seen here. http://www.sjsu.edu/faculty/wa... 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.

0 votes
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.
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3 vertical planes + 3 horizontal planes= 6 planes
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