D01-40

Question

How many total triangles and quadrilaterals can be created using the vertices of a 7-sided  regular  polygon? (A regular polygon is a polygon with equal sides and angles.)

A. 35
B. 40
C. 50
D. 65
E. 70

Bunuel · Accepted Answer

Official Solution:

How many total triangles and quadrilaterals can be created using the vertices of a 7-sided  regular  polygon? (A regular polygon is a polygon with equal sides and angles.)

A. 35
B. 40
C. 50
D. 65
E. 70

Since the given polygon is regular, it means that no three vertices of it are collinear, so no three vertices are on the same line. As a result, selecting any 3 vertices from the 7 available can create a triangle, and choosing any 4 vertices can create a quadrilateral. Therefore, the total number of different triangles and quadrilaterals that can be formed is:  C^3_7 + C^4_7 = 35 + 35 = 70 .
 
Note that this is NOT a geometry question. Rather, it's a combinations question, as no specific knowledge of geometry is required to solve it.

Answer: E

mcolbert · Answer

I think it is very hard to understand what this question is asking. I never would have been able to understand this premise regardless of time spent on it.
Is the seven sided regular polygon a regular star or a heptagon? You cannot form any triangles or quadrilaterals using only the angle of the heptagon.
Finally, triangles and quadrilaterals aren't made from vertices, but from vertices and sides. There could be infinitely many side lengths, therefore infinitely many triangles and quadrilaterals.

mvictor · Answer

it is a  regular  polygon with 7 sides. Regular polygons have equal sides, and equal angles.

for the sake of understanding, draw a 7 sides polygon and start drawing triangles and quadrilaterals.
we clearly see that from one angle, we can draw lines and form 4 triangles. How many triangles can be? well, we have 7 sides, and from 7 sides, we can have 4 triangles
4C7 = 7!/4!3! = 5*6*7/1*2*3 = 35
now, draw quadrilaterals. From each angle, we can draw lines and form 3 quadrilaterals. How many quadrilaterals can be drawn? 7 sides, 3 quadrilaterals...
3C7 = which is equal to 4C7 
now we have 35+35 = 70.

rishi02 · Answer

I think this is a  high-quality  question and I agree with explanation.

AndreiGMAT · Answer

I drew the lines and ended up with 5 triangles (not 4) and 4 quadrilaterals (not 3): Red, Purple, Green, and Blue.
Check the images that I drew.
Please, point out, where is the mistake.

Thank you.

Bunuel · Answer

As an example I'll show you 2 triangles that you missed: Untitled.png You are missing triangles and quadrilaterals that can be formed from other vertices.

AndreiGMAT · Answer

I think you missing 1 triangle and 1 quadrilateral, Please, see your quote:

In my images I stated that you can form   5 triangles   from one angle, and   4 quadrilaterals   from one angle. While You formed 4 triangles and 3 quadrilaterals.

So shouldn't be the solution:

7C5+7C4=21+35=  56   different triangles and quadrilaterals can be formed

Bunuel · Answer

Sorry I cannot draw all 35 triangles and 35 quadrilaterals but the logic in my solution explains why it is so. Any 3 distinct points from 7 will form a triangle and any 4 distinct points from 7 will form a quadrilateral.

AndreiGMAT · Answer

Bunuel, if you could explain this method just algebraically for square and pentagon (how many triangles can be formed), so I could better understand your logic.

Thanks

Bunuel · Answer

Below are cases for square. We can make only 1 (4C4 = 1) quadrilateral using the vertices of a square and we can make 4 triangles (4C3 = 4). Untitled.png

BrentGMATPrepNow · Answer

TRIANGLES  
If you choose any 3 of the 7 vertices, you can connect them with lines to create a unique triangle. 
So, the question becomes "In how many different ways can we select 3 vertices from 7 vertices?"
Since the order in which we select the 3 vertices does not matter, we can use COMBINATIONS. 
We can select 3 vertices from 7 vertices in 7C3 ways. 
7C3 =  35

ASIDE: we have a video on calculating combinations (like 7C3 and 7C4) in your head - see below

QUADRILATERALS  
If you choose any 4 of the 7 vertices, you can connect them with lines to create a unique quadrilateral. 
So, in how many different ways can we select 4 vertices from 7 vertices?
Since the order in which we select the 4 vertices does not matter, we can use COMBINATIONS. 
We can select 4 vertices from 7 vertices in 7C4 ways. 
7C4 =  35

So, the TOTAL number of triangles and quadrilaterals possible =  35  +  35  = 70

Answer: E

Cheers,
Brent

RELATED VIDEO
 https://www.youtube.com/watch?v=05a0A7vjG8I

reddyg · Answer

I think this is a  high-quality  question and I agree with explanation. I don't think this is a 600 level question, maybe a medium 600-700 level one.

RenanBragion · Answer

I think this is a  high-quality  question and I agree with explanation.

aussieMBA · Answer

I think this is a  high-quality  question and I agree with explanation.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

BottomJee · Answer

I think this is a  high-quality  question and I agree with explanation.

okrahulnambiar · Answer

Don't we need to factor in for instances where the lines are co-linear? In a sense, they're on the same side of the polygon

Bunuel · Answer

The point is that no two sides in a seven-sided regular polygon are parallel.

Atrocious12 · Answer

I think this is a  high-quality  question. Is this question a part of the GMAT focus syllabus?

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