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# D01-40

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General Discussion
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mcolbert
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.

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.
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I think this is a high-quality question and I agree with explanation.
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mvictor
mcolbert
How many triangles can be? well, we have 7 sides, and from 7 sides, we can have 4 triangles

now, draw quadrilaterals. From each angle, we can draw lines and form 3 quadrilaterals. How many quadrilaterals can be drawn? 7 sides, 3 quadrilaterals...

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.
Attachments

4 quad.jpg [ 51.26 KiB | Viewed 63547 times ]

File comment: 5 triangles

tri.jpg [ 44.07 KiB | Viewed 63543 times ]

Math Expert
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AndreiGMAT
mvictor
mcolbert
How many triangles can be? well, we have 7 sides, and from 7 sides, we can have 4 triangles

now, draw quadrilaterals. From each angle, we can draw lines and form 3 quadrilaterals. How many quadrilaterals can be drawn? 7 sides, 3 quadrilaterals...

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.

As an example I'll show you 2 triangles that you missed:
Attachment:

Untitled.png [ 187.22 KiB | Viewed 92220 times ]
You are missing triangles and quadrilaterals that can be formed from other vertices.
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Bunuel
Any 3 vertices from 7 can form a triangle and any 4 vertices from 7 can form a quadrilateral, so total of C37+C47=35+35=70C73+C74=35+35=70 different triangles and quadrilaterals can be formed.

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
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Bunuel
Any 3 vertices from 7 can form a triangle and any 4 vertices from 7 can form a quadrilateral, so total of C37+C47=35+35=70C73+C74=35+35=70 different triangles and quadrilaterals can be formed.

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

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.
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Bunuel

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.

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
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AndreiGMAT
Bunuel

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.

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

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).

Attachment:

Untitled.png [ 5.1 KiB | Viewed 91452 times ]
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Bunuel
How many triangles and quadrilaterals altogether can be formed using the vertices of a 7-sided regular polygon?

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

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

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

Cheers,
Brent

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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.
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I think this is a high-quality question and I agree with explanation.
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I think this is a high-quality question and I agree with explanation.
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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I think this is a high-quality question and I agree with explanation.
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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
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okrahulnambiar
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

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