How many triangles can be inscribed in the heptagon pictured

Question

heptagon.png How many triangles can be inscribed in the heptagon pictured, where the three vertices of the triangle are also vertices of the heptagon? A. 9 B. 21 C. 35 D. 140 E. 210

Bunuel · Accepted Answer

Generally in a plane if there are n points of which no three are collinear , then: 1. The number of triangles that can be formed by joining them is C^3_n . 2. The number of quadrilaterals that can be formed by joining them is C^4_n . 3. The number of polygons with k sides that can be formed by joining them is C^k_n . Since no 3 vertices in given heptagon are collinear, then the number of triangles possible is C^3_7=35 . Answer: C. Similar questions to practice: if-4-points-are-indicated-on-a-line-and-5-points-are-132677.html abcde-is-a-regular-pentagon-with-f-at-its-center-how-many-d-86284.html abcde-is-a-regular-pentagon-with-f-at-its-center-how-many-133328.html the-sides-bc-ca-ab-of-triangle-abc-have-3-4-5-interior-109690.html gmat-diagnostic-test-question-79373.html m03-71107.html how-many-triangles-with-positive-area-can-be-drawn-on-the-98236.html right-triangle-abc-is-to-be-drawn-in-the-xy-plane-so-that-88958.html how-many-circles-can-be-drawn-on-the-line-segment-128149.html Hope it helps.

Zarrolou · Answer

You have 7 vertices, you have to pick three vertices to form a triangle.

You can pick the first one in 7 ways, the second in 6 and the third in 5. 
 7*6*5 , then you divide by  3!  because the order does not matter (triangle ABC=CAB for example)

Tot ways  \frac{7*6*5}{3!}=35 , or if you want you can solve it with the formula  7C3=35

arpanpatnaik · Answer

I guess another way to look at it is that if we are to consider any line under the heptagon, we can see there are 5 other vertices left with which the line can form a traingle. For example for a line composed of AB, there are vertices C,D,E,F,G with which traingles can be constructed.

In total there are 7 possible lines fo the heptagon. Hence total traingles = 7*5 = 35. Hence   .

Regards,
A

fozzzy · Answer

so in the same question if we were asked how many quadrilaterals to be formed it would be 7C4?

Bunuel · Answer

Yes, that's correct.

ajay2121988 · Answer

Bunuel ,

I don't seem to remember formulas during exam. Can you please point out the flaw in my approach ?

We are required to make triangles inside the heptagon such that three vertices of the triangle are also vertices of the heptagon.
This means we are simply drawing lines to connect the vertices of the heptagon,  connecting 2 vertices of the heptagon at a time  (that's how we form the triangles as directed in the problem)

That means we have to choose 2 points at a time from 7 points. That is  C^2_7=21 .

Bunuel · Answer

Sorry don't follow you. We need three vertices for a triangle not two.

ajay2121988 · Answer

It's alright, I was confusing this with another question. All clear now. Thanks.

Shrey9 · Answer

and that will also give the same answer as number of triangles.. as 35 ?
7C4

khalid9338 · Answer

things become interesting when specific number of sides of heptagon is asked 2 include

KanishkM · Answer

Since you need 3 non collinear points to make a triangle

7C3 = 35

Answer C

ScottTargetTestPrep · Answer

Since there are 7 vertices on the heptagon and any of its 3 vertices will form a triangle, then the number of possible triangles is 7C3 = 7!/(3! x 4!) = (7 x 6 x 5)/3! = (7 x 6 x 5)/(3 x 2) = 35.

Answer: C

vivek920368 · Answer

what about the repeating Triangles? Should we not take that into consideration?
Side GA can make 5 triangles, and seven sides can produce 7*5= 35, but GAB and ABG are the same triangles.

please advise on this.

bumpbot · Answer

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How many triangles can be inscribed in the heptagon pictured

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How many triangles can be inscribed in the heptagon pictured

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