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Re: How many triangles can be inscribed in the heptagon pictured
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12 Jun 2013, 22:58
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fozzzy wrote:
Attachment:
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
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 \(C^3_7=35\).
Re: How many triangles can be inscribed in the heptagon pictured
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12 Jun 2013, 21:45
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fozzzy wrote:
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
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 \(7C3=35\)
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Re: How many triangles can be inscribed in the heptagon pictured
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12 Jun 2013, 23:09
fozzzy wrote:
Attachment:
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
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 [C].
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 wrote:
fozzzy wrote:
Attachment:
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
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 \(C^3_7=35\).
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 wrote:
fozzzy wrote:
Attachment:
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
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 \(C^3_7=35\).
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 \(C^2_7=21\).
Sorry don't follow you. We need three vertices for a triangle not two.
Re: How many triangles can be inscribed in the heptagon pictured
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17 Jan 2019, 17:56
fozzzy wrote:
Attachment:
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
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
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