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find the no. of squares formed by joining the vertices of a polygon of 10 sides.

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Mohammed

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Number of triangles formed by joining the vertices of n sided polygon which has no side in common with that of the polygon.

Click here👆to get an answer to your question ✍️ Number of triangles formed by joining the vertices of n sided polygon which has no side in common with that of the polygon.

Question

A

2 n(n−3) ​

B

3! (n−4)(n−5) ​

C

3! n(n−4)(n−5) ​

D

None of these

Medium Open in App Solution Verified by Toppr

Correct option is C)

For the case we consider the equation

Total number of triangles formed

= Triangle having no side common+ triangle having exactly one side common+ triangles having exactly two sides common (with those of the polygon)

∴ Number of triangles having no sides common with that of polygon

=(Total Number of triangles i.e

n C 3 ​

)−Number of δ exactly one side common − Number of triangles having exactly two sides common.

Now Number of Δ having exactly one side common =n(n−4)

and Number of triangles having exactly two sides common.

For this we have to select three consecutive vertices of the polygon,

If vertices of polygon are marked by (A

1 ​ ,A 2 ​ .....A n ​ ) i.eA 1 ​ A 2 ​ A 3 ​ ,A 1 ​ A 2 ​ A 3 ​ A 4 ​ ......A n ​ A 1 ​ A 2 ​

which can be done by n ways.

∴ Required triangles having no sides common with the of the polygon

= n C 3 ​ −n(n−4)−n = 6 n ​ [n 2 −9n+20]= 3! n(n−4)(n−5) ​

29 5

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[Solved] Consider a regular polygon with 10 sides. What is the number

Formula used: Triangle having no side common with those of the polygon =  $$\rm \frac{n(n-4)(n-5)}{6}$$ n is the sides of the polygon. Calculation: He

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Consider a regular polygon with 10 sides. What is the number of triangles that can be formed by joining the vertices which have no common side with any of the sides of the polygon?

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Detailed Solution

Formula used:

Triangle having no side common with those of the polygon =

n(n−4)(n−5)6

n is the sides of the polygon.

Calculation:

Here n = 10

Triangle having no side common with those of the polygon

= 10(10−4)(10−5)6 = 10×6×56 = 50

∴ The number of triangles is 50.

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स्रोत : testbook.com

How many triangles can be formed from a polygon with 10 sides?

Answer (1 of 5): Ok, to answer this question, we need to pick a polygon with at least 10 sides (no less, no more). A decagon, I mean, it’s the only shape with 10 sides (Nah, I’m just kidding, how about a star)? Now, how many triangles does a decagon have? 8. 8 is the answer.

How many triangles can be formed from a polygon with 10 sides?

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Sort B.L. Srivastava

Polygon of 10 sides has ten vertices and no three of them lie on a line. A triangle can be formed by means of three different points (non-collinear ). Hence the required no. of triangles = C(10, 3) = (10!/3!·7!)= 120 .

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No. Of triangle in polygon having n side = n-2

Here n=10

So, no. of triangle= 10–2

=8.

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10 sides also means 10 vertices.

Any three of those vertices define a triangle.

So, the number of triangles that can be formed is the same as the combinations of 10 items taken 3 at a time.

C(10, 3) = 10!/[3!*(10 - 3)!] = 8*9*10/(1*2*3) = 120

Ron Lyall

10 sides means 10 vertices, so you have 10 options for the first point of a triangle followed by 9 points for the second and 8 for the third. So total 10×9×8=720 possibilities.

BUT the same triangle can occur in several ways depending on which point is chosen first. There are 3 ways to choose the first point followed by 2 for the second leaving only 1 option for the last point. So 3×2×1=6 ways of getting the same triangle.

SO overall there are 720÷6=120 different triangles that can be formed.

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Ok, to answer this question, we need to pick a polygon with at least 10 sides (no less, no more). A decagon, I mean, it’s the only shape with 10 sides (Nah, I’m just kidding, how about a star)? Now, how many triangles does a decagon have? 8. 8 is the answer.

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How many triangles can be formed by joining any three vertices of a polygon with n sides, the triangles having no sides in common with the polygon?

Let’s assume that the polygon is a regular polygon. You want to count the number of ways of picking three vertices. That’s easy, it’s just “n choose 3”.

That’s not the correct answer, however, since many of the triangles that are formed from three arbitrary vertices of the polygon will have an edge in common with the polygon.

Well, we could count the number of those triangles, and subtract that number from “n choose 3.” For each of the n edges of the polygon you could form n-2 triangles that share that edge, since there are that many left-over vertices that could be used for the third vertex of

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How many triangles can be formed joining the vertices of 12-sided polygons?

A 12 sided polygon has 12 vertices, so we are looking for the number of different triplets of 12.

This is 12! / (3! * 9!) =220

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How many triangles can be formed in an 18 sided polygon by drawing lines from a single corner to another corner?

First, thats important to note that a 18 sided polygon has 18 corners (vertex).

A triangle can be made conecting 3 corners, so if we want all the triangles, we are talking about combining 18 corners 3 to 3.

This is given by When n=18 n=18 and k=3. k=3.

So if N is the required answer, we have:

N= 18! 3!(18−3)! =816 N=18!3!(18−3)!=816

So in conclusion it is possible to have 816 diferent triangles.

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Mohammed 5 day ago

Guys, does anyone know the answer?