We take the n number of vertices times the possible diagonals (n-3), then consider just the ones that are not duplicated. This gives us [n(n-3)]*1/2 For a ABCDEF hexagram, we have (repeated diagonals are in parentheses): AC AD AE BD BE BF CE CF (CA) DF (DA) (DB) (EA) (EB) (EC) (FB) (FC) (FD)
Hello Wise. Not sure what you mean ... a formula can be derived from other principals but that would be a difficult explanation ... and I'm not sure that is what you're asking?
hi liz, good question. The pattern for the formula is n(n-3) all divided by 2. The - 3 comes from the pattern where n = the number of sides and the - 3 is the number of diagonals that cam come from any one vertex depending on the number of sides in the polygon. For example, a four-sided figure (n = 4) can only have one diagonal from any vertex ( n - 3) which equals one. Try it your self. Draw a 5 or 6 sided figure and count how many diagonals can be drawn from only one vertex. You will always see an n - 3 pattern.
You explained it so well...thank you!
thank you stcac!
Hi may i know the formula for finding the no. Of triangles formed by diagonals in a polygon
dang you came in clutch I smashed that like button
Thanks Luis. Glad it helped you out
We take the n number of vertices times the possible diagonals (n-3), then consider just the ones that are not duplicated. This gives us [n(n-3)]*1/2
For a ABCDEF hexagram, we have (repeated diagonals are in parentheses):
AC AD AE
BD BE BF
CE CF (CA)
DF (DA) (DB)
(EA) (EB) (EC)
(FB) (FC) (FD)
Thanks :)
heres a trick
triangle=0+2=quadrilateral+3=pentagon+4=hexagon+5=heptagon+6=octagon+7=nonagon+8=decagon
Thank you
You're welcome
Thankyou 💗
You’re welcome 😊
Can you explain exactly how the formula works?
Hello Wise. Not sure what you mean ... a formula can be derived from other principals but that would be a difficult explanation ... and I'm not sure that is what you're asking?
Thank you💙
Thank, that really helped me. X_X
you're welcome Neha!
Thanks
you're welcome
why do we need to subtract 3? why 3? 😁 thank you
hi liz, good question. The pattern for the formula is n(n-3) all divided by 2. The - 3 comes from the pattern where n = the number of sides and the - 3 is the number of diagonals that cam come from any one vertex depending on the number of sides in the polygon. For example, a four-sided figure (n = 4) can only have one diagonal from any vertex ( n - 3) which equals one. Try it your self. Draw a 5 or 6 sided figure and count how many diagonals can be drawn from only one vertex. You will always see an n - 3 pattern.
And also if you have a exam it will bring you the formula but the formula he choosed is 3 but if you are explaining you can say another formulated
there are 3 diagonals in a triangle.....we need to draw a vertex in the centre
Hello Anjali . . . not sure what you're saying? There are no diagonals in a triangle.