goformit100
  • goformit100
How many diagonals are there if the regular convex polygon has n sides ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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goformit100
  • goformit100
@love_jessika15
goformit100
  • goformit100
@amistre64
anonymous
  • anonymous
(n - 3) + [(n - 3) + (n - 4) + (n - 5) + . . . + 1] = (n - 3) + [(n - 3)(n - 2)/2] = (n^2 - 3n)/2

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More answers

anonymous
  • anonymous
Use a pentagon and a hexagon to get a feel for this. All good now, @goformit100 ?
goformit100
  • goformit100
ok
amistre64
  • amistre64
since a given number of sides is equal to a given number of corners; and 2 opposite corners equals 1 diagonal ..... how are we defining a diagonal?
goformit100
  • goformit100
ok then ?
amistre64
  • amistre64
then, im wondering how we define a diagonal ....
amistre64
  • amistre64
|dw:1370369669867:dw|
anonymous
  • anonymous
If you take a hexagon you will get: 3 from the first vertex, 3 from the "next" or adjacent vertex 2 from the next (because one is already drawn) 1 from the next after that (because 2 are already drawn) 0 from the next 2 because they are already drawn That equals 9 and fits the formula that was derived above.
goformit100
  • goformit100
OK
amistre64
  • amistre64
if all diagonals; then i see it as: n sides are n vertexes n + (n-1) + (n-2) + ... + 3 + 2 + 1 minus the original n sides leaves all the diagonals
goformit100
  • goformit100
........ok
amistre64
  • amistre64
pfft, mine should start at n-1
anonymous
  • anonymous
If you take a pentagon, you will have: 2 from the first vertex, 2 from the "next" or adjacent vertex 1 from the next (because one is already drawn) 0 from the next 2 because they are already drawn That equals 5 and fits the formula that was derived above.
amistre64
  • amistre64
sum of 1 to n-1\[\frac{n(1+(n-1))}{2}-n\] sum of 1 to n-1\[\frac{n^2}{2}-n\] sum of 1 to n-1\[\frac{n^2-2n}{2}\]
anonymous
  • anonymous
So, the pattern is n-3 plus n-3 and then decreasing until you hit "1". The formula I derived incorporates adding up from: 1 to n-3 Then you add another n-3. Then you simplify, and then you're done!
anonymous
  • anonymous
(n^2 - 3n)/2 not (n^2 - 2n)/2
amistre64
  • amistre64
yeah, mines off :)
anonymous
  • anonymous
np
amistre64
  • amistre64
i shoulda had n-1 and not n up there
amistre64
  • amistre64
\[\frac{(n-1)(n)}{2}-n\to \frac{n^2-3n}{2}\]\]
anonymous
  • anonymous
|dw:1370370278679:dw|2 from "A", 2 from "B", 1 from "C", none from "D" and "E" That is none that aren't drawn already.
anonymous
  • anonymous
Making sense to you now, @goformit100 ?
goformit100
  • goformit100
Ya I got THANKS
anonymous
  • anonymous
uw! Anytime.

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