polygon

[vaivish]

. If a polygon has 44 diagonals, what is the sum of degree measure of interior angles of the polygon?
(A) 1,260
(B) 1,440
(C) 1,620
(D) 1,800
(E) 1,980

[atlantic]

Vavish,

This time I'm not sure but here it goes.....

44 diagonals means 22 polygon sides.

22polygon sides * 90º = 1,980

My pick would be E.

Can you post OA? TIA

[vaivish]

no the OA is 1620..as the sides would be 11 and not 22.......hence the angle would be (sides -2)180= 1620....

[Ian Stewart]

44 diagonals means 22 polygon sides.

22polygon sides * 90º = 1,980

There are a couple of issues here. To find the number of sides, you do not divide the number of diagonals in half. And if you have n sides, the sum of the angles will be (n-2)*180.

We have 44 diagonals, and want to know the number of sides; if we know the number of sides, we just use the formula above to find the sum of the angles. If you have n sides, how many diagonals should you have? Well, from one vertex A, you can make a diagonal by connecting that corner with any other vertex except A itself, and the two points with which A shares an edge. So from A, you can draw n-3 diagonals. You can do this from each of the n vertices. That would give n*(n-3) diagonals, but that's not quite the right answer. Counting this way, we've counted every diagonal exactly twice (if A and D are two vertices, we counted AD and DA as though they were both diagonals, when really they're the same). So there should be n*(n-3)/2 diagonals in an n-sided regular polygon.

Now, we know

n*(n-3)/2 = 44
n*(n-3) = 88
n^2 - 3n - 88 = 0
(n - 11)(n + 8) = 0
n = 11, or n = -8.

The negative solution is nonsensical, so n must be 11.

[chidcguy]

Jeez! There was quite some geometry knowledge involved.

[nareshattri]

...........

Now, we know

n*(n-3)/2 = 44
n*(n-3) = 88
n^2 - 3n - 88 = 0
(n - 11)(n + 8) = 0
n = 11, or n = -8.

The negative solution is nonsensical, so n must be 11.

wow! Thanks

[tokaitalbo]

i like how the 8 ) made a smiley