A polygon has d diagonals. If the polygon had one more side, it would d + 10 diagonals. How many sides does the polygon have?
A polygon of n sides can have n-1 diagonal from each vertex. From n vertices, the polygon can have n(n-1)/2 diagonals. The number has to be divided by 2 because we have counted the diagonal from both sides.
So using the formula
d(n)=n(n-1)/2
you can find n.
Hint: n is between 5 and 15.
Do me so hard MM!
Let's say the polygon has n sides.
To determine the number of diagonals in a polygon with n sides, we need to use the formula:
Number of diagonals = (n * (n - 3)) / 2
According to the problem, if the polygon had one more side (n + 1), it would have (d + 10) diagonals.
Using the formula for the polygon with n + 1 sides:
(d + 10) = ((n + 1) * (n + 1 - 3)) / 2
Simplifying this equation:
2(d + 10) = (n + 1)(n - 2)
Expanding and simplifying further:
2d + 20 = n^2 - n - 2
Rearranging the equation:
n^2 - n - 2 - 2d - 20 = 0
Now we have a quadratic equation in terms of n. To find the value of n, we can either factorize the quadratic equation or use the quadratic formula. Let's use the quadratic formula:
n = (-b ± √(b^2 - 4ac)) / (2a)
For the quadratic equation n^2 - n - 2 - 2d - 20 = 0,
a = 1, b = -1, and c = -2 - 2d - 20
Plugging in these values into the quadratic formula:
n = (1 ± √((-1)^2 - 4(1)(-2 - 2d - 20))) / (2(1))
Simplifying further:
n = (1 ± √(1 + 8 + 8d + 80)) / 2
n = (1 ± √(89 + 8d)) / 2
Since the number of sides should be a positive integer, we can discard the negative value.
Therefore, the polygon has n = (1 + √(89 + 8d)) / 2 sides.