Question

Lucero dibujo las diagonales de un polígono regular y se dio cuenta que tiene que trazar 7 diagonales más que el numero de lados. ¿cuanto es la suma de ángulos internos de ese polígono?

Answer 1

Answer:

The sum of all internal angles of this polygon is: $900^o$

Step-by-step explanation:

Recall that the formula for the number of diagonals of a polygon with n-sides is given by the formula:

$diagonals=\frac{n\,(n-3)}{2}$

and since Lucero noticed that the number of digonals was 7 more that the number of sides, then, these two quantities (n + 7) and the number of diagonals must be equal. we can set an equation stating such, and solve for "n" (number of sides of the polygon):

$n+7=\frac{n\,(n-3)}{2} \\2(n+7)=n(n-3)\\2\,n+14=n^2-3\,n\\n^2-3\,n-2\,n-14=0\\n^2-5\,n-14=0\\(n-7)\,(n+2)=0$

therefore n = 7 or n = -2

The only answer with meaning is the first one (7 sides), since a polygon must have a positive number of sides.

Now that we know the number of sides of the polygon, we can find what the sum of its internal angles should give, based on the equation;

Addition of internal angles of a polygon is = $(n-2)\,180^o$

Which in our case renders:

$(n-2)\,180^o=(7-2)\,180^o= 5\,(180^o) = 900^o$