The sum of the interior angles of the two polygons is 1440° and The sum of the sum of their diagonals is 19.

a. What is the sum of the sides of the 2 polygons?

Also show the Solution please.

[tex] \large \bold{\blue{PROBLEM:}} [/tex]The sum of the interior angles of the two polygons is 1440° and The sum of the sum of their diagonals is 19.

a. What is the sum of the sides of the 2 polygons?

[tex] \large \bold{\blue{SOLUTION:}} [/tex] There are two polygons, refer a and b as the number of their sides. And by using the formula of how to find the sum of interior angles and the number of diagonals of a polygon, we can find the number of sides of a polygon and its sum.

• Sum of all interior angles. (Refer S as the sum and n as the number of sides)

[tex] \boxed{S = (n - 2) \cdot 180 \degree} [/tex]

• Number of diagonals. (Refer D as the number of diagonals and n as the number of sides)

[tex] \boxed{D = \frac{n(n - 3)}{2}} [/tex]

» Make equations on the given statements.

[tex] \begin{cases} \small (a-2) \cdot 180 \degree + (b-2) \cdot 180 \degree = 1440 \degree \\ \frac{a(a-3)}{2} + \frac{b(b-3)}{2} = 19 \end{cases} \: \begin{align} \red{(eq. \: 1)} \\ \red{(eq. \: 2)} \end{align} [/tex]

» Find the value of a in the first equation in terms of b.

• First Equation:

[tex] \small (a-2) \cdot 180 \degree + (b-2) \cdot 180 \degree = 1440 \degree [/tex]

[tex] \small 180\degree a - 360\degree + 180\degree b - 360\degree = 1440 \degree [/tex]

[tex] \small 180\degree a + 180\degree b = 1440 \degree + 360 \degree + 360 \degree [/tex]

[tex] 180\degree a + 180\degree b = 2160 \degree [/tex]

[tex] 180\degree a = 2160 \degree - 180 \degree b [/tex]

[tex] a = 12 - b [/tex]

[tex] \begin{cases} a = 12 - b\\ \frac{a(a-3)}{2} + \frac{b(b-3)}{2} = 19 \end{cases} [/tex]

» Substitute a to the second equation to find b, but first, simplify the second equation.

• Second Equation:

[tex] \frac{a^2-3a}{2} + \frac{b^2-3b}{2} = 19 \\ [/tex]

[tex] \frac{(12-b)^2 - 3(12-b)}{2} + \frac{b^2-3b}{2} = 19 \\ [/tex]

[tex] \frac{108-21b+b^2}{2} + \frac{b^2-3b}{2} = 19 \\ [/tex]

[tex] \small \Bigg( \frac{108-21b+b^2}{2} + \frac{b^2-3b}{2} \Bigg) \cdot 2 = 19 \cdot 2 \\ [/tex]

[tex] 108-21b+b^2 + b^2-3b = 38 [/tex]

[tex] 108 - 24b + 2b^2 = 38 [/tex]

[tex] 108 - 24b + 2b^2 - 38 = 0 [/tex]

[tex] 70 - 24b + 2b^2 = 0 [/tex]

[tex] 2b^2 - 24b + 70 = 0 [/tex]

[tex] b^2 - 12b + 35 = 0 [/tex]

» Use only one solution. It could be any since a and b aren't specified. I prefer addition.

[tex] b = \frac{-(-12) + \sqrt{(-12)^2 - 4(1)(35)}}{2(1)} \\ [/tex]

[tex] b = \frac{12 + \sqrt{144 - 140}}{2} \\ [/tex]

[tex] b = \frac{12 + \sqrt{4}}{2} \\ [/tex]

[tex] b = \frac{12 + 2}{2} \\ [/tex]

[tex] b = \frac{14}{2} \\ [/tex]

[tex] b = 7 [/tex]

[tex] \begin{cases} a = 12 - b\\ b = 7 \end{cases} [/tex]

» The other polygon has 7 sides aka Heptagon. Substitute it to the first equation to find a aka the number of sides of the other polygon.

[tex] \begin{cases} a = 12 - 7 \\ b = 7 \end{cases} [/tex]

[tex] \begin{cases} a = 5 \\ b = 7 \end{cases} [/tex]

» Thus, the two polygons are Pentagon (5 sides) and Heptagon (7 sides). Identify their sum.

[tex] 5 + 7 = 12 [/tex]

[tex] \large \therefore \underline{\boxed{\tt \purple{The \: sum \: of \: their \: sides \: is \: 12}}} [/tex]

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