Regular polygons

Two regular polygons, x and y, are such that the number of sides of x is three more than the number of the sides of y. If the sum of the exterior angles of x and y is 117°, how many sides have x?

Final Answer:

x =  8

Step-by-step explanation:

$$\begin{gathered} \alpha =\frac{360°}{n} \\ x=3+y \\ \frac{360°}{x}+\frac{360°}{y}=117° \\ 360°/x+360°/(x-3)=117° \\ 360(x-3)+360x=117\cdot x\cdot (x-3) \\ 360(x-3)+360x=117\cdot x\cdot (x-3) \\ -117x^2+1071x-1080=0 \\ 117x^2-1071x+1080=0 \\ 117=3^2\cdot 13 \\ 1071=3^2\cdot 7\cdot 17 \\ 1080=2^3\cdot 3^3\cdot 5 \\ \text{GCD}(117,1071,1080)=3^2=9 \\ 13x^2-119x+120=0 \\ a=13;b=-119;c=120 \\ D=b^2-4ac=119^2-4\cdot 13\cdot 120=7921 \\ D>0 \\ {x}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{119\pm \sqrt{7921}}{26} \\ {x}_{1,2}=\frac{119\pm 89}{26} \\ {x}_{1,2}=4.576923\pm 3.423077 \\ {x}_1=8 \\ {x}_2=1.153846154 \\ x=x_1=8 \\ y=x-3=8-3=5 \end{gathered}$$

Our quadratic equation calculator calculates it.

Try another example