For a better understanding of the solution please find the diagram in the attached file.
The diagram is that of a triangular part of the regular polygon. The triangle [tex] \Delta OAB [/tex] is an isosceles triangle because the sides OA and OB are equal. Now, The angle [tex] \angle OAB=\frac{360^{\circ}}{16}=22.5^{\circ} [/tex]. This is because the angle around the central angle of a polygon is 360 degrees in total.
Now, because [tex] \Delta OAB [/tex] is an isosceles triangle, angle OAP will be half of angle OAB. Thus, [tex] \angle OAP=11.25^{\circ} [/tex].
AP will be half of AB. Therefore, AP=2.
Now, the apothem, OP can thus be found as:
[tex] \frac{AP}{OP}=tan(11.25^{\circ}) [/tex]
[tex] \therefore OP=\frac{AP}{tan(11.25^{\circ})}=\frac{2}{tan(11.25^{\circ})}\approx10.1 [/tex]
Thus the area of one triangle is [tex] \frac{1}{2}\times 4\times 10.1=20.2 [/tex]
Therefore, the area of all the 16 triangles that make up the regular polygon is: [tex] 16\times 20.2=323.2 [/tex]
Thus the area of the polygon is 323.2 squared inches.
