Answer:
A = 752.4 square units
Step-by-step explanation:
The decagon comprised of 10 congruent triangles ΔOAB:
[tex]\boxed{decagon's\ area\ (A)=10\times\Delta OAB's\ area}[/tex]
[tex]\boxed{360^o=10\times\angle AOB}[/tex]
[tex]\angle AOB =360^o\div10[/tex]
[tex]=36^o[/tex]
[tex]\angle AOO'=\frac{1}{2} \angle AOB[/tex]
[tex]=\frac{1}{2} \times36^o[/tex]
[tex]=18^o[/tex]
In order to find ΔOAB's area, we need to find the base (AB) and height (OO') of the triangle. Since we are given the length of the hypotenuse and the angle of the triangle, we can find the length of AB and OO' using trigonometry identities.
For ΔAOO':
[tex]\displaystyle sin\angle AOO'=\frac{AO'}{OA}[/tex]
[tex]\displaystyle sin18^o=\frac{AO'}{16}[/tex]
[tex]AO'=16\times sin18^o[/tex]
[tex]=4.9443\ units[/tex]
[tex]AB = 2\times AO'[/tex]
[tex]=2\times4.9443[/tex]
[tex]=9.8886\ units[/tex]
[tex]\displaystyle cos\angle AOO'=\frac{OO'}{OA}[/tex]
[tex]\displaystyle cos18^o=\frac{OO'}{16}[/tex]
[tex]OO'=16\times cos18^o[/tex]
[tex]=15.2169\ units[/tex]
[tex]Area\ of\ \Delta OAB=\frac{1}{2} \times AB\times OO'[/tex]
[tex]=\frac{1}{2}\times9.8886\times15.2169[/tex]
[tex]=75.2369\ units^2[/tex]
[tex]A=10\times\Delta OAB's\ area[/tex]
[tex]=10\times75.2369[/tex]
[tex]\approx752.4\ units^2[/tex]
