Problem:
Find the area of a pentagon which is circumscribing a circle having an area of 420.60 cm²
Solution:
Area of Circle = 420.60 cm²
Area of Circle = πr²
420.60 = πr²
[tex]\[\begin{array}{l}{r^2} = \frac{{420.6}}{\pi }\\\\r = \sqrt {133.88113812890235644878502174896} \\\\r = 11.570701712899799598cm\end{array}\][/tex]
A Pentagon is a 5 sided polygon
[tex]\[\begin{array}{l}\theta = \frac{{360}}{5}\\\\\theta = 72\\\\\frac{{72}}{2} = 36^\circ \end{array}\][/tex]
[tex]\[\begin{array}{l}\tan \theta = \frac{y}{{11.5707}}\\\\\tan 36 = \frac{y}{{11.5707}}\\\\y = 11.5707\tan 36\\\\y = 8.40660562879cm\\\end{array}\][/tex]
Area of 1 triangle = 1/2(b)(h)
Area of 1 triangle = 1/2(8.40660562879 + 8.40660562879)(11.57070171289)
Area of 1 triangle = 1/2(16.8132112576 )(11.57070171289)
Area o 1 triangle = 97.27cm²
Area of 5 triangles = 97.27(5)
Area of 5 triangles = 486.35 cm²
Answer:
The area of the pentagon circumscribing a circle is 486.35cm²
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