Find the area of a pentagon which is circumscribing a circle having an area of 420.60 cm²

[tex]\[\begin{array}{l}{r^2} = \frac{{420.6}}{\pi }\\\\r = \sqrt {{\rm{133}}{\rm{.88113812890235644878502174896}}} \\\\r = {\rm{11}}{\rm{.570701712899799598cm}}\end{array}\][/tex]

A Pentagon is a 5 sided polygon

[tex]\[\theta = \frac{{360}}{5}\][/tex]

[tex]\[\theta = 72\][/tex]

[tex]\[\frac{{72}}{2} = 36^\circ \][/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

Find the area of a pentagon which is circumscribing a circle having an area of 420.60 cm²

Sagot :

Problem: