Given:
The radius of the circle is 6.
Required:
We need to find the area of the pentagon.
Explanation:
We divided the pentagon into five equal triangle parts.
Consider the triangle AOB.
Divide 360 degrees by 5 to find the angle AOB.
[tex]\angle AOB=\frac{360\degree}{5}=72\degree[/tex]
We need to find the angle AOC.
Divide the angle AOB by 2 since AC is the bisector.
[tex]\angle AOC=\frac{\angle AOB}{2}[/tex]
[tex]\angle AOC=\frac{72\degree}{2}=36\degree[/tex]
Consider the right angle triangle AOC.
We have the opposite side = AC and hypotenuse = AO=6.
Use sine formula.
[tex]sin36\degree=\frac{AC}{6}[/tex][tex]AC=6\times sin36\degree[/tex][tex]AC=3.5267[/tex]
We know that AB=AC+BC and also AC+BC.
[tex]AB=3.5267+3.5267=7.0534[/tex]
Use the Pythagorean theorem to find the apothem of the pentagon.
[tex]AO^2=AC^2+OC^2[/tex]
Substitute AO=6 and AC=3.5267 in the formula.
[tex]6^2=3.5267^2+OC^2[/tex]
[tex]OC^2=6^2-3.5267^2[/tex][tex]OC^2=23.5624[/tex][tex]\sqrt{OC^2}=\sqrt{23.5624}[/tex]
[tex]OC=4.8541[/tex]
Use the area of the triangle formula for triangle AOB.
Height, OC=4.8541, and base AB=7.0534.
[tex]A=\frac{1}{2}(heigh\text{t }\times base)[/tex]
[tex]A=\frac{1}{2}\times4.8541\times7.0534[/tex]
[tex]A=17.11\text{90 units}^2[/tex]
The area of the pentagon is 5 times A.
[tex]Area\text{ of pentagon}=85.5947\text{ units}^2[/tex]
[tex]Area\text{ of pentagon}=85.59\text{ units}^2[/tex]
The area of the circle is
[tex]Area\text{ of circle}=\pi r^2=3.14\times36=113.04\text{ units}^2[/tex]
Subtract the area of the pentagon from the area of the circle,
[tex]Area\text{ of shaded region}=113.04-85.59[/tex]
[tex]Area\text{ of shaded region}=27.45\text{ units}^2[/tex]
[tex]Area\text{ of shaded region}=27.5\text{ units}^2[/tex]
The shaded region is
Final answer:
The area of the shaded region is 27.5 square units.
