The perimeter of a regular hexagon is 48 inches. What is the number of square inches in the positive difference between the areas of the circumscribed and the inscribed circles of the hexagon? Express your answer in terms of pi.
##16pi ” in”^2##
Refer to the image below to give us visual on how to solve this.
Given that ##P = 48 ” in”##, means we have the measurement of each side equal to ##P/6 =(48″ in”)/6 = 8 ” in”##. Since the figure is a regular six-sided polygon.
To find the positive difference in area of the circumscribed and inscribed circles, we need to find of course each area of the circle, but find these area, we need their respective radius.
The figure shows us that the ##color(red)”radius”## of the circumscribed circle is the also the radius of the hexagon, and the ##color(blue)”radius”## of the inscribed circle is the apothem of the hexagon.Apothem is the distance from the center perpendicular to one of the side of a regular polygon.
Note that, apothem is also the perpendicular bisector of the side of a regular polygon. These means that the side is divided into two, and equals ##4 ” in”##.
Another important thing, a regular hexagon is also made up of 6 equilateral triangles, and these triangles have interior angles of ##60^o##.
Using the radii and one side of the hexagon, we can form a right “special” triangle which is a 30-60-90 triangle.
We know the measure of a side and an angle, therefore we can use trigonometric functions to determine the other legs.
Solving for ##color(red)R##:
##cos theta=(adj)/(hyp)##
##cos 60^o=(4 “in”)/color(red)R##
##color(red)R=(4″in”)/(cos60^o) = color(red)(8 ” in”)##
Solving for ##color(blue)r##:
##tan theta=(opp)/(adj)##
##tan 60^o=color(blue)(r)/(4 “in”)##
##color(blue)r=4″in”*tan60^o = color(blue)(4sqrt3 ” in”)##
NOW WE HAVE THE RADII, WE COMPUTE FOR THE AREAS:
Let:##A_c =>## area of the circumscribed circle##A_i =>##area of the inscribed circle##A_s =>##area of the shaded ##->##one we are after
##A_s=A_c-A_i##
##A_s=picolor(red)R^2-picolor(blue)r^2##
##A_s=pi(color(red)R^2-color(blue)r^2)##
##A_s=pi[color(red)((8))^2-color(blue)((4sqrt3))^2]##
##A_s=pi(64-48)##
##A_s=16pi ” in”^2##
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