Question

A hexagon is inscribed in a circle. If the difference between the area of the circle and the area of the hexagon is 36 m2, use the formula for the area of a sector to approximate the radius r of the circle. (Round your answer to three decimal places.)

Answer 1

Well, do it! Area of a circle? Area of inscribed hexagon? Let's see what you get.

Answer 2

area of polygon (from known side length)
= (s^2 N ) / (4 tan (180/N)

n = number of sides
s = side length

area of polygon (from known radius length)
= ( r^2 * N * sin [360/N] ) / (2)

n = number of sides
r = radius of circle aka length from center of hexagon to any corner of hexagon

area of circle
= pi * r^2

Answer 3

as per attached drawing, how semicircle many segments are there that aren't part of the hexagon?

Answer 4

@jlangley ?

Answer 5

area of a sector of a circle = [(theta)/360] * pi * r^2

Answer 6

area of the triangle part of the sector = 1/2 x base x vert ht
= 1/2 * [ sin {1/2 * theta} * radius ] * [ cos {1/2 * theta} * radius ]

as [ sin {1/2 * theta} * radius ] = base
and [ cos {1/2 * theta} * radius ] = height

Answer 7

...done, now use any of the above equations to solve... even though they ask for sector theory, you can check ur answer by using the first 3 eqns

Answer 8

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Answer 9

That would be great if I knew how it worked . . .
A buddy is in the same class and we both had trig in high school, so we are wondering how everyone else is doing since we're having so much trouble. I managed to finally figure out how to solve using sectors only it requires knowing the relationship between the sides of a 30-60-90 triangle, which is a couple chapters down the road.

What I've gotten with sector theory is this:
\[A=\frac{ 1 }{ 2 } r ^{2} \theta\]
\[A=\frac{ 1 }{ 2 }rh\]
\[\frac{ 1 }{ 2 } r ^{2} \theta-\frac{ 1 }{ 2 }rh=36\]
\[\theta=\frac {\Pi}{3}\]
\[\frac{ 1 }{ 2 } r ^{2} (\frac {\Pi}{3})-\frac{ 1 }{ 2 }rh=36\]
\[h=\frac{ \sqrt{3} }{ 2 }r\]
\[\frac{ 1 }{ 2 } r ^{2} (\frac {\Pi}{3})-\frac{ 1 }{ 2 }r \frac{ \sqrt{3} }{ 2 }r=36\]
\[\frac{ 1 }{ 2 } r ^{2} (\frac {\Pi}{3})-\frac{ 1 }{ 2 }( \frac{ \sqrt{3} }{ 2 })r ^{2}=36\]
\[\frac{ 1 }{ 2 }r^{2}[(\frac{ \Pi }{ 3 })-(\frac {\sqrt {3}} {2})]=36\]
\[6(\frac{ 1 }{ 2 }r^{2}[(\frac{ \Pi }{ 3 })-(\frac {\sqrt {3}} {2})])=36\]
\[3r^{2}[(\frac{ \Pi }{ 3 })-(\frac {\sqrt {3}} {2})])=36\]
\[r^{2}[(\frac{ \Pi }{ 3 })-(\frac {\sqrt {3}} {2})])=12\]
\[r=\sqrt {12\div[(\frac{ \Pi }{ 3 })-(\frac {\sqrt {3}} {2})])}\]
\[r \approx 8.139 \]
While this works, is there a way to do it with trigonometric functions, but still apply sector theory?

Answer 10

Area of segment = area of circle/6
difference = area of segment - area of triangle
therefore:
sum of all differences = 36m^2
so 1 difference = 36/6 = 6 m^2

therefore: as
difference = area of segment - area of triangle

area of a segment of a circle = [(theta)/360] * pi * r^2

area of the triangle = 1/2 x base x vert ht
= 1/2 * [ sin {1/2 * theta} * radius ] * [ cos {1/2 * theta} * radius ]

so final eqn is
6 = {[(60/360)] * pi * r^2} - {1/2 * [ sin {1/2 * 60} * r ] * [ cos {1/2 * 60} * r ]}

so r