A Circle Circumscribed about a hexagon

**arenkun** · January 12th 2010, 05:30 AM

i need your help again !
help me to solve this...please

God bless you more .

*A circle is circumscribed about a hexagon. The area outside the hexagon but inside the circle is 15 square meters.

a) compute the radius of the circle.
b)compute the area of the hexagon.

**mathaddict** · January 12th 2010, 06:37 AM

Originally Posted by arenkun

i need your help again !
help me to solve this...please

God bless you more .

*A circle is circumscribed about a hexagon. The area outside the hexagon but inside the circle is 15 square meters.

a) compute the radius of the circle.
b)compute the area of the hexagon.

HI

I assume its a regular hexagon .

So the angle each angle at the centre would be 60 degree .

The formula for area of a segment , A=\frac{1}{2}r^2(\theta-\sin \theta)

Since there are 6 segments , $15=6(\frac{1}{2}\times r^2(\frac{\pi}{3}-\frac{\sqrt{3}}{2}))$
evaluate r which is approximately 5 m .

Then the area of the hexagon would be the sum of all the 6 triangles ,

Area $=6(\frac{1}{2}\times 5\times 5\times \sin 60)$

**Soroban** · January 12th 2010, 07:36 AM

Hello, arenkun!

A circle is circumscribed about a hexagon.

The area outside the hexagon but inside the circle is 15 m².

(a) Compute the radius of the circle.

(b) Compute the area of the hexagon.

Code:

                A
              * o *
          *           *
        *               *
     F o                 o B

      *                   *
      *         *         *
      *         O         *

     E o                 o C
        *               *
          *           *
               * o *
                 D

The center of the circle is $O.$
The vertices of the hexagon are: . $A,B,C,D,E,F.$
Draw chords: . $AB, BC, CD, DE, EF, FA$
. . .and radii: . $OA, OB, OC, OD, OE, OF$

Assuming it is a regular hexagon, all the chords and radii are equal to the radius $r.$

The area of the circle is: . $\pi r^2$

The hexagon is comprised of six congruent equilateral triangles of side $r.$
The area of one triangle is: . $\frac{\sqrt{3}}{4}r^2$
The area of the hexagon is: . $6 \times\frac{\sqrt{3}}{4}r^2 \:=\:\frac{3\sqrt{3}}{2}\,r^2$

The difference of the areas is 15 m²: . $\pi r^2 - \frac{3\sqrt{3}}{2}r^2 \:=\:15$

. . $\left(\pi - \frac{3\sqrt{3}}{2}\right)r^2 \:=\:15 \quad\Rightarrow\quad \left(\frac{2\pi - 3\sqrt{3}}{2}\right)r^2 \:=\:15 <br />$

Hence: . $r^2 \:=\:\frac{30}{2\pi-3\sqrt{3}} \quad\Rightarrow\quad r \;=\;\sqrt{\frac{30}{2\pi-3\sqrt{3}}}$

Therefore: . $r \;=\;5.253385683\text{ m} \;\;(a)$

The area of the hexagon is: . $\frac{3\sqrt{3}}{2}\,r^2 \;=\;\frac{3\sqrt{3}}{2}\left(\frac{30}{2\pi-3\sqrt{3}}\right) \;=\;71.70186611\text{ m}^2\;\;(b)$

**arenkun** · January 12th 2010, 06:57 PM

TO : Math Addict and Soroban

I really thank you guys for helping me.
I Love this site.
It's a big help to my assignment.
thank you so much..

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