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1.A,B,C,D, and E are points on a circle of radius  2 in counterclockwise order. We know AB=BC=DE=2 and CD = EA. Find [ABCDE]

 

Enter your answer in the form x + y sqrt(z) in simplest radical form.

 

2. ABCDEF is a regular hexagon with area 1. The intersection of ACE and BDF is a smaller hexagon. What is the area of the smaller hexagon?

abhinav  Jan 28, 2018
 #1
avatar+92622 
+2

Here's the first one:

 

See the way I've set it up

 

 

angle ABC  =  120°  ......so the area of triangle ABC  =  (1/2)*2^2 sin (120) =  √3 units^2

 

And ACDE forms an isosceles trapezoid

Let the bottom base be  DE  = 2

 

A  = (0,2)    and C  =  (-√ (3) , -1)    E  = (2, 0)    D =  ( 1, -√3)

And  AC forms the top base  = √  [   3 + 9  }  =  √12 units

Call G the mid-point of  AC  =  (-√3/2 , 1/2 )

And  F  is the midpoint  of  DE  =  ( 3/2, -√3/2 )

 

And FG  is the height of the trapezoid  =  √ [  ( 3 + √3)^2  + (1 + √3)^2 ]  /  2  = 

 

√ [  4  ( 4  + 2√3) ]/ 2  =  √ (4 + 2√3)  =  √[ 1 + √3 ]^2  =  1 + √3 units

 

So.....the area of the trapezoid  is

 

(1/2) (height)  (sum of the bases)  =

 

(1/2) ( 1 + √3)  [ 2  + √12 ]  =

 

(1/2) (1 + √3) [ 2 + 2√3 ]  =

 

( 1 + √3)^2  =   [ 4 + 2√3 ] units^2

 

 

So.....the total area is  

 

[ √3  + 4 + 2√3 ]    =    4 + 3√3  units^2

 

 

 

cool  cool  cool 

CPhill  Jan 28, 2018
 #2
avatar+92622 
+2

2)

 

Here's a pic of the way I set this up :

 

 

 

 

The radius  of the larger  hexagon can be found as follows :

Area   =  1            ....so....

 

3 r^2 sin (60)  =  1

3r^2 * √3/2 = 1

r^2  =    2  / √27

r  =  √  [ 2 / √27 ]

 

The y coordinate of  B  is      r * sin (60°) =  √  [ 2 / √27 ] * [ √3 / 2 ]  

And the y coordinate  of F  is  r * sin (-60°)   - √  [ 2 / √27 ] * [ √3 / 2 ]

 

And the distance between them, BF,  is just    2 * √  [ 2 / √27 ] * [ √3 / 2 ]    =  

√  [ 2 / √27 ] *  √3 =

√ [ 6 / √27]

 

Note that because they span equal arcs, chords AC and BF are equal

And by SSS, triangle CBA is congruent to triangle BAF  and both are isosceles

And angle ACB  = angle BAC =  angle BAF = angle FBA

So  angle FBA  =  angle BAC....so BH  = HA

 

And angle CBA  = angle BAE  =  90°

So...subtracting equal angles , then angle HBI  = angle HAM

And angle BHI  = angle AHM

So...by  SAS, triangle BHI  is congruent to triangle AHM

 

But  since HI  is a transversal that cuts parallels  DB  and AE....then angle HAM  = angle HIB

But angle HAM  =  angle HBI......so  angle HBI  = angle HIB

And angle HIB  =  angle HMA.....so angle HAM  = angle HMA

 

So  HM   = AH      and  HI  = HB

So.....HM = HI = AH = HB

So .....since triangles BHI and AHM are congruent.....then HM  = HB

 

And by similar logic, we can prove that HM  = MF

 

So..... HB  = HM = HF

 

So....side HM of hexagon HIJKLM  is 1/3  of BF

 

So....the area of the hexagon is

 

3 * [ 6 / √27] / 9 * √3 / 2  =

 

√3 / √27  =

 

√3 /  [ 3 √3]  =

 

1/3 units^2

 

 

cool cool cool

CPhill  Jan 28, 2018
edited by CPhill  Jan 28, 2018
edited by CPhill  Jan 28, 2018
edited by CPhill  Jan 28, 2018

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