Processing math: 100%
 
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avatar+351 

Here's an interesting challenge:

 

 

 

 

 

 

a. What percentage of the area of ABCD is occupied by AEF?

b. Prove that COE is isosceles

 

 

 

 

 

 

 

Not too easy, not too difficult. laugh

 Aug 12, 2018
 #1
avatar+15073 
+1

a. What percentage of the area of  ABCD is occupied by EFA?

 

Hi Mathhemathh!

 

¯EF=¯CD2¯CD¯EF=10210

 

A=¯EF2(¯CD2¯EF2)

 

A=102102(102102102)A=(10210)2

 

A=17.15729

 

A:A=100%:xx=A100%Ax=17.15729100%100x=17.15729%

 

a: EFA has 17.15729% of ABCD

 

laugh  !

 Aug 12, 2018
edited by asinus  Aug 12, 2018
edited by asinus  Aug 12, 2018
 #2
avatar+130466 
+1

Let  the center of the circle with a radius of 5 be centered at  (0,0)  and  let  A  = (-5,5)

The midpoint of  EF  is  (5 cos (45°), -5sin(45°) )  = ( 5/√2, -5/√2)

Let this midpoint be  = G

So   the height of triangle AEF    =  GA  =  √ [( 5/√2 + 5) ^2+ ( 5 +5/√2)^2 ]    = 5 + 5√2 units =

5 ( 1 + √2)   units   =  5 (√2 + 1) units

 

Let H  be the point of tangency where BD touches the circle

And  FG  = FH since these are tangents drawn to a circle from the same exterior point, F

And angle HOF  = 22.5°  and angle OFH  = 67.5°...so...using the Law of  Sines

 

FH / sin (22.5°)  = OF / sin (67.5°)

FH  = 5 sin (22.5°) / sin (67.5°)

FH = 5 sin (22.5°) / cos (22.5°)

FH  = 5 * tan (22.5°)  = 5  * [  1  - cos (45°) ] /[ sin (45°) ]  =  5 *  [ 2 - √2] / [ √2]  units  =

5 (√2 -1)    = FG

 

So...the area of  AEF  =   (GA) ( FG) =  5 (√2 + 1) * 5 ( √2 -1)  units^2  =

25 (2 - 1)  =  

25 units^2

 

The area of  square ABCD   =10*10  =  100 units^2

 

So  area   ΔAEF  =   25 / 100   =  25%    of    square ABCD

 

Here's a pic :

 

  

 

 

cool cool cool

 Aug 18, 2018
edited by CPhill  Aug 18, 2018

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