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avatar+349 

Here's an interesting challenge:

 

 

 

 

 

 

a. What percentage of the area of \(\square ABCD\) is occupied by \(\triangle AEF\)?

b. Prove that \(\triangle COE\) is isosceles

 

 

 

 

 

 

 

Not too easy, not too difficult. laugh

Mathhemathh  Aug 12, 2018
 #1
avatar+7521 
+1

a. What percentage of the area of  ABCD is occupied by \(\triangle\)EFA?

 

Hi Mathhemathh!

 

\(\overline{EF}=\overline{CD}\cdot\sqrt{2}-\overline{CD}\\ \overline{EF}=10\cdot\sqrt{2}-10\\ \)

 

\(A_\triangle=\frac{\overline{EF}}{2}\cdot (\overline{CD}\cdot \sqrt{2}-\frac{\overline{EF}}{2}) \)

 

\(A_\triangle=\frac{10\cdot\sqrt{2}-10}{2}\cdot (10\cdot\sqrt{2}-\frac{10\cdot\sqrt{2}-10}{2})\\ A_\triangle=(10\cdot\sqrt{2}-10)^2\)

 

\(A_\triangle=17.15729\)

 

\(A_\square : A_\triangle=100\% :x\\ x=\frac{ A_\triangle\cdot 100\%}{A_\square}\\ x=\frac{17.15729\cdot 100\%}{100}\\ \color{blue}x=17.15729\%\)

 

\(a:\ \triangle EFA\ has\ 17.15729\%\ of\ \square ABCD \)

 

laugh  !

asinus  Aug 12, 2018
edited by asinus  Aug 12, 2018
edited by asinus  Aug 12, 2018
 #2
avatar+91160 
+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

CPhill  Aug 18, 2018
edited by CPhill  Aug 18, 2018

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