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

Hi Mathhemathh!

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

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

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

\(A_\triangle=17.15729\)

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

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

  !

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 :