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.
a. What percentage of the area of ABCD is occupied by △EFA?
Hi Mathhemathh!
¯EF=¯CD⋅√2−¯CD¯EF=10⋅√2−10
A△=¯EF2⋅(¯CD⋅√2−¯EF2)
A△=10⋅√2−102⋅(10⋅√2−10⋅√2−102)A△=(10⋅√2−10)2
A△=17.15729
A◻:A△=100%:xx=A△⋅100%A◻x=17.15729⋅100%100x=17.15729%
a: △EFA has 17.15729% of ◻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 :