My way of solution is ➡ HC=x EH= y ⇒ FE= 20-y BH= 20-x GE= BH A(HCE)= 80 square units 80= HC*EH/2 160= x*y xy= 160 AC would be equal to the radius of the circle, which is also the side length of the square: a= 20 length units R= 20 according to the Pythagorean theorem: x²+y²= R² x²+y²= 20² x²+y²= 400 we know that: (x+y)²= x²+y²+2xy (x+y)²= 400+2*160 (x+y)²= 720 x+y= √720 x+y= √144*5 x+y= 12√5 the area of the green rectangle: A(GEFA)= (20-x)*(20-y) A(GEFA)= 400-20y-20x+xy = 400-20(x+y)+xy x+y= 12√5 xy= 160 ⇒ A(GEFA)= 400-20*12√5+160 A(GEFA)= 400-240√5+160 A(GEFA)= 560-240√5 square units
My way of solution is ➡
HC=x
EH= y
⇒
FE= 20-y
BH= 20-x
GE= BH
A(HCE)= 80 square units
80= HC*EH/2
160= x*y
xy= 160
AC would be equal to the radius of the circle, which is also the side length of the square:
a= 20 length units
R= 20
according to the Pythagorean theorem:
x²+y²= R²
x²+y²= 20²
x²+y²= 400
we know that:
(x+y)²= x²+y²+2xy
(x+y)²= 400+2*160
(x+y)²= 720
x+y= √720
x+y= √144*5
x+y= 12√5
the area of the green rectangle:
A(GEFA)= (20-x)*(20-y)
A(GEFA)= 400-20y-20x+xy
= 400-20(x+y)+xy
x+y= 12√5
xy= 160
⇒
A(GEFA)= 400-20*12√5+160
A(GEFA)= 400-240√5+160
A(GEFA)= 560-240√5 square units