Let OD=a and DE=b (E on the a quarter circle) OE=OA=OC=5cm in ∆ ODE OD^2+DE^2=OE^2 a^2+b^2=25 (1) and 1/2ab=5 So ab=10 (2) (2): b=10/a (1): a^2+(10/a)^2=25 so a=√5 ; b=2√5 AD=OA-OD=5-√5 EF=5-2√5 So red rectangle area=(5-√5)(5-3√5)=(35-15√5) cm^2=1.46cm^2.❤
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Just assume the point on the curve is [5cos(a) , 5 sin(a) ]. It reduces the calculation
Let OD=a and DE=b
(E on the a quarter circle)
OE=OA=OC=5cm
in ∆ ODE
OD^2+DE^2=OE^2
a^2+b^2=25 (1)
and 1/2ab=5
So ab=10 (2)
(2): b=10/a
(1): a^2+(10/a)^2=25
so a=√5 ; b=2√5
AD=OA-OD=5-√5
EF=5-2√5
So red rectangle area=(5-√5)(5-3√5)=(35-15√5) cm^2=1.46cm^2.❤