Bunuel
Knewton Brutal Challenge
In the below diagram, points A and B lie on the circle with center O, and rectangle ABCD and triangle ODC have the same area. If the area of the circle with center O is pi times the area of rectangle ABCD, what is the ratio of the length of the radius of the circle to the length of segment DC?

(A) 2 : 1
(B) \(2 : \sqrt{2}\)
(C) 1 : 1
(D) \(1 : \sqrt{2}\)
(E) 1 : 2
Kudos for a correct solution.Let OF be the altitude on DC cutting AB at E.
Here AE=EB and DF=FC (perpendicular drawn from the centre to the chord bisects the chord.)
Let R b the radius. OA=OB=R
It is given that area of triangle ODC = area of rectangle ABCD
So,
1/2 ( OF x DC ) = BC x DC
or OF = 2 BC
or OE = BC
In triangle OEB,
OE^2 + EB^2 = OB^2
or BC^2 + (1/2 DC)^2 = R^2
or BC^2 + 1/4 DC^2 = R^2 ...(1)
It is also given that Area of circle is equal to pi times area of ABCD
So,
pi x R x R = pi x BC x DC
or R^2 = BC x DC ...(2)
From (1) & (2)
BC x DC = BC^2 +1/4 DC^2
or 4 x BC x DC = 4 x BC^2 + DC^2
or DC^2 - 4 x BC x DC + 4 x BC^2 = 0
or (DC - 2BC)^2=0
or DC = 2BC
or BC = DC/2
putting this value in (2)
R^2 = (DC^2) / 2
or (R^2) / (DC^2) =1/2
or R/DC = 1/ \(\sqrt{2}\)
Answer:- D