A square has two of its vertices on a circle and the other two are on the tangent to the circle. Find the area of the square if the radius of the circle is 5 units.
Area of the square: (x + 5)2
If area is 50, then x = √50 - 5; So EF = √50, OE = √50 - 5 and EA = √502
If area is 36 then x = 1; So EF = 6, OE = 1, OA = 5 and EA = 3
∠OEA = 90
If area is 64 then x = 3, EF = 8, EA = 4, OA = 5
OE, EA, and OA did not form Pythagoras triplet in first two cases.
Alternatively,
Clearly, from the figure, the side of the square is (5 + x) units.
Let the point of intersection of AD and the line parallel to the side of the square passing through the centre O of the circle be E.
Considering right-angled triangle ODE,
OD2 = OE2 + ED2
(5)2 = x2 + ED2
ED2 = 25 − x2
ED = √(25−x2)
AD = 2 ED = 2 √(25−x2) ------- 1)
But AD = 5 + x -------- 2)
Equating the above two, we get x = 3 , -5
x = 3 as -5 is not possible.
Side of the square = 5 + 3 = 8
Area = 64