Given: The parallelogram ABCD is inscribed in the circle with centre O as shown.
Asked: What is the area of shaded region?
Since diagonals of parallelogram ABCD intersect at centre of the circle at O , AD & BC are diameters of the circle and angle CAB = angle CDB = angle ABD = angle ACD =90
ABCD is a rectangle.
If radius of the circle r and angles AOC = x are known ;
Area of the shaded region = \(\pi r^2 * (x/360) - 1/2 r^2 sin x\)
1) Angle OAB = 30 deg and BD = 8
Since angle OAB = 30 ; angle AOB = 180 - 2*30 = 120; angle AOC = 180 - angle AOB = 180 - 120 = 60 ; Triangle AOC is equilateral triangle ; BD = AC = 8 = r = AO = )C
Since r = 8 and angle AOC = 60
Area of the shaded region = \(\pi 8^2 * (\frac{60}{360}) - \frac{1}{2}* 8^2 sin 60 = 64 (\frac{\pi}{6} - \frac{\sqrt{3}}{4})\)
SUFFICIENT
2) ABCD is a rectangle with area 64 root3
\(AC * AB = 64 \sqrt{3}\)
Let x & y be sides of the rectangle
\(x*y = 64\sqrt{3}\)
\(x^2 + y^2 = 4r^2\)
It is NOT possible to derive r and angle AOC from information provided since various combinations are possible
NOT SUFFICIENT
IMO A
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Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com