Two mutually perpendicular chords AB and CD meet at a point P inside the circle such that AP=6 units PB=4 units and DP=3 units. What is the area of the circle?

Options:

A . 125 π4

B . 100 π7

C . 125 π8

D . 52 π43

E . 1

Answer: Option A : A As AB and CD are two chords that intersect at P, AP ∗ PB = CP ∗ PD 6∗4=CP∗3⇒CP=8 From centre O draw OM ⊥ AB and ON ⊥ CD We get AM = MB = 5 (perpendicular from center bisects the chord) Therefore MP = 1, ON = 1, CD = 11 CN = ND = 5.5 (perpendicular from centre bisects the chord) ON2+CN2+OC2 1+(5.5)2=r2 Area of circle = πr2=π(1+(5.5)2)=125π4

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