Question

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?

Solution

The correct option is **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

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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