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?
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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:
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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