Geometry & Mensuration
Aptitude

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Q.

Two mutually perpendicular chords $AB$ and $CD$ meet at a point $P$ inside the circle such that $AP = 6$ cms, $PB = 4$ units and $DP = 3$ units. 

What is the area of the circle?

 A.

$\dfrac{125\pi}{4}$ sq cms

 B.

$\dfrac{100\pi}{7}$ sq cms

 C.

$\dfrac{125\pi}{8}$ sq cms

 D.

$\dfrac{52\pi}{3}$ sq cms

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Solution:
Option(A) is correct

Consider the figure shown below.

Answer image for Geometry & Mensuration, Aptitude:2198-1

$AP \times PB = CP \times PD$

$6 \times 4 = CP \times 3$

$CP = 8$

From center $O$ draw $OM  \perp r AB$ and $ON  \perp r CD$.

From the center a line $\perp r$ to a chord bisects the chord.

So, we have $AM = MB = 5$ cm

$MP = 1$ cm, $ON = 1$ cm, $CD = 11$ cm, $CN = 5.5$ cm

$ON^2 + CN^2 = OC^2$

$1^2 + 5.5^2 = r^2$

$1 + 30.25 = r^2$

Area $=  \pi r^2$

$=\pi \times 31.25$

$=31.25 \pi  = \dfrac{125\pi}{4} \text{ sq cms}$ 


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