# Difficult Geometry & Mensuration Solved QuestionAptitude Discussion

 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

Solution:
Option(A) is correct

Consider the figure shown below.

$AP \times PB = CP \times PD$

$6 \times 4 = CP \times 3$

$CP = 8$

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 $= \pi r^2$

$=\pi \times 31.25$

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