Question

In the circle with centre O as shown chord AB and CD intersect at P and are perpendicular to each other. If AP = $4$ cm, PB = $6$cm and PC = $2$ cm, then the area of the circle is

• A. $45\pi$
• B. $49\pi$
• C. $50\pi$
• D. $41\pi$

Answer 1

Answer: (C)

We join OA and drop perpendiculars OM & ON from O to CD & AB respectively.

So CM$=\frac{1}{2}$ CD & AN$=\frac{1}{2}$ BD.........(i) (since the perpendicular

from the centre of a circle to its chord bisects the lattar).

Also AB=AP+PB$=(4+6)$ cm$=10$cm.

Now the chords AB & CD intersect at P within the given circle.

$\therefore$ PC$\times$ PD$=$AP$\times$ BP

$⟹$ PD$=\frac{AP\times BP}{PC}=\frac{4\times 6}{2}$ cm$=12$ cm.

$\therefore$ CD$=$PD$+$PC$=(12+2)$ cm$=14$ cm.

$\therefore$ CM $=\frac{1}{2}\times 14$cm$=7$ cm and

AN$=\frac{1}{2}\times 10$ cm $=5$ cm (from i).

So PM$=$CM$-$PC$=(7-2)$ cm$=5$ cm.

Now the quadrilateral OMPN has three of its angles $={90}^o$ each.

$\therefore$ The quadrilateral OMPN is a rectangle.

i.e ON$=$PM$=5$ cm.

Let us consider the triangle OAN.

$\angle$ ANO$={90}^o$, ON$=5$ cm and AN$=5$ cm.

$\therefore \Delta$ OAN is a right one with OA, which is the radius$=r

of the circle, as its hypotenuse.

So, by Pythagoras theorem, we get ${OA}^2=({ON}^2+{AN}^2)=(5^2+5^2)cm=50cm=r^2$.

So $ar.\left(circle\right)=\pi r^2=\pi \times 50{cm}^2=50\pi {cm}^2$.

Ans- Option C.