Question

Let r be the radius of the given circle centered at O. If PD =10, & PC =15, P is the midpoint
of OB, what is r
^2?​

Answer 1

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Required answer-:

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=10cm.

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

∴   PC× PD=AP× BP

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

∴  CD=PD+PC=(12+2) cm=14 cm.

∴  CM =$\frac{1}{2}$×14cm=7 cm and

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

So PM=CM−PC=(7−2) cm=5 cm. 

Now the quadrilateral OMPN has three of its angles =90o each.

∴  The quadrilateral OMPN is a rectangle.

i.e ON=PM=5 cm.

 Let us consider the triangle OAN.

∠ ANO=90o, ON=5 cm and AN=5 cm.

∴Δ OAN is a right one with OA, which is the radius=r 

 of the circle, as its hypotenuse.

So, by Pythagoras theorem, we get OA²

=(ON²+AN²)=(5²+5²)cm=50 cm²=50πr²

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