Question

In the given figure, O is the centre of the circle, AB and CD are the chords of the circle which intersect at P. If PA=4 cm, PB = 6 cm, OB 7 cm and CD is 1 cm greater than twice of OP, then the difference between the length of PD and PC is​

Answer 1

Answer:

ANSWER

Given: AB and AC are two equal chords of a circle with centre O.

OP⊥AB and OQ⊥AC.

To prove: PB=QC

Proof: OP⊥AB

⇒AM=MB .... (perpendicular from centre bisects the chord)....(i)

Similarly, AN=NC....(ii)

But, AB=AC

⇒

2

AB

=

2

AC

⇒MB=NC ...(iii) ( From (i) and (ii) )

Also, OP=OQ (Radii of the circle)

and OM=ON (Equal chords are equidistant from the centre)

⇒OP−OM=OQ−ON

⇒MP=NQ ....(iv) (From figure)

In ΔMPB and ΔNQC, we have

∠PMB=∠QNC (Each =90

∘

)

MB=NC ( From (iii) )

MP=NQ ( From (iv) )

∴ΔPMB≅ΔQNC (SAS)

⇒PB=QC (CPCT)

Answer 2

4 cm

IS the ANSWER.................