Two circles with centre O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through B intersecting the circles at P and Q. Prove that PQ = 2 OO'. -Maths 9th

Two circles with centre O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through B intersecting the circles at P and Q. Prove that PQ = 2 OO'. -Maths 9th
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1 Answer

Answer :

Solution :- Construction: Draw two circles having centres O and O'  intersecting at points                            A and B.                         Draw a parallel line PQ to OO'                        Join OO',OP,O'Q,OM and O'N To Prove: PQ = 2OO' Proof:    In △OPB,              BM = MP    ....(i)     (Perpendicular from the centre to the circle bisect the chord)    Similarly in △O'BQ,            BN = NQ      ....(ii)   (Perpendicular from the centre to the circle bisects the chord)    Adding (i) and (ii),   BM + BN = PM + NQ  Adding BM + BN to both the sides  BM + BN + BM + BN = BM + PM + NQ + BN               2BM + 2BN = PQ               2(BM + BN) = PQ  ....(iii) Again,         OO' = MN   [As OO' NM is a rectangle] ...(iv) ⇒  2OO' = PQ Hence proved.   
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In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Description : In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Last Answer : Given In a parallelogram PSDA, points 0 and R are on PS such that PQ = QR = RS and PA || QB || RC. To prove ar (PQE) = ar (CFD) Proof In parallelogram PABQ, and PA||QB [given] So, ... = ΔDCF [by ASA congruence rule] ∴ ar (ΔPQE) = ar (ΔCFD) [since,congruent figures have equal area] Hence proved.

In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Description : In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Last Answer : Given In a parallelogram PSDA, points 0 and R are on PS such that PQ = QR = RS and PA || QB || RC. To prove ar (PQE) = ar (CFD) Proof In parallelogram PABQ, and PA||QB [given] So, ... = ΔDCF [by ASA congruence rule] ∴ ar (ΔPQE) = ar (ΔCFD) [since,congruent figures have equal area] Hence proved.