Question

In a circle with centre o seg OP parpendicular chord AB, A B = 48cm, op=7cm Then find radius of the circle. ​

Answer 1

seg OP ⊥ chord CD … [Given] ∴ l(PD) = (1/2) l(CD) … [Perpendicular drawn from the centre of a circle to its chord bisects the chord] ∴ l(PD) = (1/2) x 48 …[∵ l(CD) = 48 cm] ∴ l(PD) = 24 cm …(i) In ∆OPD, m∠OPD = 90° ∴ [l(OD)]2 = [l(OP)]2 + [l(PD)]2 … [Pythagoras theorem] ∴ (25)2 = [l(OP)]2 + (24)2 … [From (i) and l(OD) = 25 cm] ∴ (25)2 – (24)2 = [l(OP)]2 ∴ (25 + 24) (25 – 24) = [l(OP)]2 …[∵ a2 – b2 = (a + b) (a – b)] ∴ 49 x 1 = [l(OP)]2 ∴ [l(OP)]2 = 49 ∴ l(OP) = √49 …[Taking square root of both sides] ∴ l(OP) = 7 cm ∴The distance of the chord from the centre of the circle is 7 cm.

____________________________

Let AB=12 cm be the chord of the circle with radius AO=10 cm

Draw OP⊥AB Join OA

AP=

2

1

AB=

2

1

×12=6cm

In △APO,∠P=90

∘

∴AO

2

=AP

2

+OP

2

⇒10

2

=6

2

+OP

2

⇒OP

2

=100−36

⇒OP

2

=64

⇒OP=8cm

Hence the distance of the chord from the circle is 8 cm.

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Answer By shreyas