radius of the circle is 0 is 41 unit length of a chord PQ is 80 unit find the distance of the chord from the centre of the circle About the author Athena
Step-by-step explanation: Given: In a circle with centre O, OP is radius and PQ is its chord, seg OM ⊥ chord PQ, P-M-Q OP = 41 units, PQ = 80 units, To Find: Distance of the chord from the centre of the circle(OM) i. (1/2) PM = (PQ) [Perpendicular drawn from the centre of a circle to the chord bisects the chord.] = (1/2) (80) = 40 Units ….(i) ii. In ∆OMP, ∠OMP = 90° ∴ OP2 = OM2 + PM2 [Pythagoras theorem] ∴ 412 = OM2 + 402 [From (i)] ∴ OM2 = 412 – 402 = (41 – 40) (41 + 40) [a2 – b2 = (a – b) (a + b)] = (1)(81) ∴ OM2 = 81 OM = √81 = 9 units [Taking square root on both sides] [From (i)] ∴ The distance of the chord from the centre of the circle is 9 units. Reply
Answer:
9 unit
Step-by-step explanation:
Hope it will help you
Step-by-step explanation:
Given: In a circle with centre O, OP is radius and PQ is its chord, seg OM ⊥ chord PQ, P-M-Q OP = 41 units, PQ = 80 units,
To Find: Distance of the chord from the centre of the circle(OM)
i. (1/2) PM = (PQ) [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
= (1/2) (80) = 40 Units ….(i)
ii. In ∆OMP, ∠OMP = 90°
∴ OP2 = OM2 + PM2 [Pythagoras theorem]
∴ 412 = OM2 + 402 [From (i)]
∴ OM2 = 412 – 402 = (41 – 40) (41 + 40) [a2 – b2 = (a – b) (a + b)] = (1)(81)
∴ OM2 = 81
OM = √81 = 9 units [Taking square root on both sides] [From (i)]
∴ The distance of the chord from the centre of the circle is 9 units.