Practice Set 6.1 Geometry 9th Std Maths Part 2 Answers Chapter 6 Circle
Solution & Step-by-Step Answer:
i. AP = AB [Perpendicular drawn from the centre of a circle to the chord bisects the chord.] ∴ AP = x 12 = 6 cm ….(i)
ii. In ∆OPA, ∠OPA = 90°
∴ OA2= OP2+ AP2[Pythagoras theorem]
= 82+ 62[From (i)]
= 64 + 36
∴ OA2= 100
∴ OA = [Taking square root on both sides]
= 10 cm
iii. Radius (r) = 10 cm
∴ Diameter = 2r = 2 x 10 = 20 cm
∴ The diameter of the circle is 20 cm.
Solution & Step-by-Step Answer:
Given: In a circle with centre O, PO is radius and PQ is its chord, seg OR ⊥ chord PQ, P-R-Q PQ = 24 cm, diameter (d) = 26 cm To Find: Distance of the chord from the centre (OR) Solution: Radius (OP) = = = 13 cm ……(i) ∴ PR = PQ [Perpendicular drawn from the centre of a circle to the chord bisects the chord.] = x 24 = 12 cm …..(ii)
ii. In ∆ORP, ∠ORP = 90°
∴ OP2= OR2+ PR2[Pythagoras theorem]
∴ 132= OR2+ 122[From (i) and (ii)]
∴ 169 = OR2+ 144
∴ OR2= 169 – 144
∴ OR2= 25
∴ OR = √25 = 5 cm [Taking square root on both sides]
∴ The distance of the chord from the centre of the circle is 5 cm.
Solution & Step-by-Step Answer:
I. In ∆AMP, ∠AMP = 90° ∴ AP2 = AM2 + PM2 [Pythagoras theorem] 342 = 302 + PM2 ∴ PM2 = 342 – 302 ∴ PM2 (34 – 30)(34 + 30) [a2 – b2 = (a – b)(a + b)] = 4 x 64 ∴ PM = ………(i) [Taking square root on both sides] = 2 x 8 = 16cm
ii. Now, PM = (PQ) [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
16 = (PQ) [From (i)]
∴ PQ = 16 x 2
= 32cm
∴ The length of the chord of the circle is 32cm.
Solution & Step-by-Step Answer:
i. PM = (PQ) [Perpendicular drawn from the centre of a circle to the chord bisects the chord.] = (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.
Solution & Step-by-Step Answer:
Proof: For smaller circle, seg OM ⊥ chord PQ [Construction, A-P-M, M-Q-B] ∴ PM = MQ …..(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.] For bigger circle, seg OM ⊥ chord AB [Construction] ∴ AM = MB [Perpendicular drawn from the centre of the circle to the chord bisects the chord.] ∴ AP + PM = MQ + QB [A-P-M, M-Q-B] ∴ AP + MQ = MQ + QB [From (i)] ∴ AP = BQ
Solution & Step-by-Step Answer:
Given: O is the centre of the circle. seg PQ is the diameter. Diameter PQ bisects the chords AB and CD in points M and N respectively. To prove: chord AB || chord CD. Proof: Diameter PQ bisects the chord AB in point M [Given] ∴ seg AM ≅ seg BM ∴ seg OM ⊥ chord AB [Segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord, P-M-O, O-N-Q] ∴ ∠OMA = 90° …..(i) Also, diameter PQ bisects the chord CD in point N [Given] ∴ seg CN ≅ seg DN seg ON ⊥ chord CD [Segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord, P-M-O, O-N-Q] ∴ ∠ONC = 90° …..(ii) Now, ∠OMA + ∠ONC = 90° + 90° [From (i) and (ii)] = 180° But, ∠OMA and ∠ONC form a pair of interior angles on lines AB and CD when seg MN is their transversal. ∴ chord AB || chord CD [Interior angles test]