Circle class 9

Maharashtra State Board Class 9 Maths Solutions Chapter 6 Circle Practice Set 6.1

Question 1.
Distance of chord AB from the centre of a circle is 8 cm. Length of the chord AB is 12 cm. Find the diameter of a circle.
Given: In a circle with centre O,
OA is radius and AB is its chord,
seg OP ⊥ chord AB, A-P-B
AB = 12 cm, OP =8 cm
To Find: Diameter of the circle
Solution:

i. AP = 1/2 AB [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
∴ AP = 1/2 x 12 = 6 cm ….(i)

ii. In ∆OPA, ∠OPA = 90°
∴ OA² = OP² + AP² [Pythagoras theorem]
= 8² + 6² [From (i)]
= 64 + 36
∴ OA² = 100
∴ OA = 100−−−√ [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.

Question 2.
Diameter of a circle is 26 cm and length of a chord of the circle is 24 cm. Find the distance of the chord from the centre.
Solution:

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) = d/2 = 26/2 = 13 cm ……(i)
∴ PR = 12 PQ [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
= 1/2 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.

Question 3.
Radius of a circle is 34 cm and the distance of the chord from the centre is 30 cm, find the length of the chord.
Given: in a circle with centre A,
PA is radius and PQ is chord,
seg AM ⊥ chord PQ, P-M-Q
AP = 34 cm, AM = 30 cm
To Find: Length of the chord (PQ)
Solution:

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 = √4×√64 ………(i) [Taking square root on both sides]
= 2 x 8 = 16cm

ii. Now, PM = 12(PQ) [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
16 = 1/2(PQ) [From (i)]
∴ PQ = 16 x 2
          = 32cm
∴ The length of the chord of the circle is 32cm.

Question 4.
Radius of a circle with centre O is 41 units. Length of a chord PQ is 80 units, find the distance of the chord from the centre of the circle.
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)
Solution:

i. 12PM = (PQ) [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
= 12(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.

Question 5.
In the adjoining figure, centre of two circles is O. Chord AB of bigger circle intersects the smaller circle in points P and Q. Show that AP = BQ.

Given: Two concentric circles having centre O.
To prove: AP = BQ
Construction: Draw seg OM ⊥ chord AB, A-M-B
Solution:
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

Question 6.
Prove that, if a diameter of a circle bisects two chords of the circle then those two chords are parallel to each other.
Solution:

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]

Maharashtra State Board Class 9 Maths Solutions Chapter 6 Circle Practice Set 6.2

Question 1.
Radius of circle is 10 cm. There are two chords of length 16 cm each. What will be the distance of these chords from the centre of the circle ?
Given: In a circle with centre O,
OR and OP are radii and RS and PQ are its congruent chords.
PQ = RS= 16 cm,
OR = OP = 10 cm
seg OU ⊥ chord PQ, P-U-Q
seg OT ⊥ chord RS, R-T-S
To find: Distance of chords from centre of the circle.
Solution:

i. PU = 12(PQ) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ PU= 12 x 16 = 8 cm …(i)

ii. In ∆OUP, ∠OUP = 90°
∴ OP² = OU² + PU² [Pythagoras theorem]
∴ 10² = OU² + 8² [From (i)]
∴ 100 = OU² + 64
∴ OU² = 100 – 64 = 36
∴ OU = √36 [Taking square root on both sides]
∴ OU = 6 cm

iii. Now, OT = OU [Congruent chords of a circle are equidistant from the centre.]
∴ OT = OU = 6cm
∴ The distance of the chords from the centre of the circle is 6 cm.

Question 2.
In a circle with radius 13 cm, two equal chords are at a distance of 5 cm from the centre. Find the lengths of chords.
Given: In a circle with cente O,
OA and OC are the radii and AB and CD are its congruent chords,
OA = OC = 13cm
0E = OF = 5 cm
seg 0E ⊥ chord CD, C-E-D
seg OF ⊥ chord AB. A-F-B
To find: length of the chords
Solution:

i. In ∆AFO, ∠AFO = 90°
∴ AO² = AF² + FO² [Pythagoras theorem]
∴ 13² = AF² + 5²
∴ 169 = AF² + 25
∴ AF² = 169-25
∴ AF² = 144
∴ AF = 144−−−√ [Taking square root on both sides]
∴ AF = 12 cm …..(i)

ii. Now AF = 12AB [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ 12 = 12 (AB) [From (i)]
∴ AB = 12 x 2 = 24 cm
∴ CD = AB = 24 cm [chord AB ≅ chord CD]
∴ The lengths of the two chords are 24 cm each.

Question 3.
Seg PM and seg PN are congruent chords of a circle with centre C. Show that the ray PC is the bisector of ∠NPM.
Given: Point C is the centre of the circle.
chord PM ≅ chord PN
To prove: Ray PC is the bisector of ∠NPM.
Construction: Draw seg CR ⊥ chord PN, P-R-N
seg CQ ⊥ chord PM, P-Q-M
Proof:

chord PM chord PN [Given]
seg CR ⊥ chord PN
seg CQ ⊥ chord PM [Construction]
∴ segCR ≅ segCQ ….(i) [Congruent chords are equidistant from the centre]
In ∆PRC and ∆PQC,
∠PRC ≅ ∠PQC [Each is of 90°]
segCR ≅ segCQ [From (i)]
seg PC ≅ seg PC [Common side]
∴ ∆PRC ≅ ∆PQC [Hypotenuse side test]
∴ ∠RPC ≅ ∠QPC [c. a. c. t.]
∴ ∠NPC ≅ ∠MPC [N- R-P, M-Q-P]
∴ Ray PC is the bisector of ∠NPM.

Maharashtra Board Class 9 Maths 

Chapter 6 Circle

 Practice Set 6.2

 Intext Questions and Activities

Question 1.
Prove the following two theorems for two congruent circles. (Textbook pg. no. 81)
i. Congruent chords in congruent circles are equidistant from their respective centres.
ii. Chords of congruent circles which are equidistant from their respective centres are congruent.
Write ‘Given’. ‘To prove’ and the proofs of these theorems.
Solution:
(i) Congruent chords in congruent circles are equidistant from their respective centres.
Given: Point P and point Q are the centres of congruent circles.
chord AB ≅ chord CD
seg PM ⊥ chord AB, A-M-B
seg QN ⊥ chord CD, C-N-D
To prove: PM = QN

Construction: Draw seg PA and seg QC.
Proof:
seg PM ⊥ chord AB, seg QN ⊥ chord CD [Given]
∴ AM = 12(AB) ………(i) [Perpendicular drawn from the centre of the circle to the
∴ CN = 12(CD) ……..(ii) chord bisects the chord.]
But, AB = CD ………(iii) [Given]
∴ AM = CN [From (i), (ii) and (iii)]
i.e., segAM ≅ segCN ….(iv) [Segments of equal lengths]
In ∆PMA and ∆QNC,
∠PMA ≅ ∠QNC [Each is of 90°]
hypotenuse PA ≅ hypotenuse QC [Radii of congruent circles]
seg AM ≅ seg CN [From (iv)]
∴ ∆PMA ≅ ∆QNC [Hypotenuse side test]
∴ segPM ≅ segQN [c. s. c. t.]
∴ PM ≅ QN [Length of congruent segments]

(ii) Chords of congruent circles which are equidistant from their respective centres are congruent.

Given: Point P and point Q are the centres of congruent circles.
seg PM ⊥ chord AB, A-M-B
seg QN ⊥ chord CD, C-N-D
PM = QN
To prove: chord AB ≅ chord CD
Construction: Draw seg PA and seg QC.
Proof:
In ∆PMA and ∆QNC,
∴ ∠PMA ≅ ∠QNC [Each is of 90°]
seg PM ≅ seg QN [Given]
hypotenuse PA ≅ hypotenuse QC [Radii of the congruent circles]
∴ ∆PMA ≅ ∆QNC [Hypotenuse side test]
∴ seg AM ≅ seg CN [c. s. c. t.]
∴ AM = CN ….(i) [Length of congruent segments]
Now, seg PM ⊥ chord AB, and seg QN ⊥ chord CD
∴ AM = 12(AB) …(ii)
∴ CN = 12 (CD) ..(iii) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ AB = CD [From (i), (ii) and (ii)]
∴ chord AB ≅ chord CD [Segments of equal lengths]