## 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 = \(\frac { 1 }{ 2 }\) AB [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]

∴ AP = \(\frac { 1 }{ 2 }\) x 12 = 6 cm ….(i)

ii. In ∆OPA, ∠OPA = 90°

∴ OA^{2} = OP^{2} + AP^{2} [Pythagoras theorem]

= 8^{2} + 6^{2} [From (i)]

= 64 + 36

∴ OA^{2} = 100

∴ OA = \(\sqrt { 100 }\) [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) = \(\frac { d }{ 2 }\) = \(\frac { 26 }{ 2 }\) = 13 cm ……(i)

∴ PR = \(\frac { 1 }{ 2 }\) PQ [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]

= \(\frac { 1 }{ 2 }\) x 24 = 12 cm …..(ii)

ii. In ∆ORP, ∠ORP = 90°

∴ OP^{2}= OR^{2} + PR^{2} [Pythagoras theorem]

∴ 13^{2} = OR^{2} + 12^{2} [From (i) and (ii)]

∴ 169 = OR^{2} + 144

∴ OR^{2} = 169 – 144

∴ OR^{2} = 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°

∴ AP^{2} = AM^{2} + PM^{2} [Pythagoras theorem]

34^{2} = 30^{2} + PM^{2}

∴ PM^{2} = 34^{2} – 30^{2}

∴ PM^{2} (34 – 30)(34 + 30) [a^{2} – b^{2} = (a – b)(a + b)]

= 4 x 64

∴ PM = \(\sqrt { 4\times64 }\) ………(i) [Taking square root on both sides]

= 2 x 8 = 16cm

ii. Now, PM = \(\frac { 1 }{ 2 }\)(PQ) [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]

16 = \(\frac { 1 }{ 2 }\)(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. \(\frac { 1 }{ 2 }\)PM = (PQ) [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]

= \(\frac { 1 }{ 2 }\)(80) = 40 Units ….(i)

ii. In ∆OMP, ∠OMP = 90°

∴ OP^{2} = OM^{2} + PM^{2} [Pythagoras theorem]

∴ 41^{2} = OM^{2} + 40^{2} [From (i)]

∴ OM^{2} = 41^{2} – 40^{2}

= (41 -40) (41 +40) [a^{2} – b^{2} = (a – b) (a + b)]

= (1)(81)

∴ OM^{2} = 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]