Jharkhand Board JAC Class 9 Maths Important Questions Chapter 10 Circles Important Questions and Answers.

## JAC Board Class 9th Maths Important Questions Chapter 10 Circles

Question 1.

In figure, AB = CB and is the centre of the circle. Prove that BO bisects ∠ABC.

Solution :

Given: In figure, AB = CB and O is the centre of the circle.

To Prove : BO bisects ∠ABC.

Construction: Join OA and OC.

Proof: In ΔOAB and ΔOCB,

OA = OC [Radii of the same circle]

AB = CB [Given]

OB = ОВ (Common)

∴ ΔOAB ≅ ΔOCB [By SSS]

∴ ∠ABO = ∠CBO [By CPCT]

⇒ BO bisects ∠ABC. Hence, Proved.

Question 2.

In figure, AB = AC and O is the centre of the circle. Prove that OA is the perpendicular bisector of BC.

Solution :

Given: In figure, AB = AC and O is the centre of the circle.

To Prove: OA is the perpendicular bisector of BC.

Construction: Join OB and OC.

Proof: AB = AC [Given]

∴ chord AB = chord AC.

[∵ If two arcs of a circle are congruent, then their corresponding chords are equal.]

∴ ∠AOB = ∠AOC ……(i) [∵ Equal chords of a circle subtend equal angles at the centre]

In ΔOBD and ΔOCD,

∠DOB = ∠DOC [From (i)]

OB = OC [Radii of the same circle]

OD = OD (Common)

∴ ΔOBD ≅ ΔOCD [By SAS]

∠ODB = ∠ODC …(ii) (By CPCT)

And BD = CD ……..(iii) [By CPCT]

∴ ∠ODB + ∠ODC = 180° [Linear pair]

⇒ ∠ODB + ∠ODB = 180°

[From equation (ii)]

⇒ 2∠ODB = 180°

⇒ ∠ODB = 90°

∴ ∠ODB = ∠ODC=90°….(iv) [From(ii)]

So, by (iii) and (iv), OA is the perpendicular bisector of BC. Hence, proved.

Question 3.

Prove that the line joining the midpoints of the two parallel chords of a circle passes through the centre of the circle.

Solution :

Let AB and CD be two parallel chords of a circle whose centre is O.

Let L and M be the mid-points of the chords AB and CD respectively. Join OL and OM. Draw OX || AB or CD.

As L is the mid-point of the chord AB and O is the centre of the circle

∴ ∠OLB = 90°

But, OX || AB

∴ ∠LOX = 90° ………….(i)

[∵ Sum of the consecutive interior angles on the same side of a transversal is 180°]

As, M is the mid-point of the chord CD and O is the centre of the circle.

∴ ∠OMD = 90° [∵ The perpendicular drawn from the centre of a circle to a chord bisects the chord]

But OX || CD ………….(ii)

[∵ Sum of the consecutive interior angles on the same side of a transversal is 180°]

∴ ∠MOX = 90°

From above equations, we get

∠LOX + ∠MOX = 90° + 90° = 180°

⇒ ∠LOM = 180°

⇒ LM is a straight line passing through the centre of the circle.

Hence, proved.

Question 4.

PQ and RS are two parallel chords of a circle whose centre is O and radius is 10 cm. If PQ = 16 cm and RS = 12 cm, find the distance between PQ and RS, if they lie

(i) on the same side of the centre O.

(ii) on opposite sides of the centre O.

Solution :

(i) Draw the perpendicular bisectors OL and OM of PQ and RS respectively.

∵ PQ || RS

∴ OL and OM are in the same line.

⇒ O, L and M are collinear.

Join OP and OR

In right triangle OLP,

OP^{2} = OL^{2} + PL^{2}

[By Pythagoras Theorem]

⇒ (10)^{2} = OL^{2} + (\(\frac {1}{2}\) × PQ)^{2}

[∵ The perpendicular drawn from the centre of a circle to a chord bisects the chord]

⇒ 100 = OL^{2} + (\(\frac {1}{2}\) × 16)^{2}

⇒ 100 = OL^{2} + (8)^{2}

⇒ 100 = OL^{2} + 64

⇒ OL^{2} = 100 – 64 = 36 = (6)^{2}

⇒ OL = 6 cm

In right triangle OMR,

OR^{2} = OM^{2} + RM^{2}

[By Pythagoras Theorem]

⇒ OR^{2} = OM^{2} + (\(\frac {1}{2}\)× RS)^{2}

[∵ The perpendicular drawn from the centre of a circle to a chord bisects the chord]

⇒ (10)^{2} = OM^{2} + (\(\frac {1}{2}\) × 12)^{2}

⇒ (10)^{2} = OM^{2} + (6)^{2}

⇒ OM^{2} = (10)^{2} – (6)^{2} = 100 – 36 = 64

⇒ OM = 8 cm

∴ LM = OM – OL = 8 – 6 = 2 cm

Hence, the distance between PO and RS, if they lie on the same side of the centre O, is 2 cm.

(ii) Draw the perpendicular bisectors OL and OM to PQ and RS respectively,

∵ PQ || RS

∴ OL and OM are in the same line

⇒ L, O and M are collinear. Join OP and OR.

In right triangle OLP,

OP^{2} = OL^{2} + PL^{2}

[By Pythagoras Theorem]

⇒ OP^{2} = OL^{2} + (\(\frac {1}{2}\) × PQ)^{2}

[∵ The perpendicular drawn from the centre of a circle to a chord bisects the chord]

⇒ (10)^{2} = OL^{2} + (\(\frac {1}{2}\) × 16)^{2}

⇒ 100 = OL^{2} + (8)^{2}

⇒ 100 = OL^{2} + 64

⇒ OL^{2} = 100 – 64

⇒ OL^{2} = 36 = (6)^{2}

⇒ OL = 6 cm

In right triangle OMR,

OR^{2} = OM^{2} + RM^{2}

[By Pythagoras Theorem]

⇒ OR^{2} = OM^{2} + (\(\frac {1}{2}\) × 12)^{2}

[∵ The perpendicular drawn from the centre of a circle to a chord bisects the chord]

⇒ (10)^{2} = OM^{2} + (6)^{2}

⇒ OM^{2} = (10)^{2} – (6)^{2} = (10 – 6)(10 + 6)

⇒ (4)(16) = 64 = (8)^{2}

⇒ OM = 8 cm

∴ LM = OL + OM = 6 + 8 = 14 cm

Hence, the distance between PQ and RS, if they lie on the opposite sides of the centre O, is 14 cm.

Question 5.

Bisector AD of ∠BAC of ΔABC passed through the centre of the circumcircle of ΔABC. Prove that AB = AC.

Solution :

Given: Bisector AD of ∠BAC of ΔABC passed through the centre of the circumcircle of ΔABC,

To Prove: AB = AC.

Construction: Draw OP ⊥ AB and OQ ⊥ AC.

Proof: In ΔAPO and ΔAQO,

∠OPA = ∠OQA

[Each = 90° (by construction)]

∠OAP = ∠OAQ

[Given]

OA = OA (Common)

∴ ΔAPO ≅ ΔAQO

(By AAS congruence crieterion)

∴ OP = OQ [By CPCT]

∴ AB = AC. [∵ Chords equidistant from the centre are equal] Hence, proved.

Question 6.

In figure, ∠ABC = 79°, ∠ACB = 41°, find ∠BDC.

Solution :

In ΔABC.

∠BAC + ∠ABC + ∠ACB = 180°

[Sum of all the angles of a triangle is 180°]

⇒ ∠BAC + 79° + 41° = 180°

⇒ ∠BAC + 120° = 180°

⇒ ∠BAC = 180° – 120° = 60°

Now, ∠BDC = ∠BAC = 60°

[Angles in the same segment of a circle are equal]

Question 7.

ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 80°, ∠BAC = 40°, find ∠BCD. Further, if AB = BC, find ∠ECD.

Solution :

∠CDB = ∠BAC = 40° ………….(i)

[Angles in the same segment of a circle are equal]

∠DBC = 80° ………….(ii)

In ΔBCD

∠BCD + ∠DBC + ∠CDB = 180°

[Sum of all the angles of a triangle is 180°]

⇒ ∠BCD + 80° + 40° = 180°

[Using (i) and (ii)]

⇒ ∠BCD + 120° = 180°

⇒ ∠BCD = 180° – 120°

⇒ ∠BCD = 60° ………………(ii)

In ΔABC,

AB = BC

∴ ∠BCA = ∠BAC = 40° …(iv)

[Angles opposite to equal sides of a triangle are equal]

Now, ∠BCD = 60° [From (iii)]

⇒ ∠BCA + ∠ECD = 60°

⇒ 40° + ∠ECD = 60°

⇒ ∠ECD = 60° – 40°

⇒ ∠ECD = 20°

Question 8.

Find the area of a triangle, the radius of whose circumcircle is 3 cm and the length of the altitude drawn from the opposite vertex to the hypotenuse is 2 cm.

Solution :

We know that the hypotenuse of a right-angled triangle is the diameter of its circumcircle.

∴ BC = 2(OB) = 2 × 3 = 6 cm

Let, AD ⊥ BC

AD = 2 cm [Given]

∴ Area of ΔABC = \(\frac {1}{2}\)(BC)(AD)

= \(\frac {1}{2}\)(6)(2) = 6 cm^{2}.

Question 9.

In figure, PQ is a diameter of a circle with centre O. If ∠PQR = 65°, ∠SPR = 40°, ∠PQM = 50°, find ∠QPR, ∠PRS and ∠QPM.

Solution :

(i) ∵ PQ is a diameter

∴ ∠PRQ = 90°

[Angle in a semi-circle is 90°]

In ΔPQR,

∠QPR + ∠PRQ + ∠PQR = 180°

[Angle sum property of a triangle]

⇒ ∠QPR + 90° + 65° = 180°

⇒ ∠QPR = 180° – 155° = 25°

(ii) PQRS is a cyclic quadrilateral

∴ ∠PSR + ∠PQR = 180°

[∵ Opposite angles of a cyclic quadrilateral are supplementary]

⇒ ∠PSR + 65° = 180°

⇒ ∠PSR = 180° – 65°

⇒ ∠PSR = 115°

In ΔPSR

∠PSR + ∠SPR + ∠PRS = 180°

[Angle sum property of a triangle]

⇒ 115° + 40° + ∠PRS = 180°

⇒ 155° + ∠PRS = 180°

⇒ ∠PRS = 180° – 155°

⇒ ∠PRS = 25°

(iii) PQ is a diameter

∴ ∠PMQ = 90°

[∵ Angle in a semi-circle is 90°]

In ΔPMQ,

∠PMQ + ∠PQM + ∠QPM = 180°

[Angle sum property of a triangle]

⇒ 90° + 50° + ∠QPM = 180°

⇒ 140° + ∠QPM = 180°

⇒ ∠QPM = 180° – 140°

⇒ ∠QPM = 40°

Multiple Choice Questions

Question 1.

If two circular wheels rotate on a horizontal road then locus of their centres will be

(a) Circles

(b) Rectangle

(c) Two straight lines

(d) Parallelogram

Solution :

(c) Two straight lines

Question 2.

In a circle of radius 10 cm, the length of chord whose distance is 6 cm from the centre is

(a) 4 cm

(b) 5 cm

(c) 8 cm

(d) 16 cm

Solution :

(d) 16 cm

Question 3.

If a chord a length 8 cm is situated at a distance of 3 cm form centre, then the diameter of circle is:

(a) 11 cm

(b) 10 cm

(c) 12 cm

(d) 15 cm

Solution :

(b) 10 cm

Question 4.

In a circle the lengths of chords which are situated at a equal distance from centre are :

(a) double

(b) four times

(c) equal

(d) three times

Solution :

(c) equal

Question 5.

In the given figure, O is the centre of the circle and ∠BDC = 42°. The ∠ACB is equal to:

(a) 48°

(b) 45°

(c) 42°

(d) 60°

Solution :

(a) 48°

Question 6.

In the given figure, ∠CAB = 80°, ∠ABC = 40°. The sum of ∠DAB and ∠ABD is equal to:

(a) 80°

(b) 100°

(c) 120°

(d) 140°

Solution :

(c) 120°

Question 7.

In the given figure, if C is the centre of the circle and ∠PQC = 25° and ∠PRC = 15°, then ∠QCR is equal to:

(a) 40°

(b) 60°

(c) 80°

(d) 120°

Solution :

(c) 80°

Question 8.

In a cyclic quadrilateral if ∠B – ∠D = 60°, then the smaller of the angles B and D is:

(a) 30°

(b) 45°

(c) 60°

(d) 75°

Solution :

(c) 60°

Question 9.

Three wires of length l_{1}, l_{2}, l_{3}, form a triangle surmounted by another circular wire. If l_{3} is the diameter and l_{3} = 2l_{1}, then the angle between l_{1} and l_{3} will be

(a) 30°

(b) 60°

(c) 45°

(d) 90°

Solution :

(b) 60°

Question 10.

In a circle with centre O, OD ⊥ chord AB. If BC is the diameter, then:

(a) AC = BC

(b) OD = BC

(c) AC = 2OD

(d) None of these

Solution :

(c) AC = 2OD

Question 11.

The sides AB and DC of cyclic quadrilateral ABCD are produced to meet at P, the sides AD and BC are produced to meet at Q. If ∠ADC = 85° and ∠BPC = 40°, then ∠CQD equals:

(a) 30°

(b) 45°

(c) 60°

(d) 75°

Solution :

(a) 30°

Question 12.

In the given figure, if ∠ACB = 40°, ∠DPB = 120°, then ∠CBD is equal to

(a) 40°

(b) 20°

(c) 0°

(d) 60°

Solution :

(a) 40°

Question 13.

Any cyclic parallelogram is a:

(a) rectangle

(b) rhombus

(c) trapezium

(d) square

Solution :

(a) rectangle

Question 14.

The locus of the centre of all circles of given radius r, in the same plane, passing through a fixed point is:

(a) A point

(b) A circle

(c) A straight line

(d) Two straight lines

Solution :

(b) A circle

Question 15.

In a cyclic quadrilateral if ∠A – ∠C = 70°, then the greater of the angles A and C is equal to:

(a) 95°

(b) 105°

(c) 125°

(d) 115°

Solution :

(c) 125°

Question 16.

The length of a chord of a circle is equal to the radius of the circle. The angle which this chord subtends on the longer segment of the circle is equal to :

(a) 30°

(b) 45°

(c) 60°

(d) 90°

Solution :

(a) 30°

Question 17.

If a trapezium is cyclic then,

(a) Its parallel sides are equal.

(b) Its non-parallel sides are equal.

(c) Its diagonals are not equal.

(d) None of these

Solution :

(b) Its non-parallel sides are equal.