## Class 9 Maths Chapter 10 Circles

Ex 10.1 Class 9 Maths Question 1.

Fill in the blanks.

(i) The centre of a circle lies in ___ of the circle. (exterior/interior)

(ii) A point, whose distance from the centre of a circle is greater than its radius lies in ____ of the circle, (exterior/interior)

(iii) The longest chord of a circle is a ____ of the circle.

(iv) An arc is a ____ when its ends are the ends of a diameter.

(v) Segment of a circle is the region between an arc and ____ of the circle.

(vi) A circle divides the plane, on which it lies, in ____ parts.

Solution:

(i) interior

(ii) exterior

(iii) diameter

(iv) semicircle

(v) the chord

(vi) three

Ex 10.1 Class 9 Maths Question 2.

Write True or False. Give reason for your answers.

(i) Line segment joining the centre to any point on the circle is a , radius of the circle.

(ii) A circle has only finite number of equal chords.

(iii) If a circle is divided into three equal arcs, each is a major arc.

(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.

(v) Sector is the region between the chord and its corresponding arc.

(vi) A circle is a plane figure.

Solution:

(i) True [∵ All points on the circle are equidistant from the centre]

(ii) False [ ∵ A circle can have an infinite number of equal chords]

(iii) False [∵ Each part will be less than a semicircle]

(iv) True [ ∵ Diameter = 2 x Radius]

(v) False [ ∵ The region between the chord and its corresponding arc is a segment]

(vi) True [ ∵ A circle is drawn on a plane]

Ex 10.1 Ex 10.2 Class 9 Maths Question 1.

Recall that two circles are congruent, if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres

Solution:

Given: Two congruent circles with centres O and O’ and radii r, which have chords AB and CD respectively such that AB = CD.

To Prove: ∠AOB = ∠CO’D

Proof: In ∆AOB and ∆CO’D, we have

AB = CD [Given]

OA = O’C [Each equal to r]

OB = O’D [Each equal to r]

∴ ∆AOB ≅ ∆CO’D [By SSS congruence criteria]

⇒ ∠AOB = ∠CO’D [C.P.C.T.]

Ex 10.2 Class 9 Maths Question 2.

Prove that, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Solution:

Given: Two congruent circles with centres O & O’ and radii r which have chords AB and CD respectively such that ∠AOB = ∠CO’D.

To Prove: AB = CD

Proof: In ∆AOB and ∆CO’D, we have

OA = O’C [Each equal to r]

OB = O’D [Each equal to r]

∠AOB = ∠CO’D [Given]

∴ ∆AOB ≅ ∆CO’D [By SAS congruence criteria]

Hence, AB = CD [C.P.C.T.]

Ex 10.3 Class 9 Maths Question 1.

Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Solution:

Let us draw different pairs of circles as shown below:

We have

Figure | Maximum number of common points |

(i) | nil |

(ii) | one |

(«i) | two |

Thus, two circles can have at the most two points in common.

Ex 10.3 Class 9 Maths Question 2.

Suppose you are given a circle. Give a construction to find its centre.

Solution:

Steps of construction :

Step I : Take any three points on the given circle. Let these points be A, B and C.

Step II : Join AB and BC.

Step III : Draw the perpendicular bisector, PQ of AB.

Step IV: Draw the perpendicular bisector, RS of BC such that it intersects PQ at O.

Thus, ‘O’ is the required centre of the given drcle.

Ex 10.3 Class 9 Maths Question 3.

If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Solution:

We have two circles with centres O and O’, intersecting at A and B.

∴ AB is the common chord of two circles and OO’ is the line segment joining their centres.

Let OO’ and AB intersect each other at M.

∴ To prove that OO’ is the perpendicular bisector of AB,

we join OA, OB, O’A and O’B. Now, in ∆QAO’ and ∆OBO’,

we have

OA = OB [Radii of the same circle]

O’A = O’B [Radii of the same circle]

OO’ = OO’ [Common]

∴ ∆OAO’ ≅ ∆OBO’ [By SSS congruence criteria]

⇒ ∠1 = ∠2 , [C.P.C.T.]

Now, in ∆AOM and ∆BOM, we have

OA = OB [Radii of the same circle]

OM = OM [Common]

∠1 = ∠2 [Proved above]

∴ ∆AOM = ∆BOM [By SAS congruence criteria]

⇒ ∠3 = ∠4 [C.P.C.T.]

But ∠3 + ∠4 = 180° [Linear pair]

∴∠3=∠4 = 90°

⇒ AM ⊥ OO’

Also, AM = BM [C.P.C.T.]

⇒ M is the mid-point of AB.

Thus, OO’ is the perpendicular bisector of AB.

Ex 10.4 Class 9 Maths Question 1.

Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

Solution:

We have two intersecting circles with centres at O and O’ respectively. Let PQ be the common chord.

∵ In two intersecting circles, the line joining their centres is perpendicular bisector of the common chord.

∴∠OLP = ∠OLQ = 90° and PL = LQ

Now, in right ∆OLP, we have

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

⇒ PL^{2} + (4 – x)^{2} = 5^{2}

⇒ PL^{2} = 5^{2} – (4 – x)^{2}

⇒ PL^{2} = 25 -16 – x^{2} + 8x

⇒ PL^{2} = 9 – x^{2} + 8x …(i)

Again, in right ∆O’LP,

PL^{2} = PO^{‘2} – LO^{‘2}

= 3^{2} – x^{2} = 9 – x^{2} …(ii)

From (i) and (ii), we have

9 – x^{2} + 8x = 9 – x^{2}

⇒ 8x = 0

⇒ x = 0

⇒ L and O’ coincide.

∴ PQ is a diameter of the smaller circle.

⇒ PL = 3 cm

But PL = LQ

∴ LQ = 3 cm

∴ PQ = PL + LQ = 3cm + 3cm = 6cm

Thus, the required length of the common chord = 6 cm.

Ex 10.4 Class 9 Maths Question 2.

If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Solution:

Given: A circle with centre O and equal chords AB and CD intersect at E.

To Prove: AE = DE and CE = BE

Construction : Draw OM ⊥ AB and ON ⊥ CD.

Join OE.

Proof: Since AB = CD [Given]

∴ OM = ON [Equal chords are equidistant from the centre]

Now, in ∆OME and ∆ONE, we have

∠OME = ∠ONE [Each equal to 90°]

OM = ON [Proved above]

OE = OE [Common hypotenuse]

∴ ∆OME ≅ ∆ONE [By RHS congruence criteria]

⇒ ME = NE [C.P.C.T.]

Adding AM on both sides, we get

⇒ AM + ME = AM + NE

⇒ AE = DN + NE = DE

∵ AB = CD ⇒ 12AB = 12DC

⇒ AM = DN

⇒ AE = DE …(i)

Now, AB – AE = CD – DE

⇒ BE = CE …….(ii)

From (i) and (ii), we have

AE = DE and CE = BE

Ex 10.4 Class 9 Maths Question 3.

If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Solution:

Given: A circle with centre O and equal chords AB and CD are intersecting at E.

To Prove : ∠OEA = ∠OED

Construction: Draw OM ⊥ AB and ON ⊥ CD.

Join OE.

Proof: In ∆OME and ∆ONE,

OM = ON

[Equal chords are equidistant from the centre]

OE = OE [Common hypotenuse]

∠OME = ∠ONE [Each equal to 90°]

∴ ∆OME ≅ ∆ONE [By RHS congruence criteria]

⇒ ∠OEM = ∠OEN [C.P.C.T.]

⇒ ∠OEA = ∠OED

Ex 10.4 Class 9 Maths Question 4.

If a line intersects two concentric circles (circles with the same centre) with centre 0 at A, B, C and D, prove that AB = CD (see figure).

Solution:

Given : Two circles with the common centre O.

A line D intersects the outer circle at A and D and the inner circle at B and C.

To Prove : AB = CD.

Construction:

Draw OM ⊥ l.

Proof: For the outer circle,

OM ⊥ l [By construction]

∴ AM = MD …(i)

[Perpendicular from the centre to the chord bisects the chord]

For the inner circle,

OM ⊥ l [By construction]

∴ BM = MC …(ii)

[Perpendicular from the centre to the chord bisects the chord]

Subtracting (ii) from (i), we have

AM – BM = MD – MC

⇒ AB = CD

Ex 10.4 Class 9 Maths Question 5.

Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?

Solution:

Let the three girls Reshma, Salma and Mandip be positioned at R, S and M respectively on the circle with centre O and radius 5 m such that

RS = SM = 6 m [Given]

Equal chords of a circle subtend equal angles at the centre.

∴ ∠1 = ∠2

In ∆POR and ∆POM, we have

OP = OP [Common]

OR = OM [Radii of the same circle]

∠1 = ∠2 [Proved above]

∴ ∆POR ≅ ∆POM [By SAS congruence criteria]

∴ PR = PM and

∠OPR = ∠OPM [C.P.C.T.]

∵∠OPR + ∠OPM = 180° [Linear pair]

∴∠OPR = ∠OPM = 90°

⇒ OP ⊥ RM

Now, in ∆RSP and ∆MSP, we have

RS = MS [Each 6 cm]

SP = SP [Common]

PR = PM [Proved above]

∴ ∆RSP ≅ ∆MSP [By SSS congruence criteria]

⇒ ∠RPS = ∠MPS [C.P.C.T.]

But ∠RPS + ∠MPS = 180° [Linear pair]

⇒ ∠RPS = ∠MPS = 90°

SP passes through O.

Let OP = x m

∴ SP = (5 – x)m

Now, in right ∆OPR, we have

x^{2} + RP^{2} = 5^{2}

RP^{2} = 5^{2} – x^{2}

In right ∆SPR, we have

(5 – x)^{2} + RP^{2} = 6^{2}

⇒ RP^{2} = 6^{2} – (5 – x)^{2} ……..(ii)

From (i) and (ii), we get

⇒ 5^{2} – x^{2} = 6^{2} – (5 – x)^{2}

⇒ 25 – x^{2} = 36 – [25 – 10x + x^{2}]

⇒ – 10x + 14 = 0

⇒ 10x = 14 ⇒ x = 1410 = 1.4

Now, RP^{2} = 5^{2} – x^{2}

⇒ RP^{2} = 25 – (1.4)^{2}

⇒ RP^{2} = 25 – 1.96 = 23.04

∴ RP = 23.04−−−−√= 4.8

∴ RM = 2RP = 2 x 4.8 = 9.6

Thus, distance between Reshma and Mandip is 9.6 m.

Ex 10.4 Class 9 Maths Question 6.

A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Solution:

Let Ankur, Syed and David are sitting at A, S and D respectively in the circular park with centre O such that AS = SD = DA

i. e., ∆ASD is an equilateral triangle.

Let the length of each side of the equilateral triangle be 2x.

Draw AM ⊥ SD.

Since ∆ASD is an equilateral triangle.

∴ AM passes through O.

⇒ SM = 12 SD = 12 (2x)

⇒ SM = x

Now, in right ∆ASM, we have

AM^{2} + SM^{2} = AS^{2} [Using Pythagoras theorem]

⇒ AM^{2}= AS^{2} – SM^{2}= (2x)^{2} – x^{2}

= 4x^{2} – x^{2} = 3x^{2}

⇒ AM = 3x−−√m

Now, OM = AM – OA= (3x−−√ – 20)m

Again, in right ∆OSM, we have

OS^{2} = SM^{2} + OM^{2} [using Pythagoras theorem]

20^{2} = x^{2} + (3x−−√ – 20)^{2}

⇒ 400 = x^{2} + 3x^{2} – 403x−−√ + 400

⇒ 4x^{2} = 40 3x−−√

⇒ x = 10√3 m

Now, SD = 2x = 2 x 10√3 m = 20√3 m

Thus, the length of the string of each phone = 20√3 m

Ex 10.5 Class 9 Maths Question 1.

In figure A,B and C are three points on a circle with centre 0 such that ∠BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ ADC.

Solution:

We have a circle with centre O, such that

∠AOB = 60° and ∠BOC = 30°

∵∠AOB + ∠BOC = ∠AOC

∴ ∠AOC = 60° + 30° = 90°

The angle subtended by an arc at the circle is half the angle subtended by it at the centre.

∴ ∠ ADC = 12 (∠AOC) = 12(90°) = 45°

Ex 10.5 Class 9 Maths Question 2.

A chord of a circle is equal to the radius of the circle, find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Solution:

We have a circle having a chord AB equal to radius of the circle.

∴ AO = BO = AB

⇒ ∆AOB is an equilateral triangle.

Since, each angle of an equilateral triangle is 60°.

⇒ ∠AOB = 60°

Since, the arc ACB makes reflex ∠AOB = 360° – 60° = 300° at the centre of the circle and ∠ACB at a point on the minor arc of the circle.

Hence, the angle subtended by the chord on the minor arc = 150°.

Similarly, ∠ADB = 12 [∠AOB] = 12 x 60° = 30°

Hence, the angle subtended by the chord on the major arc = 30°

Ex 10.5 Class 9 Maths Question 3.

In figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Solution:

The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point pn the circumference.

∴ reflex ∠POR = 2∠PQR

But ∠PQR = 100°

∴ reflex ∠POR = 2 x 100° = 200°

Since, ∠POR + reflex ∠POR = 360°

⇒ ∠POR = 360° – 200°

⇒ ∠POR = 160°

Since, OP = OR [Radii of the same circle]

∴ In ∆POR, ∠OPR = ∠ORP

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

Also, ∠OPR + ∠ORP + ∠POR = 180°

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

⇒ ∠OPR + ∠ORP + 160° = 180°

⇒ 2∠OPR = 180° -160° = 20° [∠OPR = ∠ORP]

⇒ ∠OPR = 20∘2 = 10°

Ex 10.5 Class 9 Maths Question 4.

In figure, ∠ABC = 69°,∠ACB = 31°, find ∠BDC.

Solution:

In ∆ABC, ∠ABC + ∠ACB + ∠BAC = 180°

⇒ 69° + 31° + ∠BAC = 180°

⇒ ∠BAC = 180° – 100° = 80°

Since, angles in the same segment are equal.

∴∠BDC = ∠BAC ⇒ ∠BDC = 80°

Ex 10.5 Class 9 Maths Question 5.

In figure, A, B and C are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠BAC.

Solution:

∠BEC = ∠EDC + ∠ECD

[Sum of interior opposite angles is equal to exterior angle]

⇒ 130° = ∠EDC + 20°

⇒ ∠EDC = 130° – 20° = 110°

⇒ ∠BDC = 110°

Since, angles in the same segment are equal.

∴ ∠BAC = ∠BDC

⇒ ∠BAC = 110°

Ex 10.5 Class 9 Maths Question 6.

ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.

Solution:

Since angles in the same segment of a circle are equal.

∴ ∠BAC = ∠BDC

⇒ ∠BDC = 30°

lso, ∠DBC = 70° [Given]

In ∆BCD, we have

∠BCD + ∠DBC + ∠CDB = 180° [Sum of angles of a triangle is 180°]

⇒ ∠BCD + 70° + 30° = 180°

⇒ ∠BCD = 180° -100° = 80°

Now, in ∆ABC,

AB = BC [Given]

∴ ∠BCA = ∠BAC [Angles opposite to equal sides of a triangle are equal]

⇒ ∠BCA = 30° [∵ ∠B AC = 30°]

Now, ∠BCA + ∠BCD = ∠BCD

⇒ 30° + ∠ECD = 80°

⇒ ∠BCD = 80° – 30° = 50°

Ex 10.5 Class 9 Maths Question 7.

If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Solution:

Since AC and BD are diameters.

⇒ AC = BD …(i) [All diameters of a circle are equal]

Also, ∠BAD = 90° [Angle formed in a semicircle is 90°]

Similarly, ∠ABC = 90°, ∠BCD = 90°

and ∠CDA = 90°

Now, in ∆ABC and ∆BAD, we have

AC = BD [From (i)]

AB = BA [Common hypotenuse]

∠ABC = ∠BAD [Each equal to 90°]

∴ ∆ABC ≅ ∆BAD [By RHS congruence criteria]

⇒ BC = AD [C.P.C.T.]

Similarly, AB = DC

Thus, the cyclic quadrilateral ABCD is such that its opposite sides are equal and each of its angle is a right angle.

∴ ABCD is a rectangle.

Ex 10.5 Class 9 Maths Question 8.

If the non – parallel sides of a trapezium are equal, prove that it is cyclic.

Solution:

We have a trapezium ABCD such that AB ॥ CD and AD = BC.

Let us draw BE ॥ AD such that ABED is a parallelogram.

∵ The opposite angles and opposite sides of a parallelogram are equal.

∴ ∠BAD = ∠BED …(i)

and AD = BE …(ii)

But AD = BC [Given] …(iii)

∴ From (ii) and (iii), we have BE = BC

⇒ ∠BCE = ∠BEC … (iv) [Angles opposite to equal sides of a triangle are equal]

Now, ∠BED + ∠BEC = 180° [Linear pair]

⇒ ∠BAD + ∠BCE = 180° [Using (i) and (iv)]

i.e., A pair of opposite angles of a quadrilateral ABCD is 180°.

∴ ABCD is cyclic.

⇒ The trapezium ABCD is cyclic.

Ex 10.5 Class 9 Maths Question 9.

Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A,D and P, Q respectively (see figure). Prove that ∠ ACP = ∠QCD.

Solution:

Since, angles in the same segment of a circle are equal.

∴ ∠ACP = ∠ABP …(i)

Similarly, ∠QCD = ∠QBD …(ii)

Since, ∠ABP = ∠QBD …(iii) [Vertically opposite angles]

∴ From (i), (ii) and (iii), we have

∠ACP = ∠QCD

Ex 10.5 Class 9 Maths Question 10.

If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Solution:

We have ∆ABC, and two circles described with diameter as AB and AC respectively. They intersect at a point D, other than A.

Let us join A and D.

∵ AB is a diameter.

∴∠ADB is an angle formed in a semicircle.

⇒ ∠ADB = 90° ……(i)

Similarly, ∠ADC = 90° ….(ii)

Adding (i) and (ii), we have

∠ADB + ∠ADC = 90° + 90° = 180°

i. e., B, D and C are collinear points.

⇒ BC is a straight line. Thus, D lies on BC.

Ex 10.5 Class 9 Maths Question 11.

ABC and ADC are two right angled triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.

Solution:

We have ∆ABC and ∆ADC such that they are having AC as their common hypotenuse and ∠ADC = 90° = ∠ABC

∴ Both the triangles are in semi-circle. Case – I: If both the triangles are in the same semi-circle.

⇒ A, B, C and D are concyclic.

Join BD.

DC is a chord.

∴ ∠CAD and ∠CBD are formed in the same segment.

⇒ ∠CAD = ∠CBD Case – II : If both the triangles are not in the same semi-circle.

⇒ A,B,C and D are concyclic. Join BD. DC is a chord.

∴ ∠CAD and ∠CBD are formed in the same segment.

⇒ ∠CAD = ∠CBD

Ex 10.5 Class 9 Maths Question 12.

Prove that a cyclic parallelogram is a rectangle.

Solution:

We have a cyclic parallelogram ABCD. Since, ABCD is a cyclic quadrilateral.

∴ Sum of its opposite angles is 180°.

⇒ ∠A + ∠C = 180° …(i)

But ∠A = ∠C …(ii)

[Opposite angles of a parallelogram are equal]

From (i) and (ii), we have

∠A = ∠C = 90°

Similarly,

∠B = ∠D = 90°

⇒ Each angle of the parallelogram ABCD is 90°.

Thus, ABCD is a rectangle.

Ex 10.6 Class 9 Maths Question 1.

Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Solution:

Given : Two circles with centres O and O’ respectively such that they intersect each other at P and Q.

To Prove: ∠OPO’ = ∠OQO’.

Construction : Join OP, O’P, OQ, O’Q and OO’.

Proof: In ∆OPO’ and ∆OQO’, we have

OP = OQ [Radii of the same circle]

O’P = O’Q [Radii of the same circle]

OO’ = OO’ [Common]

∴ AOPO’ = AOQO’ [By SSS congruence criteria]

⇒ ∠OPO’ = ∠OQO’ [C.P.C.T.]

Ex 10.6 Class 9 Maths Question 2.

Two chords AB and CD of lengths 5 cm and 11 cm, respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.

Solution:

We have a circle with centre O.

AB || CD and the perpendicular distance between AB and CD is 6 cm and AB = 5 cm, CD = 11 cm.

Let r cm be the radius of the circle.

Let us draw OP ⊥ AB and OQ ⊥ CD such that

PQ = 6 cm

Join OA and OC.

Let OQ = x cm

∴ OP = (6 – x) cm

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

Ex 10.6 Class 9 Maths Question 3.

The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre ?

Solution:

We have a circle with centre O. Parallel chords AB and CD are such that the smaller chord is 4 cm away from the centre.

Let r cm be the radius of the circle and draw OP ⊥ AB and join OA and OC.

∵ OP ⊥ AB

∴ P is the mid-point of AB.

⇒ AP = 12AB = 12(6cm) = 3 cm

Similarly, CQ = 12CD = 12(8cm)= 4 cm

Now in ∆OPA, we have OA^{2} = OP^{2} + AP^{2}

⇒ r^{2} = 4^{2} + 3^{2}

⇒ r^{2} = 16 + 9 = 25

⇒ r = 25−−√ =5

Again, in ∆CQO, we have OC^{2} = OQ^{2} + CQ^{2}

⇒ r^{2} = OQ^{2} + 4^{2}

⇒ OQ^{2} = r^{2} – 4^{2}

⇒ OQ^{2} = 5^{2} – 4^{2} = 25 – 16 = 9 [∵ r = 5]

⇒ OQ

⇒ √9 = 3

The distance of the other chord (CD) from the centre is 3 cm.

Note: In case if we take the two parallel chords on either side of the centre, then

In ∆POA, OA^{2} = OP^{2} + PA^{2}

⇒ r^{2} = 4^{2} + 3^{2} = 5^{2}

⇒ r = 5

In ∆QOC, OC^{2} = CQ^{2} + OQ^{2}

⇒ OQ^{2} = 4^{2} + OQ^{2}

⇒ OQ^{2} = 5^{2} – 4^{2} = 9

⇒ OQ = 3

Ex 10.6 Class 9 Maths Question 4.

Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Solution:

Given : ∠ABC is such that when we produce arms BA and BC, they make two equal chords AD and CE.

To prove: ∠ABC = 12 [∠DOE – ∠AOC]

Construction: Join AE.

Proof: An exterior angle of a triangle is equal to the sum of interior opposite angles.

∴ In ∆BAE, we have

∠DAE = ∠ABC + ∠AEC ……(i)

The chord DE subtends ∠DOE at the centre and ∠DAE in the remaining part of the circle.

⇒ ∠ABC = 12 [(Angle subtended by the chord DE at the centre) – (Angle subtended by the chord AC at the centre)]

⇒ ∠ABC = 12 [Difference of the angles subtended by the chords DE and AC at the centre]

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