NCERT Solutions for Class 9 Subjectwise · Class 9 Science – Chemistry Class 9 Science –...

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#464018

Topic: Chords of Circle

Fill in the blanks:

(1) The centre of a circle lies in ___________ of the circle (exterior/interior)

(2) A point whose distance from the centre of a circle is greater than its radius lies in ___________ of the circle

(3) The longest chord of a circle is a ____________ of the circle.

(4) An arc is a ____________ when its ends are the ends of a diameter.

(5) Segment of a circle is the region between an arc and __________ of the circle.

(6) A circle divides the plane, on which it lies in ______________ parts.

Solution

From the figure , we can see that

(1) Interior

Explanation : Centre O lies in interior of the circle.

(2) Exterior

Explanation : Let a point R be such that OR>OP(radius), then R lies in exterior of the circle.

(3) Diameter.

Explanation : AB , the diameter is the longest chord of the circle

(4) Semi- circle

Explanation : We can see, the arc AB with its' end points being the ends of diameter is the semicircle.

(5) Chord

Explanation : segment of a circle is the region between an arc and chord

(6) Three

Explanation : three parts are interior, exterior and circumference.

#464025

Topic: Sector, Arc and Segment of a Circle

Write true or false: Give reasons for your answers.

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

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

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

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

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

(6) A circle is a plane figure.

Solution



(i) True : Radius is the length of line segment from centre to any point on circumference of circle.

(ii) False : We can draw infinite number of chord by joining any two points on circle since infinite number of points lie on the circle, .

(iii) False : These arcs are called equal arc since these arcs are equal in length.

(iv) True : Diameter is twice the radius of the circle and it is the longest chord.

(v) False : Sector is the region between an arc and the two radii.

(vi) True : Circle lies in a plane since a circle is the locus of a point in a plane which moves in such a way that its distance from a given fixed point is always constant.

#464031


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

Let us consider a circle with center O and two equal chords of a circle AB and CD.

We need to prove that ∠AOB = ∠COD

In △ AOB and COD, we have

AO = CO (Radius of the circle)

BO = DO (Radius of the circle)

AB = CD (Equal chords)

By SSS criterion of congruence, we have

△AOB ≅ △COD

⇒ ∠AOB = ∠COD

#464032


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

Solution



Take two congruent circles, and let their centres be O1 and O2

Let, the chords making equal angles at centres be AB and CD respectively for two circles.

Let the radius of both circles be r.

We know that O1A = O1B = O2C = O2D.

Since, O1A = O1B = O2C = O2D, we get,

∠O1AB = ∠O1BA∠O2CD = ∠O2DC.

So, we get ∠AO1B = ∠CO2D.

So, by SAS congruency, we get triangle AO1B is congruent to triangle CO2D.

∴ AB = CD, which implies that the length of chords are equal.

#464033

Topic: Introduction

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

Solution

As we can see from the figure, that two circles have two points in common.

Two circles cannot intersect each other at more than two points.

Let us assume that two circles cut each other at three points.

But we already know that through three points only one circle can pass.

So, two circles if intersect each other will intersect at maximum two points.

#464034

Topic: Introduction

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

Solution



1. Let us consider any three points P, Q and R on the circumference of the circle.

2. Make chords PQ and QR by joining P, Q andR.

3. Draw AB and CD, the perpendicular bisectors of PQ and QR. They will intersect each other at O.

4. O will be the center of the circle.

#464035


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

Solution



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Let the two circles have centres O and Q and they intersect at two points A and B.

Then, AB is the common chord of the two circles.

OQ is the line joining the centres of the two circles. Let OQ cuts AB at P.

We need to prove OQ is the perpendicular bisector of AB.

Draw line segment OA, OB, QA and QB.

In △OAQ and △OBQ,

OA = OB ....(Radius of the circle)

QA = QB ...(Radius of the circle)

and OQ = OQ ....(common side)

△OAQ ≅ △OBQ ....SSS test of congruence

⇒ ∠AOQ = ∠BOQ ....c.a.c.t. ....(1)

[Since ∠AOQ = ∠AOP and ∠BOP = ∠BOQ]

In triangles AOP and BOP,

OA = OB ....(Radius of the circle)

∠AOP = ∠BOP .....From (1)

OP = OP .....Common side

△AOP ≅ △BOP ....SAS test of congruence,

⇒ AP = BP and ∠APO = ∠BPO

But, ∠APO + ∠BPO = 180o

2∠APO = 180o ⇒ ∠APO = 90o

AP = BP and ∠APO = ∠BPO = 90o

Hence, OQ is the perpendicular bisector of AB.

#464036

Topic: Non-intersecting, Intersecting, Touching circles

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

Solution

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Let the common chord be AB and P and Q be the centers of the two circles.

∴ AP = 5cm and AQ = 3cm.

PQ = 4cm ....given

Now, segPQ ⊥ chord AB

∴ AR = RB =1

2AB ....perpendicular from center to the chord, bisects the chord

Let PR = xcm, so RQ = (4 − x)cm

In △ARP,

AP2 = AR2 + PR2

AR2 = 52 − x2 ...(1)

In △ARQ,

AQ2 = AR2 + QR2

AR2 = 32 − (4 − x)2 ...(2)

∴ 52 − x2 = 32 − (4 − x)2 ....from (1) & (2)

25 − x2 = 9 − (16 − 8x + x2)

25 − x2 = − 7 + 8x − x2

32 = 8x

∴ x = 4

Substitute in eq(1) we get,

AR2 = 25 − 16 = 9

∴ AR = 3cm.

∴ AB = 2 × AR = 2 × 3

∴ AB = 6 cm.

So, length of common chord AB is 6 cm.

#464037


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

Solution



Drop a perpendicular from O to both chords AB and CD

In △OMP and △ONP

As chords are equal, perpendicular from centre would also be equal.

OM = ON

OP is common.

∠OMP = ∠ONP = 90o

△OMP ≅ △ONP (RHS Congruence)

PM = PN ......................(1)

AM = BM (Perpendicular from centre bisects the chord)

Similarly ,CN = DN

As AB = CD

AB − AM = CD − DN

BM = CN .........................(2)

From (1) and (2)

BM − PM = CN − PN

PB = PC

AM = DN(Half the length of equal chords are equal)

AM + PM = DN + PN

AP = PD

Therefore , PB = PC and AP = PD is proved.

#464038


It 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 chord.

Solution



Let us consider a circle with centre O and AB and CD be the chords which intersect at M

According to the question, AB = CD.

We need to prove ∠OME = ∠OMF

Drop perpendiculars OE and OF on AB and CD respectively and join OM.

In triangles OEM and OFM ,

OE = OF (equal chords of a circle are equidistant from the centre)

OM = OM (common side)

∠OEM = ∠OFM (both equal to 90o)

By RHS criterion of congruence,

△OEM ≅ △OFM

⇒ ∠OME = ∠OMF (by C.P.C.T.)

#464039


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

Solution

We know that, OA = OD and OB = OC.

They are radius of respective circles.

In ΔOBC, we know that OB = OC, so ∠OBC = ∠OCB

∴ ∠OCD = ∠OBA

In ΔOAD, we know that OA = OD, so ∠OAD = ∠ODA

Since, ∠OCD = ∠OBA and ∠OAD = ∠ODA, we get ∠AOB in ΔOAB is equal to ∠COD in ΔOCD.

∴ From SAS congruency, we can say that ΔOAB and ΔOCD are congruent.

So, AB = CD.



#464040


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 6m each, what is the distance between Reshma and Mandip?

Solution

Let the three girls Reshma, Salma and Mandip are standing on the circle of radius 5 cm at points R, S and M respectively.

Here, chord RS = chord SM.

∴ center O lies on bisector of ∠RSM i.e. ray OS

Now, RS = SM = 6 m.

Let OA ⊥ chord RM

RA = AM ....perpendicular from center to the chord, bisects the chord.

In △SAR, ∠A = 90o

SR2 = SA2 + RA2 ....Pyhtagoras theorem

⇒ 36 = SA2 + RA2

⇒ RA2 = 36 − SA2 ... (1)

In the △OAR,

RO2 = AO2 + RA2 ....Pythagoras theorem

⇒ 25 = (OS − AS)2 + RA2

⇒ RA2 = 25 − (OS − AS)2

RA2 = 25 − (5 − AS)2 ... (2)

From (1) and (2) , we get

36 − SA2 = 25 − (5 − SA)2

⇒ 11 − SA2 + (5 − SA)2 = 0

⇒ 11 − SA2 + 25 − 10SA + AS2 = 0

⇒ 10SA = 36

⇒ SA = 3.6

Putting SA = 3.6 in (1) we get

RA2 = 3.6 − (3.6)2 = 36 − 12.96

⇒ 36 − 12.96 = 23.04 m

∴ RA = 4.8m

⇒ RM = 2RA = 2 × 4.8 = 9.6 m

Hence, the distance between Reshma and Mandip = 9.6 m



#464042


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

Solution

Here,

∠AOC = ∠AOB + ∠BOC

⇒ ∠AOC = 60° + 30°

⇒ ∠AOC = 90°

We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.

∠ADC =1

2∠AOC =

1

2× 90o = 445o

#464043

Topic: Cyclic Polygons

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

Given,

AB is equal to the radius of the circle.

In △OAB,

OA = OB = AB = radius of the circle.

Thus, △OAB is an equilateral triangle.

∠AOC = 60°

Also, ∠ACB =1

2∠AOB =

1

2× 60° = 30°

ACBD is a cyclic quadrilateral,

∠ACB + ∠ADB = 180° ∣ Opposite angles of cyclic quadrilateral

⇒ ∠ADB = 180° − 30° = 150°

Thus, angle subtend by the chord at a point on the minor arc and also at a point on the major arc are 150° and 30° respectively.

#464045




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

Solution

Reflex ∠POR = 2 × ∠PQR = 2 × 100° = 200°

∴ ∠POR = 360° − 200° = 160°

In ΔOPR,

OP = OR ...Radii of the circle

∠OPR = ∠ORP

Now,

∠OPR + ∠ORP + ∠POR = 180° ...Sum of the angles in a triangle

⇒ ∠OPR + ∠OPR + 160° = 180°

⇒ 2∠OPR = 180° − 160°

⇒ ∠OPR = 10°

#464046


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

Solution

∠BAC = ∠BDC ...Angles in the segment of the circle

In ΔABC,

∠BAC + ∠ABC + ∠ACB = 180° ...Sum of the angles in a triangle

⇒ ∠BAC + 69° + 31° = 180°

⇒ ∠BAC = 180° − 100°

⇒ ∠BAC = 80°

Thus, ∠BDC = 80°

#464047


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

Solution



∠BAC = ∠CDE ...Angles in the segment of the circle

In ΔCDE,

∠CEB = ∠CDE + ∠DCE ...Exterior angles of the triangle

⇒ 130° = ∠CDE + 20°

⇒ ∠CDE = 110°

Thus, ∠BAC = 110°

#464048


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

Solution

For chord CD,

∠CBD = ∠CAD ...Angles in same segment

∠CAD = 70°

∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°

∠BCD + ∠BAD = 180° ...Opposite angles of a cyclic quadrilateral

⇒ ∠BCD + 100° = 180°

⇒ ∠BCD = 80°

In △ABC

AB = BC (given)

∠BCA = ∠CAB ...Angles opposite to equal sides of a triangle

∠BCA = 30°

Also, ∠BCD = 80°

∠BCA + ∠ACD = 80°

⇒ 30° + ∠ACD = 80°

∠ACD = 50°

∠ECD = 50°

#464050


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

Solution



Let ABCD be a cyclic quadrilateral and its diagonal AC and BD are the diameters of the circle through the vertices of the quadrilateral.

∠ABC = ∠BCD = ∠CDA = ∠DAB = 90° ...Angles in the semi-circle

Thus, ABCD is a rectangle as each internal angle is 90°

#464051


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

Solution

Given:

ABCD is a trapezium where non-parallel sides AD and BC are equal.

Construction:

DM and CN are perpendicular drawn on AB from D and C respectively.

To prove:

ABCD is cyclic trapezium.

Proof:

In △DAM and △CBN,

AD = BC ...Given

∠AMD = ∠BNC ...Right angles

DM = CN ...Distance between the parallel lines

△DAM ≅ △CBN by RHS congruence condition.

Now,

∠A = ∠B ...by CPCT

Also, ∠B + ∠C = 180° ....Sum of the co-interior angles

⇒ ∠A + ∠C = 180°

Thus, ABCD is a cyclic quadrilateral as sum of the pair of opposite angles is 180°.



#464052


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

Chords AP and DQ are joined.

For chord AP,

∠PBA = ∠ACP ...Angles in the same segment --- (i)

For chord DQ,

∠DBQ = ∠QCD ...Angles in same segment --- (ii)

ABD and PBQ are line segments intersecting at B.

∠PBA = ∠DBQ ...Vertically opposite angles --- (iii)

By the equations (i), (ii) and (iii),

∠ACP = ∠QCD

#464053


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

Given,

Two circles are drawn on the sides AB and AC of the triangle ABC as diameters. The circles intersected at D.

Construction: AD is joined.

To prove: D lies on BC. We have to prove that BDC is a straight line.

Proof:

∠ADB = ∠ADC = 90° ...Angle in the semi circle

Now,

∠ADB + ∠ADC = 180°

⇒ ∠BDC is straight line.

Thus, D lies on the BC.

#464054


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

Solution



Given,

AC is the common hypotenuse. ∠B = ∠D = 90°.

To prove,

∠CAD = ∠CBD

Proof:

Since, ∠ABC and ∠ADC are 90°.

These angles are in the semi-circle.

Thus, both the triangles are lying in the semi-circle and AC is the diameter of the circle.

⇒ Points A, B, C and D are concyclic.

Thus, CD is the chord.

⇒ ∠CAD = ∠CBD ...Angles in the same segment of the circle

#464055


Prove that a cyclic parallelogram is a rectangle.

Solution

Given,

ABCD is a cyclic parallelogram.

To prove,

ABCD is a rectangle.

Proof:

∠1 + ∠2 = 180° ...Opposite angles of a cyclic parallelogram

Also, Opposite angles of a cyclic parallelogram are equal.

Thus,

∠1 = ∠2

⇒ ∠1 + ∠1 = 180°

⇒ ∠1 = 90°

One of the interior angle of the parallelogram is right angled. Thus, ABCD is a rectangle.



#464056


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

Solution

Given: Two circles with centers A and B, which intersect each other at C and D.

To prove : ∠ACB = ∠ADB

Construction : Join AC, AD, BD and BC.

Proof : In triangles ACB and ADB, we have

AC = AD ...Radii of the same circle

BC = BD ...Radii of the same circle

AB = AB ...Common

Therefore, by SSS criterion of congruence,

△ACB ≅ △ADB

⇒ ∠ACB = ∠ADB ...CPCT

#464057


Two chords AB and CD of lengths 5cm and 11cm 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 6cm, find the radius of the circle.

Solution



Join OA and OC.

Let the radius of the circle be r cm and O be the centre

Draw OP⊥AB and OQ⊥CD.

We know, OQ⊥CD, OP⊥AB and AB ∥ CD.

Therefore, points P, O and Q are collinear. So, PQ = 6 cm.

Let OP = x.

Then, OQ = (6– x) cm.

And OA = OC = r.

Also, AP = PB = 2.5 cm and CQ = QD = 5.5 cm.

(Perpendicular from the centre to a chord of the circle bisects the chord.)

In right triangles QAP and OCQ, we have

OA2 = OP2 + AP2 and OC2 = OQ2 + CQ2

∴ r2 = x2 + (2.5)2 ..... (1)

and r2 = (6 − x)2 + (5.5)2 ..... (2)

⇒ x2 + (2.5)2 = (6 − x)2 + (5.5)2

⇒ x2 + 6.25 = 36 − 12x + x2 + 30.25

12x = 60

∴ x = 5

Putting x = 5 in (1), we get

r2 = 52 + (2.5)2 = 25 + 6.25 = 31.25

⇒ r2 = 31.25 ⇒ r = 5.6

Hence, the radius of the circle is 5.6 cm

#464058


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



There are two parallel chords of length 8 cm and 6 cm.

The distance between center and shortest chord (6cm chord) is 4cm.

So, the perpendicular distance from center to shortest chord is 4cm.

The perpendicular line from center to chord divides chord into two equal parts.

So, the radius of circle is √32 + 42 = 5.

Since, the radius is 5.

The distance from center to largest chord is √52 − 42 = 3.

#464060


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

In △BDC,

∠ADC is the exterior angle

∴ ∠ADC = ∠DBC + ∠DCB .... (1)

(The measure of exterior angle is equal to sum of the remote interior angles)

By inscribed angle theorem,

∠ADC =1

2∠AOC and ∠DCB =

1

2∠DOE ....... (2)

From (1) and (2), we have

1

2∠AOC = ∠ABC +

1

2∠DOE ...[Since ∠DBC = ∠ABC]

⇒ ∠ABC =1

2(∠AOC − ∠DOE)

Hence, ∠ABC is equal to half the difference of angles subtended by the chords AC and DE at the centre.

#464061


Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Solution



Let us consider rhombus ABCD with the two diagonals as AC and BD. Consider O as the point of intersection of diagonals.

Diagonals of a rhombus bisect each other at right angles.

We need to prove that circle drawn on AB as diameter will pass through O.

Draw lines MN and XY such that MN ∥ AD and XY ∥ AB such that M and N are mid-points of AB and CD.

AB = DC

⇒1

2AB =

1

2DC

AM = DN (M and N are mid-points of AB and CD)

Similarly, AX = OM

⇒ AN = ON = NB

A circle drawn with N as centre and radius AN passes through A, O and B.

Circle obtained is the required circle.

#464062


ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD

Solution



From the figure,

⇒ ∠ADE + ∠ADC = 180o (Linear pair) ....... (i)

Also, ◻ABCE is a cyclic quadrilateral.

So, ∠AED + ∠ABC = 180o ....... (ii) (opposite angles of cyclic quadrilateral sum upto 180o)

∠ADC = ∠ABC ....... (iii) (Opposite angles of a parallellogram are equal)

Using (i), (ii) and (iii),

∠AED + ∠ABC = ∠ADE + ∠ABC

⇒ ∠AED = ∠ADE

Now, In △AED,

∠AED = ∠ADE

⇒ AD = DE (Sides opposite to equal angles)

#464063


AC and BD are chords of a circle which bisect each other. Prove that

(i) AC and BD are diameters (ii) ABCD is a rectangle.

Solution



(i) It is given that AB and CD are the two chords of a circle

Let the point of intersection be O.

Join AB, BC, CD and AD.

In triangles AOB and COD,

∠AOB = ∠COD ...(Vertically opposite angles)

OB = OD ....(O is the mid-point of BD)

OA = OC ....(O is the mid-point of AC)

△AOB ≅ △COD ....SAS test of congruence

∴ AB = CD ....c.s.c.t.

Similarly, we can prove △AOD ≅ △BOC, then we get

AD = BC ....c.s.c.t.

So, ◻ABCD is a parallelogram, since opposite sides are equal in length.

So, opposite angles are equal as well.

So, ∠A = ∠C

Also, for a cyclic quadrilateral opposite angles add up to 180o

So, ∠A + ∠C = 180o

∠A + ∠A = 180o

∠A = 90o

So, BD is the diameter. Similarly, AC is also the diameter.

ii) Since AC and BD are diameters,

∴ ∠A = ∠B = ∠C = ∠D = 90o ...Angle inscribed in a semi circle is a right angle

Hence, parallelogram ABCD is a rectangle

#464064


AC and BD are chords of a circle which bisect each other. Prove that ABCD is a rectangle.

Solution



AO = CO ....AC and BD bisect each other

BO = DO ....AC and BD bisect each other

∠ABO = ∠COD ...Vertically opposite angles

△AOB ≅ △COD ...[SAS test]

⇒ ∠BAO i.e., ∠BAC = ∠ABO i.e., ∠ABD

⇒ AC ∥ BD

Again, △AOD ≅ △COB ...[SAS test]

⇒ AD ∥ CB

⇒ ABCD is a cyclic parallelogram.

⇒ ∠DAC = ∠DBA.............. (1)

[Opp.angles of a parallelogram are equal]

Also, ACBD is a cyclic quadrilateral

Therefore ∠DAC + ∠DBA = 180 ∘ ................ (2)

From (1) and (2), we get

∠DAC = ∠DBA = 90 ∘

Hence, ABCD is a rectangle.

#464065


Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangles DEF are 90o −1

2A, 90o −

1

2B and

90o −1

2C.

Solution



In △DEF,

∠D = ∠EDF

But ∠EDF = ∠EDA + ∠FDA ....angle addition property

Now, ∠EDA = ∠EBA and ∠FDA = ∠FCA .....Angles inscribed in the same arc

∴ ∠EDF = ∠EBA + ∠FCA

=1

2∠B +

1

2∠C

[Since BE is bisector of ∠B and CF is bisector ∠C]

∴ ∠D =∠B + ∠C

2 ....(1)

Similarly, ∠E =∠C + ∠A

2 and ∠F =

∠A + ∠B

2 ...(2)

Now, ∠A + ∠B + ∠C = 180o ....angle sum property of triangle

∴ ∠B + ∠C = 180 ∘ − ∠A ...(3)

Similarly, ∠C + ∠A = 180o − ∠B and ∠A + ∠B = 180o − ∠C ...(4)

Substituting eq (3) in eq (1), we get

∴ ∠D =180o − ∠A

2

∠D = 90o −1

2∠A

Similarly, ∠E = 90o −1

2∠B and ∠F = 90o −

1

2∠C

#464066


Two congurent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.

Solution



Let point C and D be the centres of the two congruent circles.

Let AB be the common chord.

∴ arc AEB = arc AFB ....arcs of congruent chords of congruent circles

∴ ∠APB = ∠AQB ....angles subtended by congruent arcs ....(1)

In △BPQ,

∠APB = ∠AQB ...from (1)

i.e. ∠QPB = ∠PQB

∴ BQ = BP ...converse of isosceles triangle theorem

#464068


In any triangle ABC, if the angle bisector ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Solution

∠BAE = ∠CAE ....Angles inscribed in the same arc

∴ arc BE = arc CE

∴ chord BE ≅ chord CE ...(1)

In △BDE and △CDE,

BE = CE ....from (1)

BD = CD ...ED is perpendicular bisector of BC

DE = DE ...common side

∴ △BDE ≅ △CDE ...SSS test of congruence

⇒ ∠BDE = ∠CDE .....c.a.c.t.

Also, ∠BDE + ∠CDE = 180 ∘ ...angles in linear pair

∴ ∠BDE = ∠CDE = 90 ∘

Therefore, DE is the right bisector of BC.

NCERT Solutions for Class 9 Subjectwise · Class 9 Science – Chemistry Class 9 Science –...

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