# R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 9 - Congruence of Triangles and Inequalities in a Triangle

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## Chapter 9 - Congruence of Triangles and Inequalities in a Triangle Exercise MCQ

Solution 1

Correct option: (a)

SSA is not a criterion for congruence of triangles.

Solution 2

Correct option: (c)

Solution 3

Solution 4

Correct option: (c)

In ΔABC,

AB = AC

C = B (angles opposite to equal sides are equal)

C = 50°

Now, A + B + C = 180°

A + 50° + 50° = 180°

A = 80°

Solution 5

Correct option: (a)

In ΔABC,

BC = AB

A = C (angles opposite to equal sides are equal)

Now, A + B + C = 180°

A + 80° + A = 180°

2A = 100°

A = 50°

Solution 6

Correct option: (a)

In ΔABC,

C = A

AB = BC (sides opposite to equal angles are equal)

AB = 4 cm

Solution 7

Correct option: (b)

The sum of any two sides of a triangle is greater than the third side.

Since, 4 cm + 2.5 cm = 6.5 cm

The length of third side of a triangle cannot be 6.5 cm.

Solution 8

Correct option: (b)

We know that in a triangle, the greater angle has the longer side opposite to it.

In ΔABC,

C > B

AB >AC

Solution 9

Solution 10

Solution 11

Correct option: (c)

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

## Chapter 9 - Congruence of Triangles and Inequalities in a Triangle Exercise Ex. 9B

Solution 1(i)

No, it is not possible to construct a triangle with lengths of its sides given because the sum of two sides, 5 cm and 4 cm, is not greater than the third side, 9 cm.

Solution 1(ii)

Yes, it is not possible to construct a triangle with lengths of its sides given because the sum of any two sides is greater than the third side.

Solution 1(iii)

Yes, it is not possible to construct a triangle with lengths of its sides given because the sum of any two sides is greater than the third side.

Solution 1(iv)

Yes, it is not possible to construct a triangle with lengths of its sides given because the sum of any two sides is greater than the third side.

Solution 1(v)

No, it is not possible to construct a triangle with lengths of its sides given because the sum of two sides, 3 cm and 4 cm, is not greater than the third side, 8 cm.

Solution 2

In ΔABC,

A + B + C = 180°

50° + 60° + C = 180°

C = 70°

Thus, we have

A < B < C

BC < AC < AB

Hence, the longest side is AB and the shortest side is BC.

Solution 3(iii)

In ΔABC,

A + B + C = 180°

100° + B + 50° = 180°

B = 30°

Thus, we have

B < C < A

AC < AB < BC

Hence, the shortest side is AC.

Solution 3(i)

Solution 3(ii)

Solution 4

Solution 5

In ΔAOB,

B < A

AO < BO ….(i)

In ΔCOD,

C < D

DO < CO ….(ii)

AO + DO < BO + CO

Solution 6

Construction: Join AC and BD.

In ΔABC,

BC > AB

BAC > ACB ….(i)

In ΔACD,

Adding (i) and (ii), we get

BAC + CAD > ACB + ACD

A > C

In ΔBDC,

CD > BC

CBD > BDC ….(iv)

Adding (iii) and (iv), we get

ABD + CBD > ADB + BDC

B > D

Solution 7

In ΔABC,

AB + BC > AC  ….(i)

In ΔACD,

DA + CD > AC  ….(ii)

DA + AB > BD  ….(iii)

In ΔBDC,

BC + CD > BD  ….(iv)

Adding (i), (ii), (iii) and (iv), we get

AB + BC + DA + CD + DA + AB + BC + CD > AC + AC + BD + BD

2(AB + BC + CD + DA) > 2(AC + BD)

AB + BC + CD + DA > AC + BD

Solution 8

In ΔAOB,

AO + BO > AB  ….(i)

In ΔBOC,

BO + CO > BC  ….(ii)

In ΔCOD,

CO + DO > CD  ….(iii)

In ΔAOD,

DO + AO > DA  ….(iv)

Adding (i), (ii), (iii) and (iv), we get

AO + BO + BO + CO + CO + DO + DO + AO > AB + BC + CD + DA

2(AO + CO) + 2(BO + DO) > AB + BC + CD + DA

2AC + 2BD > AB + BC + CD + DA

2(AC + BD) > AB + BC + CD + DA

AB + BC + CD + DA < 2(AC + BD)

Solution 9

Solution 10

In ΔPQR,

PQ > PR

PRQ > PQR

SRQ > SQR

SQ > SR

Solution 11

In ΔABC,

AB = AC

ABC = ACB ….(i)

Now, ABC = ABD + DBC

ABC > DBC

ACB > DBC [From (i)]

DCB > DBC

BD > CD

i.e. CD < BD

Solution 12

Let PQR be the required triangle.

Let PR be the longest side.

Then, PR > PQ

Q > R ….(i)

Also, PR > QR

Q > P ….(ii)

Adding (i) and (ii), we get

2Q > R + P

2Q + Q > P + Q + R (adding Q to both sides)

3Q > 180°

Q > 60°

Solution 13(i)

In ΔCDA,

CD + DA > AC ….(i)

In ΔABC,

AC + AB > BC ….(ii)

Adding (i) and (ii), we get

CD + DA + AC + AB > AC + BC

Subtracting AC from both sides, we get

CD + DA + AB > BC

Solution 13(ii)

In ΔCDA,

CD + DA > AC ….(i)

In ΔABC,

AB + BC > AC ….(ii)

Adding (i) and (ii), we get

CD + DA + AB + BC > AC + AC

CD + DA + AB + BC > 2AC

Solution 14(i)

Given : ABC is a triangle and O is appoint insideit.

To Prove : (i) AB+AC > OB +OC

Solution 14(ii)

AB+BC+CA > OA+OB+OC

Solution 14(iii)

OA+OB+OC> (AB+BC+CA)

Proof:

(i)InABC,

AB+AC>BC.(i)

And in , OBC,

OB+OC>BC.(ii)

Subtracting (i) from (i) we get

(AB+AC)-(OB+OC)> (BC-BC)

i.e.AB+AC>OB+OC

(ii)AB+AC> OB+OC[proved in (i)]

Similarly,AB+BC > OA+OC

AndAC+BC> OA +OB

Addingboth sides of these three inequalities, we get

(AB+AC) +(AC+BC) +(AB+BC)>OB+OC+OA+OB+OA+OC

i.e.2(AB+BC+AC)> 2(OA+OB+OC)

Therefore, we have

AB+BC+AC > OA+OB+OC

(iii)InOAB

OA+OB > AB(i)

InOBC,

OB+OC > BC(ii)

And, in OCA,

OC+OA>CA

Adding (i), (ii) and (iii)we get

(OA+OB) + (OB+OC)+(OC+OA)> AB+BC+CA

i.e2(OA+OB+OC) > AB+BC+CA

OA+OB+OC> (AB+BC+CA)

Solution 15

Construction: Mark a point S on BC such that BD = SD. Join AS.

BD = SD (by construction)

AB = AS (c.p.c.t.)

Now, in ΔABS,

AB = AS

ASB = ABS ….(i)(angles opposite to equal sides are equal)

In ΔACS,

ASB > ACS ….(ii)

From (i) and (ii), we have

ABS > ACS

ABC > ACB

AC > AB

Solution 16

In ΔABC,

AB + AC > BC

AB + AC >BD + DC

AB + AC >BD + DE ….(i) [since CD = DE]

In ΔBED,

BD + DE > BE ….(ii)

From (i) and (ii), we have

AB + AC > BE

## Chapter 9 - Congruence of Triangles and Inequalities in a Triangle Exercise Ex. 9A

Solution 1

(i) In ΔAOB and ΔDOC,

BAO = CDO (AB CD, alternate angles)

AO = DO (O is the mid-point of AD)

AOB = DOC (vertically opposite angles)

ΔAOB ΔDOC (by ASA congruence criterion)

(ii) Since ΔAOB ΔDOC,

BO = CO (c.p.c.t.)

O is the mid-point of BC.

Solution 2

In ΔAOD and ΔBOC,

AOD = BOC (vertically opposite angles)

DAO = CBO (Each 90°)

ΔAOD BOC (by AAS congruence criterion)

AO = BO (c.p.c.t.)

CD bisects AB.

Solution 3

In ΔABC and ΔCDA

BAC = DCA  (alternate interior angles for p q)

AC = CA  (common)

BCA = DAC (alternate interior angles for l m)

ΔABC ΔCDA (by ASA congruence rule)

Solution 4

AB = AC (given)

BD = CD (c.p.c.t.)

Solution 5

(i) In ΔABE and ΔACF,

AEB = AFC (Each 90°)

BE = CF (given)

BAE = CAF (common A)

ΔABE ACF (by ASA congruence criterion)

(ii) Since ΔABE ≅ ΔACF,

AB = AC (c.p.c.t.)

Solution 6

(i) In ΔABD and ΔACD,

AB = AC  (equal sides of isosceles ΔABC)

DB = DC  (equal sides of isosceles ΔDBC)

ΔABD ACD (by SSS congruence criterion)

(ii) Since ΔABD ≅ ΔACD,

BAE = CAE ….(1)

Now, in ΔABE and ΔACE

AB = AC (equal sides of isosceles ΔABC)

BAE = CAE [From (1)]

AE = AE (common)

ΔABE ACE (by SAS congruence criterion)

(iii) Since ΔABD ≅ ΔACD,

BAE = CAE

Thus, AE bisects A.

In ΔBDE and ΔCDE,

BD = CD (equal sides of isosceles ΔABC)

BE = CE (c.p.c.t. since ΔABE ACE)

DE = DE (common)

ΔBDE CDE (by SSS congruence criterion)

BDE = CDE (c.p.c.t.)

Thus, DE bisects D, i.e., AE bisects D.

Hence, AE bisects A as well as D.

(iv) Since ΔBDE ≅ ΔCDE,

BE = CE and BED = CED (c.p.c.t.)

BE = CE and BED = CED = 90° (since BED and CED form a linear pair)

DE is the perpendicular bisector of BC.

AE is the perpendicular bisector of BC.

Solution 7

Solution 8

(i) In ΔAPB and ΔAQB,

APB = AQC (Each 90°)

BAP = BAQ (line l is the bisector of A)

AB = AB (common)

ΔAPB AQB (by AAS congruence criterion)

(ii) Since ΔAPB ≅ ΔAQB,

BP = BQ (c.p.c.t.)

Solution 9

BAC = DAC (AC bisects A)

AC = AC (common)

BCA = DCA (AC bisects C)

ΔABC ADC (by ASA congruence criterion)

AB = AD and CB = CD (c.p.c.t.)

Solution 10

Construction: Draw DE BC.

In ΔDAC and ΔDEC,

DAC = DEC (Each 90°)

DCA = DCE (CD bisects C)

CD = CD (common)

ΔDAC ≅ ΔDEC (by AAS congruence criterion)

DA = DE (c.p.c.t.) ….(i)

and AC = EC (c.p.c.t.)  ….(ii)

Given, AB = AC

B = C (angles opposite to equal sides are equal)

In ΔABC, by angle sum property,

A + B + C = 180°

90° + B + B = 180°

2B = 90°

B = 45°

In ΔBED,

BDE + B = 90° (since BED = 90°)

BDE + 45° = 90°

BDE = 45°

BDE = DBE = 45°

DE = BE ….(iii)

From (i) and (iii),

DA = DE = BE ….(iv)

Now, BC = BE + EC

BC = DA + AC [From (ii) and (iv)

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

ΔOAB is an equilateral triangle.

OAB = OBA = AOB = 60°

ABCD is a square.

A = B = C = D = 90°

Now, A = DAO + OAB

90° = DAO + 60°

DAO = 90° - 60° = 30°

Similarly, CBO = 30°

AD = BC  (sides of a square ABCD)

DAO = CBO = 30°

OA = OB (sides of an equilateral ΔOAB)

ΔOAD ≅ ΔOBC (by SAS congruence criterion)

OD = OC (c.p.c.t.)

Hence, ΔOCD is an isosceles triangle.

Solution 16

Solution 17

Solution 18

Solution 19

Construction: Join AN and BN.

In ΔANM and ΔBNM

AM = BM (M is the mid-point of AB)

AMN = BMN (Each 90°)

MN = MN (common)

ΔANM ΔBNM (by SAS congruence criterion)

AN = BN (c.p.c.t.) ….(i)

And, ANM = BNM (c.p.c.t.)

90° - ANM = 90° - BNM

AND = BNC ….(ii)

In ΔAND and DBNC,

AN = BN [From (i)]

AND = BNC [From (ii)]

DN = CN (N is the mid-point of DC)

ΔAND ΔBNC (by SAS congruence criterion)

Solution 20

In ΔABC, AB = AC

ABC = ACB

OBC = OCB ….(i)

Now, by exterior angle property,

MOC = OBC + OCB

MOC = 2OBC [From (i)]

MOC = ABC (OB is the bisector of ABC)

Solution 21

In ΔABC, AB = AC

ABC = ACB

OBC = OCB ….(i)

In ΔBOC, by angle sum property,

BOC + OBC + OCB = 180°

BOC + 2OBC = 180° [From (i)]

BOC + ABC = 180°

BOC + (180° - ABP) = 180° (ABC and ABP form a linear pair)

BOC + 180° - ABP = 180°

BOC - ABP = 0

BOC = ABP

Solution 22

AB PQ and BP is a transversal.

ABP = BPQ (alternate angles) ….(i)

BP is the bisector of ABC.

ABP = PBC

ABP = PBQ ….(ii)

From (i) and (ii), we have

BPQ = PBQ

PQ = BQ (sides opposite to equal angles are equal)

ΔBPQ is an isosceles triangle.

Solution 23

To prove that the image is as far behind the mirror as the object is in front of the mirror, we need to prove that AT = BT.

We know that angle of incidence = angle of reflection.

ACN = DCN ….(i)

AB CN and AC is the transversal.

TAC = ACN (alternate angles) ….(ii)

Also, AB CN and BD is the transversal.

TBC = DCN (corresponding angles) ….(iii)

From (i), (ii) and (iii),

TAC = TBC ….(iv)

In ΔACT and ΔBCT,

TAC = TBC [From (iv)]

ATC = BTC (Each 90°)

CT = CT (common)

ΔACT ΔBCT (by AAS congruence criterion)

AT = BT (c.p.c.t.)

Solution 24

Let AB be the breadth of a river. Now take a point M on that bank of the river where point B is situated. Through M draw a perpendicular and take point N on it such that point, A, O and N lie on a straight line where point O is the mid point of BM.

Solution 25

D is the mid-point of AC.

Given, BD =

BAD = ABD (i)(angles opposite to equal sides are equal)

Consider CD = BD

BCD = CBD (ii)(angles opposite to equal sides are equal)

In ΔABC, by angle sum property,

ABC + BAC + BCA = 180°

ABC + BAD + BCD = 180°

ABC + ABD + CBD = 180° [From (i) and (ii)]

ABC + ABC = 180°

2ABC = 180°

ABC = 90°

Hence, ABC is a right angle.

Solution 26

The given statement is not true.

Two triangles are congruent if two sides and the included angle of one triangle are equal to corresponding two sides and the included angle of another triangle.

Solution 27

The given statement is not true.

Two triangles are congruent if two angles and the included side of one triangle are equal to corresponding two angles and the included angle of another triangle.

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