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R S AGGARWAL AND V AGGARWAL Solutions Mathematics Class 9 Chapter 9 Congruence of Triangles and Inequalities in a Triangle

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

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

Question 1

In the given figure, AB
CD and O is the midpoint of AD.

Show that (i) Δ
AOB ≅ Δ DOC (ii) O is the midpoint of BC.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

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.

Question 2

ΔABC
is a right triangle right angled at A such that AB = AC and bisector of ∠C
intersect the side AB at D. Prove that AC + AD = BC.

Solution 2

Construction:
Draw DE
⊥ BC.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

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°

2∠B = 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)

AC + AD = BC

Question 3

In the given figure, OA = OB and OP = OQ. Prove that (i) PX = QX (ii) AX = BX.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 3

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 4

In the given figure, ABC is an equilateral triangle; PQ || AC and AC is produced to R such that CR = BP. Prove that QR bisects PC.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 4

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 5

In the given figure, ABCD is a quadrilateral in which AB || DC and P is the midpoint BC. On producing, AP and DC meet at Q. Prove that (i) AB = CQ, (ii) DQ = DC + AB.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 5

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 6

In the given figure, ABCD is square and P is a point inside it such that PB = PD. Prove that CPA is a straight line.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 6

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 7

In the given figure, O is a point
in the interior of square ABCD such that
ΔOAB
is an equilateral triangle. Show that ΔOCD is an isosceles triangle.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 7

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°

In
ΔOAD and ΔOBC,

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.

Question 8

In the adjoining figure, X and Y are respectively two points on equal sides AB and AC of R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleABC such that AX = AY. Prove that CX = BY.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 8

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 9

In R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleABC, D is the midpoint of BC. If DL R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleAB and DM R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleAC such that DL = DM, prove that AB = AC.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 9

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 10

In R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleABC, AB = AC and the bisectors of R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleB and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleC meet at a point O. Prove that BO = CO and the ray AO is the bisector R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleA.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 10

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 11

The line segments joining the
midpoints M and N of parallel sides AB and DC respectively of a trapezium
ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC.

Solution 11

Construction:
Join AN and BN.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

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)

AD = BC (c.p.c.t.)

Question 12

In the given figure, AD and BC are
equal perpendiculars to a line segment AB. Show that CD bisect AB.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 12

In
ΔAOD and ΔBOC,

AOD = ∠BOC (vertically opposite angles)

DAO = ∠CBO (Each 90°)

AD
= BC (given)

ΔAOD ≅ BOC (by AAS
congruence criterion)


AO = BO (c.p.c.t.)


CD bisects AB.

Question 13

The bisectors of ∠B
and ∠C
of an isosceles triangle with AB = AC intersect each other at a point O. BO
is produced to meet AC at a point M. Prove that ∠MOC
= ∠ABC.

Solution 13

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

In
ΔABC, AB = AC

∠ABC = ∠ACB

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

∠OBC = ∠OCB ….(i)

Now,
by exterior angle property,

MOC = ∠OBC + ∠OCB

∠MOC = 2∠OBC [From (i)]

∠MOC = ∠ABC (OB is the bisector of ∠ABC)

Question 14

The bisectors of ∠B
and ∠C
of an isosceles ΔABC with AB = AC intersect each other at a point O. Show
that the exterior angle adjacent to ∠ABC is equal to ∠BOC.

Solution 14

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

In
ΔABC, AB = AC

∠ABC = ∠ACB

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

∠OBC = ∠OCB ….(i)

In
ΔBOC, by angle
sum property,

BOC + ∠OBC + ∠OCB = 180°

∠BOC + 2∠OBC = 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

Question 15

P is a point on the bisector of ∠ABC.
If the line through P, parallel to BA meets BC at Q, prove that ΔBPQ
is an isosceles triangle.

Solution 15

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

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.

Question 16

The image of an object placed at a
point A before a plane mirror LM is seen at the point B by an observer at D,
as shown in the figure. Prove that the image is as far behind the mirror as
the object is in front of the mirror.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 16

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.)

Question 17

In the adjoining figure, explain how one can find the breadth of the river without crossing it.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 17

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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 18

In a ΔABC,
D is the midpoint of side AC such that BD = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle. Show that ∠ABC is a right angle.

Solution 18

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

D
is the mid-point of AC.

AD = CD = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Given,
BD =
R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle 

AD = CD =
BD

Consider
AD = 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°

2∠ABC = 180°

∠ABC = 90°

Hence,
∠ABC is a right
angle.

Question 19

“If two sides and an angle of one
triangle are equal to two sides and an angle of another triangle then the two
triangles must be congruent.” Is the statement true? Why?

Solution 19

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.

Question 20

“If two angles and a side of one
triangle are equal to two angles and a side of another triangle then the two
triangles must be congruent.” Is the statement true? Why?

Solution 20

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.

Question 21

In the given figure, two parallels
lines l and m are intersected by two parallels lines p and q. Show that
Δ ABC ≅
Δ CDA.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 21

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)

Question 22

AD is an altitude of an isosceles ΔABC
in which AB = AC.

Show that (i) AD bisects BC, (ii)
AD bisects
∠A.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 22

(i) In ΔBAD and ΔCAD

ADB = ∠ADC  (Each 90° as AD is an
altitude)

AB
= AC (given)

AD
= AD (common)

 ΔBAD ≅ ΔCAD (by RHS Congruence criterion)

BD = CD (c.p.c.t.)

Hence
AD bisects BC.

 

(ii) Also, ∠BAD = ∠CAD (c.p.c.t.)

 Hence, AD bisects ∠A.

Question 23

In the given figure, BE and CF are
two equal altitudes of
ΔABC.

Show that (i) ΔABE
≅ ΔACF,
(ii) AB = AC.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 23

(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.)

Question 24

ΔABC
and ΔDBC
are two isosceles triangles on the same base BC and vertices A and D are on
the same side of BC. If AD is extended to intersect BC at E, show that

(i) ΔABD
≅ ΔACD

(ii) ΔABE
≅ ΔACE

(iii) AE bisects ∠A as well as ∠D

(iv) AE is the perpendicular bisector of BC.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 24

(i) In ΔABD and ΔACD,

AB
= AC  (equal sides of isosceles
ΔABC)

DB
= DC  (equal sides of isosceles
ΔDBC)

AD
= AD (common)

ΔABD ≅ ACD (by SSS
congruence criterion)

(ii) Since
ΔABD ≅ ΔACD,

BAD = ∠CAD (c.p.c.t.)

∠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,

BAD = ∠CAD (c.p.c.t.)

∠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.

Question 25

In the given figure, if x = y and AB = CB, then prove that AE = CD.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 25

 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 26

In the given figure, line l is the
bisector of an angle
∠A and B is any point on l. If BP and BQ are perpendiculars
from B to the arms of ∠A, show that

(i) ΔAPB
≅ ΔAQB

(ii) BP = BQ, i.e., B is equidistant from the arms of ∠A.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 26

(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.)

Question 27

ABCD is a quadrilateral such that
diagonal AC bisect the angles
∠A and ∠C. Prove that AB = AD and CB = CD.

Solution 27

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

In
ΔABC and ΔADC,

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.)

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

Question 1

Is it possible to construct a
triangle with lengths of its sides as given below? Give reason for your
answer.

5 cm, 4 cm, 9 cm

Solution 1

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.

Question 2

Is it possible to construct a
triangle with lengths of its sides as given below? Give reason for your
answer.

8 cm, 7 cm, 4 cm

Solution 2

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.

Question 3

Is it possible to construct a
triangle with lengths of its sides as given below? Give reason for your
answer.

10 cm, 5 cm, 6 cm

Solution 3

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.

Question 4

Is it possible to construct a
triangle with lengths of its sides as given below? Give reason for your
answer.

2.5 cm, 5 cm, 7 cm

Solution 4

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.

Question 5

Is it possible to construct a
triangle with lengths of its sides as given below? Give reason for your
answer.

3 cm, 4 cm, 8 cm

Solution 5

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.

Question 6

In the given figure, PQ > PR
and QS and RS are the bisectors of
∠Q and ∠R respectively. Show that SQ > SR.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 6

In
ΔPQR,

PQ
> PR

∠PRQ > ∠PQR

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

∠SRQ > ∠SQR

SQ > SR

Question 7

D is any point on the side AC of ΔABC
with AB = AC. Show that CD < BD.

Solution 7

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

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

Question 8

Prove that in a triangle, other
than an equilateral triangle, angle opposite to the longest side is greater
than R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle of a right angle.

Solution 8

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

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

2∠Q > ∠R + ∠P

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

3∠Q > 180°

∠Q > 60°

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 9

In the given figure, prove that CD
+ DA + AB > BC

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 9

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

Question 10

In the given figure, prove that

CD + DA + AB + BC > 2AC.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 10

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

Question 11

If O is a point within R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleABC, show that:

AB + AC > OB + OC

Solution 11

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

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

Question 12

If O is a point within R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleABC, show that:

AB + BC + CA > OA + OB + OC

Solution 12

AB+BC+CA > OA+OB+OC

Question 13

If O is a point within R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleABC, show that:

OA + OB + OC > R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle(AB + BC + CA)

Solution 13

OA+OB+OC> R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle(AB+BC+CA)

Proof:

(i)InR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleABC,

AB+AC>BC.(i)

And in , R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleOBC,

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)InR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleOAB

OA+OB > AB(i)

InR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleOBC,

OB+OC > BC(ii)

And, in R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleOCA,

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleOA+OB+OC> R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle(AB+BC+CA)

Question 14

In the given figure, AD
BC and CD > BD. Show that AC > AB.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 14

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

In
ΔADB and ΔADS,

BD
= SD (by construction)

ADB = ∠ADS   (Each equal to 90°)

AD
= AD (common)

ΔADB ≅ ΔADS (by SAS congruence criterion)

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

Question 15

In the given figure, D is a point
on side BC of a
ΔABC and E is a point such that CD = DE. Prove that AB + AC
> BE.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 15

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

Question 16

In ΔABC, ∠A = 50° and ∠B = 60°. Determine the longest and shortest sides of the
triangle.

Solution 16

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.

Question 17

In R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleABC, if R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleA = 90o, which is the longest side?

Solution 17

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 18

In R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleABC, if R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleA = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleB = 45o, name the longest side.

Solution 18

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 19

In ΔABC,
∠A
= 100°
and ∠C
= 50°.
Which is its shortest side?

Solution 19

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.

Question 20

In R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleABC, side AB is produced to D such that BD = BC. If R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleB = 60o and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleA = 70o, prove that (i) AD > CD and (ii) AD > AC.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 20

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 21

In the given figure, ∠B
< ∠A
and ∠C
< ∠D.
Show that AD < BC.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 21

In
ΔAOB,

B
< ∠A


AO < BO ….(i)

In
ΔCOD,

C
< ∠D


DO < CO ….(ii)

Adding (i) and (ii),

AO + DO < BO + CO


AD < BC

Question 22

AB and CD are respectively the
smallest and largest sides of quadrilateral ABCD. Show that
∠A
> ∠C
and ∠B
> ∠D.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 22

Construction:
Join AC and BD.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

In
ΔABC,

BC
> AB

∠BAC
> ∠ACB ….(i)

In
ΔACD,

CD
> AD

∠CAD
> ∠ACD ….(ii)

Adding (i) and (ii), we get

BAC
+ ∠CAD
> ∠ACB
+ ∠ACD

∠A
> ∠C

In
ΔADB,

AD
> AB

∠ABD
> ∠ADB ….(iii)

In
ΔBDC,

CD
> BC

∠CBD
> ∠BDC ….(iv)

Adding (iii) and (iv), we get

ABD
+ ∠CBD
> ∠ADB
+ ∠BDC

∠B
> ∠D

Question 23

In a quadrilateral ABCD, show that

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

Solution 23

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

In
ΔABC,

AB
+ BC > AC
 ….(i)

In
ΔACD,

DA
+ CD > AC 
 ….(ii)

In
ΔADB,

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

Question 24

In a quadrilateral ABCD, show that

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

Solution 24

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

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)

Question 25

In R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleABC, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleB = 35o, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleC = 65o and the bisector of R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A TriangleBAC meets BC in X. Arrange AX, BX and CX in descending order.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 25

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

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

Question 1

Which of the following
is not a criterion for congruence of triangles?

(a) SSA

(b) SAS

(c) ASA

(d)SSS

Solution 1

Correct
option: (a)

SSA is not a criterion
for congruence of triangles.

Question 2

In ∆ABC, ∠A = 40o
and ∠B = 60o.
Then the longest side of ∆ABC is

(a) BC

(b) AC

(c) AB

(d) Cannot be
determined

Solution 2

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 3

In the given figure AB > AC. Then, which of the following is true?

(a) AB < AD

(b) AB = AD

(c) AB > AD

(d) Cannot be determined

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 3

Correct option: (c)

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 4

In the given figure AB > AC. If BO and CO are the bisectors of ∠B and ∠C respectively, then

(a) OB = OC

(b) OB > OC

(c) OB < OC

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 4

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 5

In the given figure, AB = AC and OB = OC. Then, ABO : ∠ACO = ?

(a) 1 :1

(b) 2 : 1

(c) 1 :2

(d) None of these

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 5

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 6

If the altitudes from
two vertices of a triangle to the opposite sides are equal, then the triangle
is

(a) Equilateral

(b) Isosceles

(c) Scalene

(d) Right-angled

Solution 6

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 7

In ∆ABC and ∆DEF, it is
given that AB = DE and BC = EF. In order that ∆ABC ≅ ∆DEF, we must
have

(a) ∠A = ∠D

(b) ∠B = ∠E

(c) ∠C = ∠F

(d) None of these

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 7

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 8

In ∆ABC and ∆DEF, it is
given that ∠B = ∠E and ∠C = ∠F. In order
that ∆ABC ≅ ∆DEF, we must
have

(a) AB = DF

(b) AC = DE

(c) BC = EF

(d) ∠A = ∠D

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 8

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 9

In ∆ABC and ∆PQR, it is
given that AB = AC, ∠C = ∠P and ∠B = ∠Q. Then,
the two triangles are

(a) Isosceles but
not congruent

(b) Isosceles but
congruent

(c) Congruent but
not isosceles

(d) Neither
congruent nor isosceles

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 9

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 10

Which is true ?

(a) A triangle can
have two right angles.

(b) A triangle can
have two obtuse angles.

(c) A triangle can
have two acute angles.

(d) An exterior
angle of a triangle is less than either of the interior opposite angles.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Solution 10

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 11

Fill in the blanks with

(a) (Sum of any two sides of a triangle)……(the third side)

(b) (Difference of any two sides of a triangle)…..(the third side)

(c) (Sum of three altitudes of a triangle) ……. (sum of its three sides

(d) (Sum of any two sides of a triangle)….. (twice the median to the 3rd side)

(e) (Perimeter of a triangle)……(sum of its medians)

Solution 11

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 12

If AB = QR, BC = RP and
CA = PQ, then which of the following holds?

(a) ∆ABC ≅ ∆PQR

(b) ∆CBA ≅ ∆PQR

(c) ∆CAB ≅ ∆PQR

(d) ∆BCA ≅ ∆PQR

Solution 12

Correct
option: (c)

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 13

Fill in the blanks

(a) Each angle of
an equilateral triangles measures …….

(b) Medians of an
equilateral triangle are ……….

(c) In a right
triangle the hypotenuse is the ….. side

(d) Drawing a ∆ABC with AB =
3cm, BC= 4 cm and CA = 7 cm is ……..

Solution 13

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 14

If ∆ABC ≅ ∆PQR then which of the following is not true?

(a) BC = PQ

(b) AC = PR

(c) BC = QR

(d) AB = PQ

Solution 14

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

Question 15

In Δ ABC, AB = AC and ∠B = 50°. Then, ∠A = ?

(a) 40°

(b) 50°

(c) 80°

(d) 130°

Solution 15

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°

Question 16

In Δ ABC, BC = AB and ∠B = 80°. Then, ∠A = ?

(a) 50°

(b) 40°

(c) 100°

(d) 80°

Solution 16

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°

2∠A = 100°

∠A = 50°

Question 17

In ΔABC, ∠C = ∠A, BC = 4 cm and AC = 5 cm. Then, AB = ?

(a) 4
cm

(b) 5
cm

(c) 8
cm

(d) 2.5
cm

Solution 17

Correct
option: (a)

In
ΔABC,

C = ∠A

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

AB = 4 cm

Question 18

Two sides of a triangle are of length 4 cm and 2.5 cm. The
length of the third side of the triangle cannot be

(a) 6
cm

(b) 6.5
cm

(c) 5.5
cm

(d) 6.3
cm

Solution 18

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.

Question 19

In ΔABC, if ∠C > ∠B, then

(a) BC
> AC

(b) AB
> AC

(c) AB
< AC

(d) BC
< AC

Solution 19

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

Question 20

It is given that ∆ABC ≅ ∆FDE in which AB
= 5 cm, ∠B = 40o, ∠A = 80o
and FD = 5 cm. Then which of the following is true?

(a) ∠D = 60o

(b) ∠E = 60o

(c) ∠F = 60o

(d) ∠D = 80o

Solution 20

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Congruence Of Triangles And Inequalities In A Triangle

 

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