# 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

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.

(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.

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.

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°

⇒ 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

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

_{}

_{}

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.

_{ }

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.

_{}

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.

_{}

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.

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.

In the adjoining figure, X and Y are respectively two points on equal sides AB and AC of _{}ABC such that AX = AY. Prove that CX = BY.

_{}

In _{}ABC, D is the midpoint of BC. If DL _{}AB and DM _{}AC such that DL = DM, prove that AB = AC.

_{}_{ }

In _{}ABC, AB = AC and the bisectors of _{B} and _{}C meet at a point O. Prove that BO = CO and the ray AO is the bisector _{}A.

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.

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)

⇒ AD = BC (c.p.c.t.)

In the given figure, AD and BC are

equal perpendiculars to a line segment AB. Show that CD bisect AB.

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.

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.

In

ΔABC, AB = AC

⇒ ∠ABC = ∠ACB

⇒ ∠OBC = ∠OCB ….(i)

Now,

by exterior angle property,

∠MOC = ∠OBC + ∠OCB

⇒ ∠MOC = 2∠OBC [From (i)]

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

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.

In

ΔABC, AB = AC

⇒ ∠ABC = ∠ACB

⇒ ∠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

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.

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.

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.

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.)

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

_{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. }

_{}

_{}

In a ΔABC,

D is the midpoint of side AC such that BD = . Show that ∠ABC is a right angle.

D

is the mid-point of AC.

⇒ AD = CD =

Given,

BD =

⇒ 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.

“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?

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.

“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?

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.

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.

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)

AD is an altitude of an isosceles ΔABC

in which AB = AC.

Show that (i) AD bisects BC, (ii)

AD bisects ∠A.

(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.

In the given figure, BE and CF are

two equal altitudes of ΔABC.

Show that (i) ΔABE

≅ ΔACF,

(ii) AB = AC.

(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.)

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.

(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.

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

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.

(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.)

ABCD is a quadrilateral such that

diagonal AC bisect the angles ∠A and ∠C. Prove that AB = AD and CB = CD.

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

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

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.

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

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.

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

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.

triangle with lengths of its sides as given below? Give reason for your

answer.

2.5 cm, 5 cm, 7 cm

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.

triangle with lengths of its sides as given below? Give reason for your

answer.

3 cm, 4 cm, 8 cm

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.

In the given figure, PQ > PR

and QS and RS are the bisectors of ∠Q and ∠R respectively. Show that SQ > SR.

In

ΔPQR,

PQ

> PR

⇒ ∠PRQ > ∠PQR

⇒ ∠SRQ > ∠SQR

⇒ SQ > SR

D is any point on the side AC of ΔABC

with AB = AC. Show that CD < BD.

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

Prove that in a triangle, other

than an equilateral triangle, angle opposite to the longest side is greater

than of a right angle.

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°

In the given figure, prove that CD

+ DA + AB > BC

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

In the given figure, prove that

CD + DA + AB + BC > 2AC.

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

If O is a point within _{}ABC, show that:

AB + AC > OB + OC

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

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

If O is a point within _{}ABC, show that:

AB + BC + CA > OA + OB + OC

AB+BC+CA > OA+OB+OC

If O is a point within _{}ABC, show that:

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

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

Proof:

(i)In_{}ABC,

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)In_{}OAB

OA+OB > AB(i)

In_{}OBC,

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)

In the given figure, AD ⊥

BC and CD > BD. Show that AC > AB.

Construction:

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

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

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.

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

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

triangle.

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.

In _{}ABC, if _{}A = 90^{o}, which is the longest side?

_{}

In _{}ABC, if _{}A = _{}B = 45^{o}, name the longest side.

_{}

In ΔABC,

∠A

= 100°

and ∠C

= 50°.

Which is its shortest side?

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.

In _{}ABC, side AB is produced to D such that BD = BC. If _{}B = 60^{o} and _{}A = 70^{o}, prove that (i) AD > CD and (ii) AD > AC.

_{}

In the given figure, ∠B

< ∠A

and ∠C

< ∠D.

Show that AD < BC.

In

ΔAOB,

∠B

< ∠A

⇒

AO < BO ….(i)

In

ΔCOD,

∠C

< ∠D

⇒

DO < CO ….(ii)

Adding (i) and (ii),

AO + DO < BO + CO

⇒

AD < BC

AB and CD are respectively the

smallest and largest sides of quadrilateral ABCD. Show that ∠A

> ∠C

and ∠B

> ∠D.

Construction:

Join AC and BD.

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

In a quadrilateral ABCD, show that

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

BD).

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

In a quadrilateral ABCD, show that

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

AC).

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)

In _{}ABC, _{}B = 35^{o}, _{}C = 65^{o} and the bisector of _{}BAC meets BC in X. Arrange AX, BX and CX in descending order.

_{}_{ }

_{ }

_{}

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

Which of the following

is not a criterion for congruence of triangles?

(a) SSA

(b) SAS

(c) ASA

(d)SSS

Correct

option: (a)

SSA is not a criterion

for congruence of triangles.

In ∆ABC, ∠A = 40^{o}

and ∠B = 60^{o}.

Then the longest side of ∆ABC is

(a) BC

(b) AC

(c) AB

(d) Cannot be

determined

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

Correct option: (c)

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

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

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

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

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

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

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.

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 3^{rd} side)

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

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

Correct

option: (c)

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 ……..

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

(a) BC = PQ

(b) AC = PR

(c) BC = QR

(d) AB = PQ

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

(a) 40°

(b) 50°

(c) 80°

(d) 130°

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°

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

(a) 50°

(b) 40°

(c) 100°

(d) 80°

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°

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

Correct

option: (a)

In

ΔABC,

∠C = ∠A

⇒ AB = BC (sides opposite to equal angles are

equal)

⇒ AB = 4 cm

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

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.

In ΔABC, if ∠C > ∠B, then

(a) BC

> AC

(b) AB

> AC

(c) AB

< AC

(d) BC

< AC

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

It is given that ∆ABC ≅ ∆FDE in which AB

= 5 cm, ∠B = 40^{o}, ∠A = 80^{o}

and FD = 5 cm. Then which of the following is true?

(a) ∠D = 60^{o}

(b) ∠E = 60^{o}

(c) ∠F = 60^{o}

(d) ∠D = 80^{o}