rd sharma solutions class 9 congruent triangles ex 10...rd sharma solutions class 9 congruent...

RD Sharma Solutions Class 9 Congruent Triangles Ex 10.1

RD Sharma Solutions Class 9 Chapter 10 Ex 10.1

(1) In Fig. (10).22, the sides BA and CA have been produced such that: BA = AD and CA = AE. Prove that segment DE | BC.

Solution:

Given that, the sides BA and CA have been produced such that BA = AD and CA = AE and given to prove DE || BC Consider triangle BAC and DAE,

We have

BA = AD and CA= AE [given in the data]

E ■D

B CAnd also ZB AC = ZD AE [vertically opposite angles]

So, by SAS congruence criterion, we have

ZBAC ~ ZD AEBC = DE and ZDEA = ZBCA, ZED A = ZCBA

[Corresponding parts o f congruent triangles are equal]

Now, DE and BC are two lines intersected by a transversal DB such that ZDEA = ZBCA i.e.. alternate angles are equal Therefore, DE BC BC.

(2) In a PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.

Solution: Given that.

In PQR, PQ = QR and L, M, N are midpoints o f the sides PQ, QP and RP respectively and given to prove that LN = MN

Here we can observe that PQR is an isosceles triangle

p

PQ = QR and Z QPR = Z QRP____(i)

And also, L and M are midpoints o f PQ and QR respectively

And also, PQ = QR

Now, consider A LPN and A MRN, LP = MR [From - (2)]

Z LPN = Z MRN _ [From - (1)]

Z QPR and Z LPN and Z QRP and Z MRN are same.

PN = NR [N is midpoint o f PR]

So, by SAS congruence criterion, we have A LPN = A MRN

LN = MN [Corresponding parts o f congruent triangles are equal]

(3) In fig. (10).23, PQRS is a square and SRT is an equilateral triangle. Prove that (i) PT = QT (ii) Z TQR = 15°

T

S

P e

Solution: Given that PQRS is a square and SRT is an equilateral triangle. And given to prove that

(i) PT =QT and (ii)ZTQ R=15°

Now, PQRS is a square

PQ = QR = RS = SP___ (i)And Z SPQ = Z PQR = Z QRS = Z RSP = 90 ° = right angle

And also, SRT is an equilateral triangle.

SR = RT = TS__ (ii)

And Z TSR = Z SRT = Z RTS = 60°

From (i) and (ii)

PQ = QR = SP = SR = RT = TS __ (iii)

And also,

Z TSP = Z TSR + Z RSP = 60° + 90° + 150° ZTRQ = ZTRS + Z SRQ = 60° + 90° + 150°150° ___ (iv)

SP = RQ [From (iii)]

So, by SAS congruence criterion we have

A TSP = A TRQ

PT = QT [Corresponding parts of congruent triangles are equal] Consider A TQR.

QR = TR [From (iii)]

A TQR is a isosceles triangle.

Z QTR = Z TQR [angles opposite to equal sides]

Now,

Sum of angles in a triangle is equal to 180°

=> Z QTR + Z TQR + ZTRQ = 180°

=> 2 Z TQR + 150° = 180° [From (iv)]

=> 2 Z TQR = 180°-150°

=> 2 Z TQR = 30° Z TQR = 15°]...

Flence proved

(4) Prove that the medians of an equilateral triangle are equal.

Solution:

Given,

To prove the medians of an equilateral triangle are equal.

Median: The line Joining the vertex and midpoint o f opposite side. Now, consider an equilateral triangle ABC.

Let D,E,F are midpoints o f BC, CA and AB.

Then, AD, BE and CF are medians o f ABC.

D Is midpoint o f BC => BD = DC = —

Similarly, CE = EA = 4p-

Z TSR = Z TRQ =

Since A ABC is an equilateral triangle

=> AB = BC = CA ____ (i)

=> BD = DC = CE = EA = AF = FB = ^ = ^ p = ^ ____ (ii)

And also, Z ABC = Z BCA = Z CAB = 60° ____(Hi)

Now, consider A ABD and A BCE AB = BC [From (i)]

BD = CE [From (ii)]

Now, in A TSR and A TRQ

TS = TR [From (iii)]

Z ABD = Z BCE [From (iii)] [Z ABD and Z ABC and Z BCE and ZBCA are same]

So, from SAS congruence criterion, we have

A ABD = A BCE

AD = BE ___ (iv)


Now, consider A BCE and A CAF, BC = CA [From (i)]

Z BCE = Z CAF [From (ii)]

[Z BCE and Z BCA and Z CAF and Z CAB are same]

CE = AF [From (ii)]

So, from SAS congruence criterion, we have

A BCE = A CAF

BE = CF (v)


From (iv) and (v), we have

AD = BE = CF

Median AD = Median BE = Median CF

The medians of an equilateral triangle are equal.

Flence proved

(5) In a A ABC, if Z A = 120° and AB = AC. Find Z B and Z C.

Solution:

Consider a A ABC

Given Mat Z A = 120° and AB = AC and given to find Z B and Z C.

A

b c

We can observe that A ABC is an isosceles triangle since AB = AC

Z B = Z C (i)

[Angles opposite to equal sides are equal]

We know that sum of angles in a triangle is equal to 180°

=> Z A + Z B + Z C= 180° [From (i)]

= > Z A + Z B + Z B = 1 80°

=> 120° + 2ZB = 180°

=> 2ZB = 180° - 120°

=> Z B = Z C = 30°

(6) In a A ABC, if AB = AC and Z B = 70°. Find Z A.

Solution:

Consider a A ABC, if AB = AC and Z B = 70°

Z B = Z C [Angles opposite to equal sides are equal]Z B = Z C = 70°

Since, AB = AC A ABC is an isosceles triangle

And also.

Sum of angles in a triangle = 180°

Z A + Z B + Z C = 180°Z A + 70 ° + 70 ° = 180°Z A = 180° - 140 °Z A = 40 °

(7) The vertical angle o f an isosceles triangle is (10)0 Find its base angles.

Solution:

Consider an isosceles AABC such that AB = AC

Given that vertical angle A is (10)0°

To find the base angles

Since A ABC is isosceles

Z B = Z C [Angles opposite to equal sides are equal]

And also.

Sum of interior angles of a triangle = 180°

Z A + Z B + Z C = 180°(10)0 ° + Z B Z B = 180°2 Z B = 1 8 0 °-(1 0 )0 °Z B = 40 °Z B = Z C = 40 °

(8) In Fig. (10).24, AB = AC and Z ACD = (10)5 °. Find Z BAC.

Solution:

Consider the given figure

We have,

AB = AC and Z ACD = (10)5 °

Since, Z BCD = 180° = Straight angle

Z B C A + Z A C D =180°Z B C A + (10)5° = 180°Z BCA = 180° -(10)5°ZBCA= 75°

And also,

A ABC is an isosceles triangle [AB = AC]

Z ABC = Z ACB [Angles opposite to equal sides are equal]

From (i), we have

Z ACB = 75°

Z ABC = Z ACB = 75°

And also.

Sum of Interior angles of a triangle = 180°

Z ABC = Z BCA + Z CAB =180°75° + 75° + Z CAB =180°150°+ Z BAC = 180°Z BAC = 180°-150° = 30°Z BAC = 30°

(9) Find the measure o f each exterior angle o f an equilateral triangle.

Solution:

Given to find the measure of each exterior angle o f an equilateral triangle consider an equilateral triangle ABC.

AB = BC = CA and Z ABC = Z BCA= CAB = = 60° ___ (i)

Now,

Extend side BC to D, CA to E and AB to F.

Here BCD is a straight line segment

BCD = Straight angle =180°Z BCA + Z ACD = 180° [From (i)]60° + ZACD = 180°ZACD = 120°

Similarly, we can find Z FAB and Z FBC also as 120° because ABC is an equilateral triangle

Z ACD = Z EAB - Z FBC = 120°

Flence, the median of each exterior angle o f an equilateral triangle is 120°

We know that for an equilateral triangle

(10) If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.

Solution:

ED is a straight line segment and B and C an points on it.

Z EBC = Z BCD = straight angle = 180°Z EBA+ Z ABC = Z ACB + Z ACD Z EBA = Z ACD + Z ACB - Z ABC Z EBA = Z ACD [From (i) ABC = ACD]

Z ABE = Z ACD

Flence proved

(11) In Fig. (10).2(5) AB = AC and DB = DC, find the ratio Z ABD : Z ACD.

Consider the figure

AB = AC, DB = DC and given to find the ratio

Z ABD = Z ACD

Now, A ABC and A DBC are isosceles triangles since AB = AC and DB = DC respectively

Z ABC = Z ACB and Z DBC = Z DCB [Angles opposite to equal sides are equal]

Now consider,

Z ABD: Z ACD

(Z ABC - Z DBC): (Z ACB - Z DCB)(Z ABC - Z DBC): (Z ABC - Z DBC) [Z ABC = Z ACB and Z DBC = Z DCB]1:1

ABD: ACD = 1:1

(12) Determine the measure of each of the equal angles o f a right-angled isosceles triangle.

OR

ABC is a right-angled triangle in which Z A = 90° and AB = AC. Find Z B and Z C.

Solution:

ABC is a right angled triangle

Consider on a right - angled isosceles triangle ABC such that

Z A = 90 ° and AB = AC Since,

AB = AC => Z C = Z B ___ (i)


Now, Sum of angles in a triangle = 180°

Z A + Z B + Z C = 1 80°

=> 90° + Z B+ Z B = 180°

=> 2Z B = 90°

=> Z B = 45°

Z B = 45°, Z C = 45°

Hence, the measure of each of the equal angles o f a right-angled Isosceles triangle Is 45°

(13) AB is a line segment. P and 0 are points on opposite sides o f AB such that each of them is equidistant from the points A and B (See Fig. (10).26). Show that the line PC is perpendicular bisector o f AB.

Consider the figure.

AB is a line segment and P, Q are points on opposite sides o f AB such that

AP = BP (i)

AC = B0 (ii)

We have to prove that PC is perpendicular bisector o f AB.

Now consider A PAG and A PBG,

We have

AP = BP [From (i)]

AC = B0 [From (ii)]

And PC - PC [Common site]

A PAG — A PBG _____(iii) [From SAS congruence]

Now, we can observe that APB and ABC are isosceles triangles. [From (i) and (ii)]

Z PAB = Z ABQ and Z QAB = Z QBA

Now consider A PAC and A PBC

C is the point of intersection of AB and PQ

PA = PB [From (i)]

Z APC = Z BPC [From (ii)]

PC = PC [common side]

So, from SAS congruency of triangle A PAC = A PBCAC = CB and Z PCA = Z PBC_______ (iv) [ Corresponding parts o f congruent triangles are equal]is line segment ZACP + ZBCP = 180°Z ACP = Z PCB Z ACP = Z PCB = 90°<

We have AC = CB => C is the midpoint o f AB

From (iv) and (v)

We can conclude that PC is the perpendicular bisector of AB

Since C is a point on the line PQ, we can say that PQ is the perpendicular bisector of AB.

And also, ACB



(1) In fig. (10).40, it is given that RT = T S , Z 1 = 2 Z 2 and 4 = 2 Z (3) Prove that A RBT = A SAT.

Solution:

In the figure, given that

RT = TS .... (i)

Z 1 = 2 Z 2 .... (ii)

And Z 4 = 2 Z 3 .... (iii)

To prove that A RBT = AS AT.

Let the point of intersection RB and SA be denoted by 0

Since RB and SA intersect at 0

Z AOR = Z BOS [Vertically opposite angles]

Z 1 = Z 42 Z 2 = 2 Z 3 [From (ii) and (iii)]

Z 2 = Z 3 .... (iv)

Now we have RT =TS in A TRS

A TRS is an isosceles triangle

Z TRS = Z TSR .... (v)

But we have

Z TRS = Z TRB + Z 2 .... (vi)

Z TSR = Z TSA + Z 3 .....(vii)

Putting (vi) and (vii) in (v) we get

Z TRB + Z 2 = Z TSA + Z B

=> Z TRB = Z TSA [From (iv)]

Now consider A RBT and A SAT

RT = ST [From (i)]

Z TRB = Z TSA [From (iv)] Z RTB = Z STA [Common angle]

From ASA criterion of congruence, we have

A RBT = A SAT

(2) Two lines AB and CD intersect at 0 such that BC is equal and parallel to AD. Prove that the lines AB and CD bisect at 0.

Solution: Given that lines AB and CD Intersect at 0

Such that BC || AD and BC = AD ..... (i)

We have to prove that AB and CD bisect at 0.

To prove this first we have to prove that A AOD = A BOC

(3) BD and CE are bisectors of Z B and Z C of an isosceles A ABC with AB = AC. Prove that BD = CE

Solution:

Given that A ABC is isosceles with AB = AC and BD and CE are bisectors of Z B and Z C We have to prove BD = CE

Since AB = AC

=> A ABC = A ACB.... (i)


Since BD and CE are bisectors of Z B and Z C

Z ABD = Z DBC = Z BCE = ECA =

Now,

Consider A EBC = A DCB

Z EBC = Z DCB [Z B = Z C] [From (i)]

BC = BC [Common side]

Z BCE = Z CBD [From (ii)]

So, by ASA congruence criterion, we have A EBC = A DCB

Now,

CE = BD [Corresponding parts of congruent triangles we equal]

or, BD = CE

Flence proved

Since AD || BC and transversal AB cuts at A and B respectively

Z DAO = Z OBC ..... (ii) [alternate angle]

And similarly AD || BC and transversal DC cuts at D and C respectively

Z ADO = Z OBC ....(Hi) [alternate angle]

Since AB end CD intersect at 0.

Z AOD = Z BOC [Vertically opposite angles]

Now consider A AOD and A BOD

Z DAO = Z OBC [From (ii)]

AD = BC [From (i)]

And Z ADO = Z OCB [From (iii)]

So, by ASA congruence criterion, we have

A AOD =* A BOC

Now,

A0= OB and DO = OC [Corresponding parts of congruent triangles are equal)

Lines AB and CD bisect at 0.

Flence proved



(1) In two right triangles one side an acute angle o f one are equal to the corresponding side and angle o f the other. Prove that the triangles are congruent.

Solution:

Given that, in two right triangles one side and acute angle of one are equal to the corresponding side and angles of the other.

We have to prove that the triangles are congruent.

Let us consider two right triangles such that

Z B = Z C = 90 ° ..... (i)

AB = DE ..... (ii)

Z C = Z F From(iii)

Now observe the two triangles ABC and DEF

Z C = Z F [From (iii)]

Z B = Z E [From (i)]

and AB = DE [From (ii)]

So, by AAS congruence criterion, we have A ABC = A DEF

Therefore, the two triangles are congruent

Flence proved

(2) If the bisector o f the exterior vertical angle o f a triangle be parallel to the base. Show that the triangle is isosceles.

Solution:

Given that the bisector of the exterior vertical angle of a triangle is parallel to the base and we have to prove that the triangle is isosceles. Let ABC be a triangle such that AD is the angular bisector of exterior vertical angle EAC and AD || BC

Let Z EAD = (i), Z DAC =(ii), Z ABC =(iii) and Z ACB = (iv)

(i) = (ii) [AD is a bisector of Z EAC]

(i) = (iii) [Corresponding angles]

and (ii) = (iv) [alternative angle]

(iii) = (iv)

AB = AC

Since, in A ABC, two sides AB and AC are equal we can say that A ABC is isosceles triangle.

(3) In an isosceles triangle, if the vertex angle is twice the sum o f the base angles, calculate the angles o f the triangle.

Solution:

Let A ABC be isosceles such that AB = AC.

ZB = ZC

Given that vertex angle A is twice the sum of the base angles B and C. i.e., Z A= 2(Z B + Z C)

Z A = 2(Z B + Z B)

Z A = 2(2 Z B)

Z A = 4(Z B)

Now, We know that sum of angles in a triangle =180°

Z A + Z B + Z C = 1 80°

4 Z B + Z B + Z B = 180°

6 Z B =180°

Z B = 30°

Since, Z B = Z C

Z B = Z C = 30°

And Z A = 4 Z B

Z A = 4 x 30° = 120°

Therefore, angles of the given triangle are 120°, 30° and 30°.

= 428 and LB = LC]

(4) PQR is a triangle in which PQ= PR and is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.

Solution: Given that PQR is a triangle such that PQ = PR ant S is any point on the side PQ and ST || QR.

PS = PT

Since, PQ= PR

PQR is an isosceles triangle.Z Q = Z R (or) Z PQR = Z PRQ

Now, Z PST = Z PQR and Z PTS = Z PRQ [Corresponding angles as ST parallel to QR]

Since, Z PQR = Z PRQ

Z PST = Z PTS

Now, In A PST, Z PST = Z PTS

A PST is an isosceles triangle

Therefore, PS = PT

(5) In a A ABC, it is given that AB = AC and the bisectors o f B and C intersect at O. If M is a point on BO produced, prove that Z MOC = Z ABC.

Solution:

Given that in A ABC,

AB = AC and the bisector of Z B and Z C intersect at 0. If M is a point on BO produced

Since,

AB = AC

ABC is isosceles Z B = Z C (or)Z ABC = Z ACB

Now,

BO and CO are bisectors of Z ABC and Z ACB respectively

=> ABO = Z OBC =Z ACO = Z OCB = | Z ABC = | Z ACB .... (i)

We have, in A OBC

Z OBC + Z OCB + Z BOC = 180° .... (ii)

And also

Z BOC + Z COM = 180°....(iii) [Straight angle]

Equating (ii) and (iii)

Z OBC + Z OCB + Z BOC = Z BOC + Z MOC Z OBC + Z OCB = Z MOC [From (i)]2 Z OBC = Z MOC [From (i)]

2 ( \ Z . ABC) = Z MOC [From (i)]

Z ABC = Z MOC

Therefore, Z MOC = Z ABC

(6) P is a point on the bisector o f an angle ABC. If the line through P parallel to AB meets BC at Q, prove that triangle BPQ is isosceles.

Sol:

Given that P is a point on the bisector of an angle ABC, and PC || AB.

Since,

BP is bisector of Z ABC = Z ABP = Z PBC .... (i)

Now,

PC || AB

Z BPO = Z ABP....(ii) [alternative angles]

From (i) and (ii), we get

Z BPO = Z PBC (or) Z BPC = Z PBC

Now,

In BPC,

Z BPC = Z PBC

A BPC is an isosceles triangle.

Flence proved

(7) Prove that each angle o f an equilateral triangle is 60°.

Sol:

Given to prove each angle of an equilateral triangle is 60°.

Let us consider an equilateral triangle ABC.

Such that AB = BC = CA

Now, AB = BC

Z A = Z C .... (i) [Opposite angles to equal sides are equal]

And BC = AC

Z B = Z A ....(ii)

From (i) and (ii), we get

Z A = Z B = Z C .... (iii)

Sum of angles in a triangle = 180

Z A + Z B + Z C = 180 Z A + Z A + Z A = 180 3 Z A = 180 Z A = 60

Z A = Z B = Z C = 60

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

(8) Angles A A, B, C o f a triangle ABC are equal to each other. Prove that ABC is equilateral.

Sol:

Given that angles A, B, C of a triangle ABC equal to each other.

We have to prove that A ABC is equilateral

We have, Z A = Z B = Z C

Now,

Z A = Z B

BC=AC [opposite sides to equal angles are equal]

And Z B = Z C

AC=AB

From the above we get

AB = BC = AC

A ABC is equilateral.

(9) ABC is a triangle in which Z B = 2 Z C . Disa point on BC such that AD bisects Z BAC and AB = CD. Prove that [Z BAC = 72.

Solution: Given that in ABC, Z B= 2 Z C and D is a point on BC such that AD bisectors Z BAC and

AB = CD.

Now, draw the angular bisector of Z ABC, which meets AC in P.

Join PD

Let C = Z ACB = y

Z B = Z ABC = 2 Z C = 2y and also Let Z BAD = Z DACZ BAC = 2x [ AD is the bisector of Z BAC ]

Now, in A BPC,

Z CBP = y [BP is the bisector of Z ABC ]Z PCB = y

Z CBP = Z PCB = y [PC = BP]

Consider, A ABP and A DCP, we have

A ABP = A DCP = y

AB = DC [Given]

And PC = BP [From above]

So, by SAS congruence criterion, we have A A B P = A D C P

Now,

Z BAP = Z CDF and AP = DP [Corresponding parts of congruent triangles are equal]

Z BAP = Z CDP = 2

Consider, A APD,

We have AP = DP

= Z ADP = Z DAP

But Z DAP = x

Z ADP = Z DAP = x

Now

In A ABD.

Z ABD + Z BAD + Z ADB = 180

And also Z ADB + Z ADC = 180° [Straight angle]

From the above two equations, we get

Z ABD + Z BAD + Z ADB = Z ADB + Z ADC

2y + x = Z ADP + Z PDC 2y + x = x + 2x 2y=2xy = x (or) x = y

We know.

Sum of angles in a triangle = 180°

So, In A ABC,

Z A + Z B + Z C = 1 80°2x + 2y + y = 180° [Z A = 2x, Z B = 2y, Z C = y]2(y)+3y=180°[x = y]5y = 180° y = 36°

Now, Z A = Z BAC = 2x36° = 72°

(10) ABC is a right angled triangle in which Z A = 90° and AB = AC. Find Z B and Z C.

Solution: Given that ABC is a right angled triangle such that Z A = 90° and AB = AC

Since, AB= AC

A ABC is also isosceles.

Therefore, we can say that A ABC is right angled isosceles triangle.

Z C = Z B and Z A = 90°....(i)

Now, we have sum of angled in a triangle = 180°

Z A + Z B + Z C = 180°90° + Z B + Z B = 180° [From(i)]2Z B = 180° - 90°Z B = 45°

Therefore, Z B = Z C = 45'



(1) In fig (10).9(2) It is given that AB = CD and AD = BC. Prove that A ADC = A CBA.D B

ASolution:

Given that in the figure AB = CD and AD = BC.

We have to prove AADC = A CBAD B

Now,

Consider A ADC and A CBA.

We have

AB = CD [Given]

BC = AD [Given]

And AC = AC [Common side]

So, by SSS congruence criterion, we have

A ADC 9* A CBA Hence proved

(2) In a A PQR. IF PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.

Sol: Given that in A PQR, PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively

We have to prove LN = MN.

p

Join L and M, M and N, N and L

We have PL = LQ, QM = MR and RN = NP

[Since, L, M and N are mid-points o f Pp. QR and RP respectively]

And also PQ = QR

PL = LQ = QM = MR = .... (i) Using mid-point theorem.

We have

MN || PQ and MN =

MN = PL = LQ.... (ii)

Similarly, we have

LN || QR and LN = (1/2)QR

LN = QM = M R.... (iii)

From equation (i), (ii) and (iii), we have

PL = LQ = QM = MR = MN = LN

LN = MN



(1) ABC is a triangle and D is the mid-point of BC. The perpendiculars from B to AB and AC are equal. Prove that the triangle is isosceles.

Sol:

Given that, in two right triangles one side and acute angle of one are equal to the corresponding side and angle of the other

We have to prove that the triangles are congruent

Let us consider two right triangles such that

Z B = Z E = 90 .... (i)

AB = DE....(ii)

Z C = Z F.... (iii)

Now observe the two triangles ABC and DEF

Z C = Z F[From (iii)]

Z B = Z E [From (i)]

And AB = DE[From (ii)]

So, by AAS congruence criterion, we have

A ABC =* A DEFTherefore, the two triangles are congruent

Flence proved

(2) ABC is a triangle in which BE and CF are, respectively, the perpendiculars to the sides AC and AB. If BE = CF, prove that A ABC is isosceles

Sol: Given that ABC is a triangle in which BE and CF are perpendicular to the sides AC and AS respectively such that BE = CF.

To prove, A ABC is isosceles

Now, consider A BCF and A CBE,

Z BFC = CEB = 90 [Given]

BC = CB[Common side]

And CF = BE[Given]

So, by RFIS congruence criterion, we have

A BFC = A CEBNow,

Z FBC = Z EBC [Incongruent triangles corresponding parts are equal]

Z ABC = Z ACB

AC = AB[Opposite sides to equal angles are equal in a triangle]

A ABC is isosceles

(3) If perpendiculars from any point within an angle on its arms are congruent. Prove that it lies on the bisector of that angle.

Solution:

Given that, if perpendicular from any point within, an angle on its arms is congruent, prove that it lies on the bisector of that angle.

Let us consider an angle ABC and let BP be one of the arm within the angle

Draw perpendicular PN and PM on the arms BC and BA such that they meet BC and BA in N and M respectively.

Now, in A BPM and A BPN

We have Z BMP = Z BNP = 90° [given]

BP = BP[Common side]

And MP = NP[given]

So, by RHS congruence criterion, we have

A BPM S A BPNNow, Z MBP = Z NBP [Corresponding parts of congruent triangles we equal]

BP is the angular bisector of ZABC.

Hence proved

(4) In fig. (10).99, AD _L CD and CB _L CD. If AQ = BP and DP = CQ, prove that Z DAQ = Z CBP.

Given in the fig. (10).99, AD _L CD and CB _L CD.

And AQ = BP and DP = CQ,

To prove that Z DAQ = Z CBP

Given that DP = QC

Add PQ on both sides

DP + PQ = PQ + QC DQ = PC....(i)

Now, consider triangle DAQ and CBP,

We have

Z ADQ = Z BCP = 90° [given] AQ = BP [given]

And DQ = PC[From (i)]

So, by RHS congruence criterion, we have

A DAQ =* A CBPNow,

Z DAQ = Z CBP [Corresponding parts of congruent triangles are equal]

Hence proved

(5) ABCD is a square, X and Y are points on sides AD and BC respectively such that AY = BX. Prove that BY = AX and Z BAY = Z ABX.

Solution:

Given that ABCD is a square, X and Y are points on sides AD and BC respectively such that AY = BX.

To prove BY = AX and Z BAY = Z ABX

Join B and X, A and Y.

Since, ABCD is a square

Z DAB = Z CBA = 90°Z XAB = Z YAB = 90°....(i)

Now, consider triangle XAB and YBA

We have

Z XAB = Z YBA = 90°. [From (i)] BX = AY[given]

And AB = BA[Common side]

So, by RHS congruence criterion, we have AXAB = A YBANow, we know that corresponding parts of congruent triangles are equal.

BY = AX and Z BAY = Z ABX

Hence proved

(6) Which of the following statements are true (T) and which are false (F):

(i) Sides opposite to equal angles of a triangle may be unequal.

(ii) Angles opposite to equal sides of a triangle are equal

(iii) The measure of each angle of an equilateral triangle is 60

(iv) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle may be isosceles.

(v) The bisectors of two equal angles of a triangle are equal.

(vi) If the bisector of the vertical angle of a triangle bisects the base, then the triangle may be isosceles.

(vii) The two altitudes corresponding to two equal sides of a triangle need not be equal.

(viii) If any two sides of a right triangle are respectively equal to two sides of other right triangle, then the two triangles are congruent.

(ix) Two right-angled triangles are congruent if hypotenuse and a side of one triangle are respectively equal to the hypotenuse and a side of the other triangle.

Solution:

(i) False (F)

Reason: Sides opposite to equal angles of a triangle are equal

(ii) True (F)

Reason: Since the sides are equal, the corresponding opposite angles must be equal

(iii) True (T)

Reason: Since all the three angles of equilateral triangles are equal and sum of the three angles is 180, each angle will be equal to ^ y - = 60°

(iv) False (F)

Reason: Here the altitude from the vertex is also the perpendicular bisector of the opposite side.

The triangle must be isosceles and may be an equilateral triangle.

(v) True (T)

Reason: Since it an isosceles triangle, the lengths of bisectors of the two equal angles are equal

(vi) False (F)

Reason: The angular bisector of the vertex angle is also a median

=> The triangle must be an isosceles and also may be an equilateral triangle.

(vii) False (F)

Reason: Since two sides are equal, the triangle is an isosceles triangle. The two altitudes corresponding to two equal sides must be equal.

(viii) False (F)

Reason: The two right triangles may or may not be congruent

(ix) True (T)

Reason: According to RHS congruence criterion the given statement is true.

(7) Fill the blanks In the following so that each of the following statements is true.

(i) Sides opposite to equal angles of a triangle are__

(ii) Angle opposite to equal sides of a triangle are__

(iii) In an equilateral triangle all angles are__

(iv) In A ABC, if Z A = Z C, then AB =

(v) If altitudes CE and BF of a triangle ABC are equal, then AB __

(vi) In an isosceles triangle ABC with AB = AC, if BD and CE are its altitudes, then BD is __CE.

(vii) In right triangles ABC and DEF, if hypotenuse AB = EF and side AC = DE, then A ABC = A __Solution:

(i) Sides opposite to equal angles of a triangle are equal

(ii) Angles opposite to equal sides of a triangle are equal

(iii) In an equilateral triangle all angles are equal Reason: Since all sides are equal in a equilateral triangle, the angles opposite to equal sides will be equal.

(iv) In a A ABC, if Z A = Z C, then AB = BC

(v) If altitudes CE and BF of a triangle ABC are equal, then AB = AC

(vi) In an isosceles triangle A ABC with AB = AC, if BD and CE are its altitudes, then BD is equal to CE

(vii) In right triangles ABC and DEF, if hypotenuse AB = EF and side AC = DE, then, A ABC = A EFD.



(1) In A ABC, if Z A = 40° and Z B = 60°. Determine the longest and shortest sides of the triangle.

Solution:

Given that in A ABC, Z A = 40° and Z B = 60°

We have to find longest and shortest side

We know that.

Sum of angles in a triangle 180°

Z A + Z B + Z C = 180°

40° + 60° + Z C = 180°

Z C = 180°-((10)0°) = 80°

Z C = 80°

Now,

=> 40° < 60° < 80° = Z A < Z B < Z C

=> Z C is greater angle and Z A is smaller angle.

Now, Z A < Z B < Z C

=> BC < AAC < AB [Side opposite to greater angle is larger and side opposite to smaller angle is smaller]

AB is longest and BC is smallest or shortest side.

(2) In a A ABC, if Z B = Z C = 45°, which is the longest side?

Solution: Given that in A ABC,

Z B = Z C = 45°

We know that.

Sum of angles in a triangle =180°

Z A + Z B + Z C = 180°

Z A + 45° + 45° = 180°

Z A = 180° - (45° + 45°) = 1 8 0 °- 90° = 90°

Z A = 90°

(3) In A ABC, side AB is produced to D so that BD = BC. If Z B = 60° and Z A = 70°. prove that: (i) AD > CD (ii) AD > AC

Sol: Given that, in A ABC, side AB is produced to D so that BD = BC.

Z B = 60°, and Z A = 70°

(i) AD > CD (ii) AD > AC

First join C and D

We know that.

Sum of angles in a triangle =180°

Z A + Z B + Z C = 180°

70° + 60° + Z C = 180°

Z C = 180° - (130°) = 50°

Z C = 50°

Z ACB = 50°.... (i)

And also in A BDC

Z DBC =180 - Z ABC [ABD is a straight angle]

1 8 0 - 60° = 120°

and also BD = BC[given]

Z BCD = Z BDC [Angles opposite to equal sides are equal]

Now,

Z DBC + Z BCD + Z BDC = 180° [Sum of angles in a triangle =180°]

=>120° + Z BCD + Z BCD = 180°

=> 120° + 2 Z BCD = 180°

=> 2 Z BCD = 180° - 120° = 60°

=> Z BCD = 30°

=> Z BCD = Z BDC = 30° ....(ii)

Now, consider A ADC.

Z BAC => Z DAC = 70° [given]

Z BDC => Z ADC = 30° [From (ii)]

Z ACD = Z ACB+ Z BCD

= 50° + 30°[From (i) and (ii)] =80°

Now, Z ADC < Z DAC < Z ACD

AC < DC < AD [Side opposite to greater angle is longer and smaller angle is smaller]

AD > CD and AD > AC

Flence proved

Or,

We have,

Z ACD > Z DAC and Z ACD > Z ADC

AD > DC and AD> AC [Side opposite to greater angle is longer and smaller angle is smaller]

(4) Is it possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?

Sol:

Given lengths of sides are 2cm, 3cm and 7cm.

To check whether it is possible to draw a triangle with the given lengths of sides

We know that,

A triangle can be drawn only when the sum of any two sides is greater than the third side.

So, let's check the rule.

2 + 3 ^ 7 or 2 + 3 < 7

2 + 7 > 3

and 3 + 7 > 2

Here 2 + 3 ^ 7

So, the triangle does not exit.

(5) 0 is any point in the interior of A ABC. Prove that

(i) AB + AC > OB + OC

(ii) AB + BC + CA > OA + QB + OC

(iii) OA + OB + OC > (1 /2)(AB + BC +CA)

Solution:

Given that 0 is any point in the interior of A ABC

To prove

(i) AB + AC > OB + OC

(ii) AB + BC + CA > OA + CB + OC

(iii) OA + OB + OC > (1 /2)(AB + BC +CA)

We know that in a triangle the sum of any two sides is greater than the third side.

So, we have

In A ABC

AB + BC > AC

BC + AC > AB

AC + AB > BC

In AO BC

OB + OC > BC..... (i)

In A OAC

OA + OC > AC.... (ii)

In A OAB

OA + OB > AB.... (iii)

Now, extend (or) produce BO to meet AC in D.

Now, in A ABD, we have

AB + AD > BD

AB + AD > BO + OD..... (iv) [BD = BO + OD]

Similarly in A ODC, we have

OD + DC > OC.... (v)

(i) Adding (iv) and (v), we get

AB + AD + OD + DC > BO + OD + OC

AB + (AD + DC) > OB + OC

AB + AC > OB + OC.... (vi)

Similarly, we have

BC + BA > OA + OC.... (vii)

and CA+ CB > OA + OB.... (viii)

(ii) Adding equation (vi), (vii) and (viii), we get

AB + AC + BC + BA + CA + CB > OB + OC + OA + OC + OA + OB => 2AB + 2BC + 2CA > 20A + 20B + 20C

=> 2(AB + BC + CA) > 2(OA + OB + OC)

=> AB + BC + CA > OA + OB + OC

(iii) Adding equations (i), (ii) and (iii)

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

20A + 20B + 20C > AB + BC + CA

We get = 2(OA + OB + OC) > AB + BC +CA

(OA + OB + OC) > (1 /2)(AB + BC +CA)

(6) Prove that the perimeter of a triangle is greater than the sum of its altitudes.

Solution:

Given, A ABC in which AD _L BC, BE _L AC and CF _L AB.

To prove,

AD + BE + CF < AB + BC + AC

Figure:

Proof:

We know that of all the segments that can be drawn to a given line, from a point not lying on it, the perpendicular distance i.e, the perpendicular

line segment is the shortest.

Therefore

AD _L BC

AB > AD and AC > AD

AB + AC > 2AD.... (i)

BE _LAC

BA > BE and BC > BE

BA + BC > 2BE.... (ii)

CF _L AB

CA > CF and CB > CF

CA + CB > 2CF.... (iii)

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

AB + AC + BA + BC + CA + CB > 2AD + 2BE + 2CF

2AB + 2BC + 2CA > 2(AD + BE + CF)

AB + BC + CA > AD + BE + CF

The perimeter of the triangle is greater than that the sum of its altitudes

Flence proved

(7) In Fig., prove that:

(ii) CD + DA + AB > BC

Solution:

To prove

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

(ii) CD + DA+ AB > BC

From the given figure.

We know that, in a triangle sum of any two sides is greater than the third side

(i) So,

In A ABC, we have

AB + BC > A C .... (i)

In A ADC, we have

CD + DA > AC.... (ii)

Adding (i) and (ii), we get

AB + BC + CD + DA > AC + AC

AB + BC + CD + DA > 2AC

(ii) Now, In A ABC, we have,

AB +AC > BC.... (iii)

And in A ADC, we have

CD + DA > AC

Add AB on both sides

CD + DA + AB > AC + AB

From equation (iii) and (iv), we get,

CD + DA + AB > AC + AB > BC

CD + DA + AB > BC

Hence proved

(8) Which of the following statements are true (T) and which are false (F)?

(i) Sum of the three sides of a triangle is less than the sum of its three altitudes.

(ii) Sum of any two sides of a triangle is greater than twice the median drown to the third side

(iii) Sum of any two sides of a triangle is greater than the third side.

(iv) Difference of any two sides of a triangle is equal to the third side.

(v) If two angles of a triangle are unequal, then the greater angle has the larger side opposite to it

(vi) Of all the line segments that can be drawn from a point to a line not containing it, the perpendicular line segment is the shortest one.

Solution:

(i) False (F)

Reason: Sum of these sides of a triangle is greater than sum of its three altitudes

(ii) True (T)

(iii) True (T)

(iv) False (F)

Reason: The difference of any two sides of a triangle is less than third side

(v ) True (T)

Reason: The side opposite to greater angle is longer and smaller angle is shorter in a triangle

(vi) True (T).

Reason: The perpendicular distance is the shortest distance from a point to a line not containing it.

(9) Fill in the blanks to make the following statements true.

(i) In a right triangle the hypotenuse is th e__ side.

(ii) The sum of three altitudes of a triangle is __ than its perimeter.

(iii) The sum of any two sides of a triangle is __ than the third side.

(iv) If two angles of a triangle are unequal, then the smaller angle has th e__ side opposite to it.

(v) Difference of any two sides of a triangle is __ than the third side.

(vi) If two sides of a triangle are unequal, then the larger side has__ angle opposite to it.

Solution:

(i) In a right triangle the hypotenuse is the largest side

Reason: Since a triangle can have only one right angle, other two angles must be less than 90

The right angle is the largest angle.

Hypotenuse is the largest side.

(ii) The sum of three altitudes of a triangle is less than its perimeter.

(Hi) The sum of any two sides of a triangle is greater than the third side, (iv) If two angles of a triangle are unequal, then the smaller angle has

the smaller side opposite to it.

(v) Difference of any two sides of a triangle is less than the third side.

(vi) If two sides of a triangle are unequal, then the larger side has greater angle opposite to it.

rd sharma solutions class 9 congruent triangles ex 10...rd sharma solutions class 9 congruent...

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