SELINA Solution Class 9 Chapter 9 Triangles (Congruency in Triangles) Exercise 9B

Question 1

On the sides AB and AC of triangle ABC, equilateral triangle ABD and ACE are drawn. 
Prove that:  (i) ∠CAD = ∠BAE 
                   (ii) CD = BE

Sol:

Given:  ΔABD is an equilateral triangle.
ΔACE is an equilateral triangle
We need to prove that 
(i) ∠CAD = ∠BAE 

Proof:
(i) ΔABD is equilateral
∴ Each angel = 60°
⇒ ∠BAD = 60°                            ...(1)
Similarly,
 ΔACE is equilateral
∴ Each angel = 60°
⇒ ∠CAE = 60°                             ...(2)
⇒ ∠BAD = ∠CAE      ...[ from (1) and (2) ]...(3)
Adding ∠BAC to both sides, we have
⇒ ∠BAD + ∠BAC = ∠CAE + ∠BAC
⇒ ∠CAD = ∠BAE                         ...(4)

(ii) In ΔCAD and ΔBAE
AC = AE                 ...[ ΔACE is equilateral ]
∠CAD = ∠BAE      ... [ from (4) ]
AD = AB                ...[ ΔABD is euilateral ]
∴ By Side-Angel-Side criterion of congruency, 
ΔCAD ≅ ΔBAE
The corresponding parts of the congruent
triangles are congruent.
∴ CD = BE             ...[ by c.p.c.t ]
Hence proved.

Question 2.1

In the following diagram, ABCD is a square and APB is an equilateral triangle.

(i) Prove that: ΔAPD≅ ΔBPC
(ii) Find the angles of ΔDPC.

Sol:

Given: ABCD is a Square and ΔAPB is an equilateral triangle.
We need to
(i) Prove that: ΔAPD≅ ΔBPC
(ii) Find the angles of ΔDPC

(i) Proof:
AP = PB = AB            ...[ APB is an equilateral triangle ]
Also, we have,
∠PBA = ∠PAB = ∠APB = 60°                        ...(1)
Since ABCD is a square, we have
∠A =∠ B = ∠C = ∠D = 90°                        ...(2)
Since ∠DAP = ∠A - PAB                                ...(3)
⇒ ∠DAP = 90° - 60°  
⇒ ∠DAP =30°     ...[ from (1) and (2) ]        ...(4)

Similarly ∠CBP = ∠B - ∠PBA
⇒ ∠CBP = 90°  - 60° 
⇒ ∠CBP = 30°       ...[ from (1) and (2) ] ...(5)
⇒ ∠DAP = ∠CBP  ....[ from (1) and (2) ] ...(6)
In ΔAPD and ΔBPC
AD = BC          ...[ Sides of square ABCD ]
∠DAP = ∠CBP  ...[ from(6) ]
AP= BP           [ Sides of equilateral ΔAPB ]
∴ By Side-Angel-Side Criterion of Congruence, we have,
ΔAPD ≅ ΔBPC

(ii)
AP = PB = AB  ....[ ΔAPB is an equilateral triangle ] ...(7)
AB = BC = CD = DA  ...[ Sides of square ABCD ] ...(8)
From (7) and (8), we have
AP = DA aand PB = BC                                   ... (9)
In ΔAPD,
AP = DA                    ...[ from (9) ]
∠ADP = ∠APD   ...[ Angel opposite to equal sides are equal ]                                        ...(10)
∠ADP + ∠APD+ +∠DAP + 180°  ...[ Sum of angel of a triangle = 180° ]
⇒ ∠ADP + ∠ADP + 30° = 180°    [ from (3), ∠DAP =30° from (10), ∠ADP = ∠APD ]
⇒ ∠ADP + ∠ADP = 180°  -  30°
⇒ 2∠ADP = ${\displaystyle \frac{{150}^0}{2}}$
⇒∠ADP= 75°
We have ∠PDC =∠D - ∠ADP 
⇒∠PDC = 90° - 75°
⇒∠PDC =15°                                                         ...(11)
In BPC, 
PB = BC     ...[ from (9) ]
∴  ∠PCB =∠ BPC  ...[Angel opposite to equal sides are equal ]                                                              ... (12)
∠PCB + ∠BPC + ∠CBP = 180°   ....[ Sum of angel of a triangle = 180°  ]
⇒ ∠PCB + ∠PCB + 30°  = 180°  ....[ from (5), ∠CBP =  from (12) , ∠PCB =∠BPC ]
⇒ 2∠PCB =  180°  -  30°
⇒ 2∠PCB = ${\displaystyle \frac{150°}{2}}$
⇒ ∠PCB = 75°
 We have ∠PCD = ∠C - ∠PCB 
⇒ ∠PCD = 90°  -  75°
⇒ ∠PCD = 15°                                             ... (13)
In ΔDPC, 
∠PDC = 15° 
∠PCD = 15°
∠PCD + ∠PDC + ∠DPC = 180°   ...[ Sum of angles of a triangle = 180°  ]

⇒ 15°  + 15°  + ∠DPC = 180°
⇒ ∠DPC = 180°   -  30°
⇒ ∠DPC = 150° 
∴  Angles of DPC, are:  15° , 150°  , 15° 

Question 2.2

In the following diagram, ABCD is a square and APB is an equilateral triangle.

(i) Prove that: ΔAPD ≅ ΔBPC
(ii) Find the angles of ΔDPC.

Sol:

Given: ABCD is a Square and ΔAPB is an equilateral triangle.

(i) Proof: In ΔAPB,
AP = PB = AB            ...[ APB is an equilateral triangle ]
Also, we have,
∠PBA = ∠PAB = ∠APB = 60°                         ...(1)
Since ABCD is a square, we have
∠A =∠ B = ∠C = ∠D = 90°                           ...(2)
Since ∠DAP = ∠A + ∠PAB                            ..(3)
⇒ ∠DAP = 90° + 60°  
⇒ ∠DAP = 150°     ...[ from (1) and (2) ]        ...(4)

Similarly ∠CBP = ∠B + ∠PBA
⇒ ∠CBP = 90° + 60° 
⇒ ∠CBP = 150°       ...[ from (1) and (2) ] ...(5)
⇒ ∠DAP = ∠CBP  ....[ from (1) and (2) ] ...(6)
In ΔAPD and ΔBPC
AD = BC          ...[ Sides of square ABCD ]
∠DAP = ∠CBP  ...[ from(6) ]
AP= BP           [ Sides of equilateral ΔAPB ]
∴ By Side-AAngel-SIde Criterion of Congruence, we have,
ΔAPD ≅ ΔBPC

(ii)
AP = PB = AB  ....[ ΔAPB is an equilateral triangle ] ...(7)
AB = BC = CD = DA  ...[ Sides of square ABCD ] ...(8)
From (7) and (8), we have
AP = DA aand PB = BC                                  ... (9)
In ΔAPD,
AP = DA                    ...[ from (9) ]
∠ADP = ∠APD   ...[ Angel opposite to equal sides are equal ]                                             ...(10)
∠ADP + ∠APD+ +∠DAP + 180°  ...[ Sum of angel of a triangle = 180° ]
⇒ ∠ADP + ∠ADP + 150° = 180°    [ from (3), ∠DAP =150° from (10), ∠ADP = ∠APD ]
⇒ ∠ADP + ∠ADP = 180° - 150°
⇒ 2∠ADP = 30°
⇒ ∠ADP = ${\displaystyle \frac{{30}^{\circ }}{2}}$
⇒∠ADP= 15°
We have ∠PDC =∠D - ∠ADP 
⇒∠PDC =90° - 15°
⇒∠PDC =75°                                               ...(11)
In ΔBPC, 
PB = BC     ...[ from (9) ]
∴  ∠PCB =∠ BPC  ...[Angel opposite to equal sides are equal ]                                           ... (12)
∠PCB + ∠BPC + ∠CBP = 180°   ....[ Sum of angel of a triangle = 180°  ]
⇒ ∠PCB + ∠PCB + 30°  = 180°  ....[ from (5), ∠CBP = 150° from (12) , ∠PCB =∠BPC ]
⇒ 2∠PCB =  180° - 150°
⇒ 2∠PCB = ${\displaystyle \frac{{30}^{\circ }}{2}}$
⇒ ∠PCB = 15°
 We have ∠PCD = ∠C - ∠PCB 
⇒ ∠PCD = 90° - 15°
⇒ ∠PCD = 75°                                           ... (13)
In ΔDPC, 
∠PDC = 75° 
∠PCD = 75°
∠PCD + ∠PDC + ∠DPC = 180°   ...[ Sum of angles of a triangle = 180°  ]

⇒ 75°  + 75°  + ∠DPC = 180°
⇒ ∠DPC = 180° - 150°
⇒ ∠DPC = 30° 
∴  Angles of DPC, are: 75°, 30° , 75° 

Question 3

In the figure, given below, triangle ABC is right-angled at B. ABPQ and ACRS are squares.

Prove that: 
(i) ΔACQ and ΔASB are congruent.
(ii) CQ = BS.

Sol:

Given: A(Δ ABC) is right-angled at B.
ABPQ and ACRS are squares

To Prove:
(i) ΔACQ ≅ ΔASB
(ii) CQ = BS

Proof:
(i)
∠ QAB = 90°    ...[ ABPQ is a square ] ...(1)
∠ SAC = 90°     ...[ ACRS is a square ] ...(2)
From (1) and (2) , We have
∠ QAB = ∠SAC                        ...(3)
Adding ∠BAC to both sides of (3), We have
∠ QAB + ∠BAC = ∠SAC + ∠BAC
⇒ ∠QAC = ∠SAB                   ...(4)

In ΔACQ and ΔASB,
QA = QB             ...[ Sides of a square ABPQ ]
∠QAC = ∠SAB    ...[ From(4) ]
AC = AS             ...[ sides of a square ACRS ]
∴ By Side -Angle-Side criterion of congruence,
ΔACQ ≅ ΔASB

(ii) 
The corresponding parts of the congruent triangles are congruent,
∴ CQ = BS           ...[ c.p.c.t. ]

Question 4

In a ΔABC, BD is the median to the side AC, BD is produced to E such that BD = DE.
Prove that: AE is parallel to BC.

Sol:

Given: A(ΔABC) in which BD is the median to AC.
BD is produced to E such that BD = DE,
We need to prove that AE II BC.
Construction: Join AE

Proof:
AD = DC          ...[ BD is median to AC ] ...(1)
In ΔBDC and ΔADE,
BD = DE                      ...[ Given ]
∠BDC = ∠ADE = 90°   ...[ Vertically opposite angles ]
AD = DC                     ...[ from(1) ]
∴ By Side-Angle-Side Criterion of congruence,
ΔBDC ≅ ΔADE
The corresponding parts of the congruent triangles are congruent.
∴ ∠EAD = ∠BCD     ...[ c.p.c.t. ]
But these are alternate angles and AC is the transversal.
Thus, AE || BC.

Question 5

In the adjoining figure, OX and RX are the bisectors of the angles Q and R respectively of the triangle PQR.
If XS ⊥ QR and XT ⊥  PQ ;

prove that: (i) ΔXTQ ≅ ΔXSQ. 
                   (ii) PX bisects angle P.

Sol:

Given: A( ΔPQR ) in which QX is the bisector of ∠Q. and RX is the bisector of ∠R.
XS ⊥ QR and XT ⊥  PQ.
We need to prove that
(i) ΔXTQ ≅ ΔXSQ.
(ii) PX bisects angle P.
Construction: Draw XZ ⊥ PR and join PX.

Proof: 
(i) In ΔXTQ and ΔXSQ,
∠QTX = ∠QSX = 90°    ...[ XS ⊥ QR and XT ⊥  PQ ]
∠TQX = ∠SQX              ...[ QX is bisector of ∠Q ]
QX = QX                      ...[ Common ]
∴ By Angle-Angle-Side Criterion of congruence,
ΔXTQ ≅ ΔXSQ

(ii) The corresponding parts of the congruent triangles are congruent.
∴ XT = XS           ...[ c.p.c.t. ]
In ΔXSR ≅ ΔXZR
∠XSR = ∠XZR = 90°   ...[ XS ⊥ QR and ∠XSR = 90° ]
∠SRX = ∠ZRX             ...[ RX is bisector of ∠R ]
RX = RX                      ....[ Common ]
∴ By Angle-Angle-Side criterion of congruence,
ΔXSR ≅ ΔXZR
The corresponding parts of the congruent triangles are congruent.
∴ XS = XZ             ...[ c.p.c.t. ] ...(2)
From (1) and (2)
XT = XZ                          ....(3)
In ΔXTP and ΔXZP
∠XTP = ∠XZP = 90°       ....[ Given ]
Hyp. XP = Hyp. XP         ....[ Common ]
XT = XZ                         ....[ from(3) ]           
∴ By Right angle-Hypotenuse-side criterion of congruence,
ΔXTP ≅ ΔXZP
The corresponding parts of the congruent triangles are congruent.
∴ ∠XPT = ∠XPZ          ...[ c.p.c.t. ]
∴ PX bisects ∠P.

Question 6

In the parallelogram ABCD, the angles A and C are obtuse. Points X and Y are taken on the diagonal BD such that the angles XAD and YCB are right angles.
Prove that: XA = YC.

SOl:

ABCD is a parallelogram in which ∠A and ∠C are obtuse.

Points X and Y are taken on the diagonal BD.
Such that ∠XAD = ∠YCB = 90°.
We need to prove that XA = YC
Proof:
ln ΔXAD and ΔYCB
∠XAD = ∠YCB= 90°        ...[ Given ]
AD = BC                          ...[ Opposite sides of a parallelogram ]
∠ADX = ∠CBY                 ...[ Alternate angles ]
∴ By Angle-Side-Angle criterion of congruence,
ΔXAD ≅ ΔYCB
The corresponding parts of the congruent triangles are congruent.
∴ XA = YC                    ...[ c.p.c.t. ]
Hence proved.

Question 7

ABCD is a parallelogram. The sides AB and AD are produced to E and F respectively, such produced to E and F respectively, such that AB = BE and AD = DF.
Prove that: ΔBEC ≅ ΔDCF.

Sol:

ABCD is a parallelogram, The sides AB and AD are produced to E and F respectively,
such that AB = BE and AD = DF
We need to prove that ΔBEC ≅ ΔDCF.

Proof: 
AB = DC          ...[ Opposite sides of a parallelogram ] ...(1)
AB = BE           ...[ Given ] ...(2)
From (1) and (2), We have
BF = DC           ...(3)
AD = BC          ...[ Opposite sides of a parallelogram ] ...(4)  
AD = DF          ....[Given]

From (4) and (5), we have
BC = DF                             ...(6)
Since AD II BC, the corresponding angles are equal.
∴ ∠DAB = ∠CBE                ...(7)
Since AB II DC, the corresponding angles are equal.
∴ ∠DAB = ∠FDC                ...(8)
From (7) and (8), we have
∠CBE = ∠FDC

ln ΔBEC and ΔDCF
BF = DC                            ....[ from (3) ]
∠CBE = ∠FDC                   ...[ from (9) ] 
BC = DF                            ....[ from (6) ]
∴ By Side-Angle-Side criterion of congruence,
ΔBEC ≅ ΔDCF
Hence proved.

Question 8

In the following figures, the sides AB and BC and the median AD of triangle ABC are equal to the sides PQ and QR and median PS of the triangle PQR.
Prove that ΔABC and ΔPQR are congruent.

Sol:

Since, BC = QR, We have
BD = QS and DC = SR  ....[ D is the mid-point of BC and S is the mid-point of QR ] 

In ΔABD and ΔPQS,
AB = PQ                  ...(1)
AD = PS                  ...(2)
BD = QS                 ...(3)
Thus, by Side-Side-Side criterion of congruence,
We have ΔABD ≅ ΔPQS
Similarly, in ΔADC and ΔPSR
AD = PS               ...(4)
AC = PR               ...(5)
DC = SR              ....(6)
Thus, by Side-Side-Side criterion of congruence,
We have ΔADC ≅  ΔPSR
We have
BC = BD + DC    ...[ D is the mid-point of BC ]
     = QS + SR     ...[ From (3) and (6) ]
     = QR           ....[ S is the mid-point of QR ] ...(7)
Now consider the triangles ΔABC and ΔPQR
AB = PQ              ...[ from(1) ]
BC = QR              ...[ from(7) ]
AC = PR              ...[ from(7) ]
∴ By Side-Side-Side criterion of congruence, we
have ΔABC ≅  ΔPQR
Hence proved.

Question 9

In the following diagram, AP and BQ are equal and parallel to each other. 

Prove that:  
(i) ΔAOP≅ ΔBOQ.
(ii) AB and PQ bisect each other.

Sol:

In the figure, AP and BQ are equal and parallel to each other. 
∴ AP = BQ and AP || BQ. 
We need to prove that
(i) ΔAOP≅ ΔBOQ.
(ii) AB and PQ bisect each other

(i) ∵ AP || BQ 
∴∠APO =∠BOQ           ...[ Alternate angles ] ...(1)
and ∠PAO =∠QBO       ...[ Alternate angles ] ...(2)
Now in ΔAOP and  ΔBOQ.
∠APO =∠BQO            ...[ from (1) ]
AP = BQ                      ...[ given ]
∠PAO = ∠QBO            ...[ from (1) ]
∴ By Angel-Side-Angel criterion of congruence, we have
ΔAOP≅ ΔBOQ.

(ii) The corresponding parts of the congruent triangles are congruent.
∴ OP = OQ                ...[ c. p. c .t ]
OA = OB                    ...[ c. p. c .t ]
Hence AB and PQ bisect each other.

Question 10

In the following figure, OA = OC and AB = BC.

Prove that:
(i) ∠AOB = 90o
(ii) ΔAOD ≅ ΔCOD
(iii) AD = CD

Sol:

Given:
In the figure, OA=OC, AB =BC
We need to prove that,
AOB = 90° 
(i) In ΔABO and ΔCBO,
AB = BC                   ...[given ]
AO = CO                ...[ given ]
OB = OB               ...[ common ]
∴By Side-Side-Side criterion of congruence, we have
ΔABO ≅ ΔCBO
The corresponding parts of the congruent triangles are congruent.
∴∠ABO = ∠CBO       ...[c. p.c.t. ]
⇒ ∠ABD = ∠CBD        
and ∠AOB = ∠COB   ...[c. p.c t ]
We have
∠AOB + ∠COB = 180°         .....[ linear pair ]
⇒ ∠AOB = ∠ COB= 90° and AC ⊥ BD 

(ii) In ΔAOD and ΔCOD,
OD = OD                ...[ common ]
∠AOD = ∠COD      ...[ each=90° ]
AO = CO                ...[ given]
∴By Side-Angel-Side criterion of congruence, we have
ΔAOD ≅ ΔCOD

(iii) The corresponding parts of the congruent
triangles are congruent.
∴AD = CD             ...[c. p.c t ]
Hence proved. 

Question 11.1

The following figure has shown a triangle ABC in which AB = AC. M is a point on AB and N is a point on AC such that BM = CN.
Prove that:  (i) AM = AN  (ii) ΔAMC ≅ ΔANB

Sol:

In ΔABC, AB = AC. m and N are points on
AB and AC such that BM = CN
BN and CM are joined

(i) In ΔAMC and ΔANB
AB = AC               ...[ Given ]  ...(1)
BM = CN              ....[ Given ] ...(2)
Subtracting (2) from (1), we have
AB - BM = AC - CN
⇒ AM = AN                   ...(3)

(ii) Consider the triangles AMC and ANB
AC = AB                      ...[ given ] 
∠AMC =  ∠ANB         ...[ common 90° ]
AM = AN                  ....[ from ( 3 ) ]
∴ By the Side-Angel-Side Criterion of congruence, we have ΔAMC ≅ ΔANB

Question 11.2

The following figure has shown a triangle ABC in which AB = AC. M is a point on AB and N is a point on AC such that BM = CN.

Prove that:  (i) BN = CM (ii) ΔBMC≅ΔCNB   

Sol:

In ΔABC, AB = AC. m and N are points on
AB and AC such that BM = CN
BN and CM are joined

(i) The corresponding parts of the congruent triangles are congruent.
∴ CM = BN         ....[ c.p.c.t ] ...(1)

(ii) Consider the triangles ΔBMC and ΔCNB
BM = CN      ...[ given ] 
BC = BC       ...[ common ]
Cm = BN      ..[ from (1) ]
∴ By Side-Side-Side criterion of congruence, we have ΔBMC ≅ ΔCNB 

Question 12

In a triangle, ABC, AB = BC, AD is perpendicular to side BC and CE is perpendicular to side AB.
Prove that: AD = CE.

Sol:

ln ΔABD and ΔCBE,
AB = BC                 ....( given )
∠ ADB = ∠ CEB = 90°  ....[Perpendiculars]

∠B = ∠B                 ....( Common angle )
∴ ΔABD ≅ ΔCBE    ....( by AAS congruence )
⇒ AD = CE            ...( c.p.c.t. )

Question 13

PQRS is a parallelogram. L and M are points on PQ and SR respectively such that PL = MR.
Show that LM and QS bisect each other.

Sol:

Given: PL = RM
To prove: SP = PQ and MP = PL
Proof:
Since SR and PQ are opposite sides of a parallelogram,
PQ = SR                          ...(i)
Also, PL = RM                 ...(ii)
Subtracting (ii) from (i),
PQ - PL = SR - RM
⇒ LQ = SM                     ....(3)
Now, in ΔSMP and ΔQLP,
∠MSP = ∠PQL              ....( alternate interior angles )
∠SMP = ∠PLQ              ....( alternate interior angles )
SM = LQ                       ....[ from(3) ]
∴ ΔSMP ≅ ΔQLP          ....( by ASA congruence )
⇒ SP = PQ and MP = PL     ....( c.p.c.t. )
⇒ LM and QS bisect each other.

Question 14

In the following figure, ABC is an equilateral triangle in which QP is parallel to AC. Side AC is produced up to point R so that CR = BP.

Prove that QR bisects PC.
Hint: ( Show that ∆ QBP is equilateral
⇒ BP = PQ, but BP = CR
⇒ PQ = CR ⇒ ∆ QPM ≅ ∆ RCM ).

Sol:

ΔABC is an equilateral triangle,
So, each of its angles equals 60°.
QP is parallel to AC,
⇒ ∠PQB = ∠RAQ = 60°
ln ΔQBP,
∠PQB = ∠BQP = 60°
So, ∠PBQ + ∠BQP + ∠BPQ = 180°   ....(angle sum property)
⇒ 60°+ 60° + ∠BPQ = 180°
⇒ ∠BPQ = 60°
So, ΔBPQ is an equilateral triangle.
⇒ QP = BP 
⇒ QP = CR                                ....(i)
Now, ∠QPM + ∠BPQ = 180°    ...(linear pair)
⇒ ∠QPM+ 60°= 180°
⇒ ∠QPM = 120°
Also, ∠RCM+ ∠ACB = 180°       ...(linear pair)
⇒ ∠RCM+ 60° = 180°
⇒ ∠RCM = 120°
ln ΔRCM and ΔQMP,
∠RCM = ∠QPM                        ....(each is 120°)
∠RMC = ∠QMP                   ...(vertically opposite angles)
QP= CR                                   ....(from(i))
⇒ ΔRCM ≅ ΔQMP  ....(AAS congruence criterion)
So, CM = PM
⇒ QR bisects PC.

Question 15

In the following figure, ∠A = ∠C and AB = BC.
Prove that ΔABD ≅ ΔCBE. 

SOl:

In triangles AOE and COD,
∠A = ∠C                    ...(given)
∠AOE = ∠COD       ...(vertically opposite angles)  
∴ ∠A + ∠AOE = ∠C + ∠COD
⇒ 180° - ∠AEO = 180° - ∠CDO
⇒ ∠AEO = ∠ CDO      ….(i)
Now, ∠AEO + ∠OEB = 180°   ....(linear pair)
And, ∠CDO + ∠ODB = 180°   ....(linear pair)
∴ ∠AEO + ∠OEB = ∠CDO + ∠ODB
⇒ ∠OEB = ∠ODB                     ....[ Using (i) ]
⇒ ∠CEB = ∠ADB                      ….(ii)
Now, in ΔABD and ΔCBE,
∠A = ∠C                                 ....(given)
∠ADB = ∠CEB                         ...[ From (ii) ]
AB = BC                                  ....(given)
⇒ ΔABD ≅ ΔCBE                    ....(by AAS congruence criterion).

Question 16

AD and BC are equal perpendiculars to a line segment AB. If AD and BC are on different sides of AB prove that CD bisects AB.

Sol:

In ΔAOD and ΔBOC,
∠ AOD = ∠ BOC     ....(vertically opposite angles)
∠ DAO = ∠ CBO      ....(each 90°)
AD = BC                  ....(given)
∴ ΔAOD ≅ ΔBOC    ...(by AAS congruence criterion) 
⇒ AO = BO             ...(c.p.c.t.)
⇒ O is the mid-point of AB.
Hence, CD bisects AB.

Question 17

In ΔABC, AB = AC and the bisectors of angles B and C intersect at point O.
Prove that : (i) BO = CO
                   (ii) AO bisects angle BAC.

Sol:

In ΔABC,
AB = AC
⇒ ∠B = ∠C ...( angles opposite to equal sides are equal )

⇒ ${\displaystyle \frac{1}{2}\angle \text{B}=\frac{1}{2}\angle \text{C}}$

⇒ ∠OBC = ∠OCB       ...[ ∵ OB and OC are bisectors of ∠B and ∠C respectively, ∠OBC = ${\displaystyle \frac{1}{2}\angle \text{B}\text{and}\angle \text{OCB}=\frac{1}{2}\angle \text{C}}$ ] ...(i)

⇒ OB = OC              ...( Sides opposite to equal angles are equal )  ...(ii)

Now, in ΔABO and ΔACO,
AB = AC                 ...( given )
∠OBC = ∠OCB       ...[ from(i) ]
OB = OC                ...[ from(ii) ] ...( proved )
∴ ΔABO ≅ ΔACO   ...( by SAS congruence criterion )
⇒ ∠BAO = ∠CAO   ...( c.p.c.t. )
⇒  AO bisects ∠BAC  ...(proved)

Question 18

In the following figure, AB = EF, BC = DE and ∠B = ∠E = 90°.

Prove that AD = FC.

SOl:

Given that, BC = DE
⇒ BC + CD = DE + CD  ....( Adding CD on both sides )
⇒ BD = CE                     ....(i)
Now, in ΔABD and ΔFEC,
AB = EF                         ....(given)
∠ABD = ∠FEC               ....(Each 90°)
BD = CE                        ....[ From (i) ]
⇒  ΔABD ≅  ΔFEC         ...(by SAS congruence criterion)
⇒ AD = FC                    ...(c.p.c.t.)

Question 19

A point O is taken inside a rhombus ABCD such that its distance from the vertices B and D are equal. Show that AOC is a straight line.

Sol:

In ΔAOD and ΔAOB,
AD = AB                 ...(given)
AO = AO                 ...(Common)
OD = OB                 ...(given) 
⇒ ΔAOD ≅ ΔAOB    ...(by SSS congruence criterion)
⇒ ∠AOD = ∠AOB    ...(c.p.c.t.)  ...(i)
Similarly, ΔDOC ≅ ΔBOC
⇒ ∠DOC = ∠BOC  ...(c.p.c.t.)  ...(ii)

But, ∠AOB + ∠AOD + ∠COD + ∠BOC = 4 Right angles ...[ Sum of the angles at a point is 4 Right angles ]
⇒ 2∠AOD + 2∠COD = 4 Right angles    ....[ Using (i) and (ii) ]
⇒ ∠AOD + ∠COD = 2 Right angles
⇒ ∠AOD + ∠COD = 180°
⇒ ∠AOD and ∠COD form a linear pair.
⇒ AO and OC are in the same straight line.
⇒ AOC is a straight line.

Question 20

In quadrilateral ABCD, AD = BC and BD = CA.
Prove that:
(i) ∠ADB = ∠BCA
(ii) ∠DAB = ∠CBA

Sol:

Given: In quadrilateral ABCD, AD = BC and BD = AC.

To Prove:

(i) ∠ADB = ∠BCA
(ii) ∠DAB = ∠CBA

Proof:

In ΔABD and ΔBAC,
AD = BC           ....(given)
BD = CA           ....(given)
AB = AB           ....(common)

∴ ΔABD ≅ ΔBAC ....(by SSS congruence criterion)

${\displaystyle \begin{array}{l}\angle \text{ADB}=\angle \text{BCA}\\ \angle \text{DAB}=\angle \text{CBA}\end{array}\}...\left(\text{c.p.c.t.}\right)}$

SChand Composite Mathematics Class 7 Chapter 9 Percentage and it's application Exercise 9B

  Exercise 9B

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Q1 | Ex-9B | Class 7 | SChand Composite Maths | Percentage and it's application | myhelper

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Question 1 

Fill in the blanks

(i) 23% + 47% + ___ =100%

(ii) 54% = 100% - ___ 

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Q2 | Ex-9B | Class 7 | SChand Composite Maths | Percentage and it's application | myhelper

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Question 2

18% of 650 boys in a school take commerce. How many boys take commerce?

Sol :

Boys take commerce=18% of 650

$=\frac{18}{100} \times 650=117$ boys

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Q3 | Ex-9B | Class 7 | SChand Composite Maths | Percentage and it's application | myhelper

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Question 3

Certain cereals contains 12% of protein. How many grams of protein is there in a 14g package?

Sol :

Protein in cereals = 12% of 14

$=\frac{12}{100} \times 14=\frac{168}{100}$

=1.68 g

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Q4 | Ex-9B | Class 7 | SChand Composite Maths | Percentage and it's application | myhelper

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Question 4

There are 160 pages in a book 15% of the pages have pictures on them. How many pages do not have pictures on them?

Sol :

If 15% pages have pictures then pages which do not have pictures=100%-15%=85%

Number of pages do not have picture=85% of 160

$=\frac{85}{100} \times 160=136$ pages

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Q5 | Ex-9B | Class 7 | SChand Composite Maths | Percentage and it's application | myhelper

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Question 5

When making a Journey I walked 12%, ran 16% and rode on scooter 28% of the total distance and for the remaining part, I travelled by bus. How many km did I travel by bus If the total distance covered was 50km?

Sol :

Journey by walk , ran , scooter=12% + 16%+23%

=56%

Journey by bus=100-56=44%

Distance covered by bus=44% of 50

$=\frac{44}{100} \times 50$

=22 km

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Q6 | Ex-9B | Class 7 | SChand Composite Maths | Percentage and it's application | myhelper

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Question 6

(i) Out of every 20 vehicles passing  a certain spot 7 were bus. What percentage was this?

Sol :

$=\frac{7}{20} \times 100$

=35%

(ii) Out of 400 tickets sold at a tennis matches 144 were sold to senior citizens. What percent of the tickets were sold to senior citizens ? What percentage were sold to the rest?

Sol :

Total tickets = 400

Tickets to senior citizens=144

Tickets to rest = 400 - 144 =256

Percentage of senior citizens 

$\frac{144}{400} \times 100$

=36%

Percentage of rest citizens

$\frac{256}{400} \times 100$

=64%

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Q7 | Ex-9B | Class 7 | SChand Composite Maths | Percentage and it's application | myhelper

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Question 7

You spend $7\frac{1}{2}$ hours out of 24 hours at school. What percentage of a day is this ?

Sol :

Percentage of the day$=\frac{7.5}{24} \times \frac{100}{10}=\frac{375}{12}$

=31.25%

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Q8 | Ex-9B | Class 7 | SChand Composite Maths | Percentage and it's application | myhelper

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Question 8

A certain alloy consists 3 part of tin to 5 parts of copper. What is the percentage compositing on the alloy?

Sol :

Ratio of tin to copper in alloy= 3 : 5

Total alloy = 3+5 =8

Percentage of tin in alloy $=\frac{3}{8} \times 100$

=37.5%

Percentage of copper in alloy $=\frac{5}{8} \times 100$

=62.5 %

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Q9 | Ex-9B | Class 7 | SChand Composite Maths | Percentage and it's application | myhelper

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Question 9

The price of a muffin increases from 4 to 5 . Wat is the percentage increase?

Sol :

Increase in price = 5-4 =1

Percentage $=\frac{\text{increase}}{\text{original}} \times 100$

$=\frac{1}{4} \times 100$

=25%

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Q10 | Ex-9B | Class 7 | SChand Composite Maths | Percentage and it's application | myhelper

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Question 10

The number of people employed in firm decreased from 240 to 210. Find the percentage decrease.

Sol :

Decrease = 240- 210=30

Decrease percentage $=\frac{30}{240} \times 100$

= 12.5%

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Q11 | Ex-9B | Class 7 | SChand Composite Maths | Percentage and it's application | myhelper

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Question 11

The speed of a train is 80 km/hr .

(i) It is increased to 130 km/hr . Find percentage increase.

Sol :

Increase = 130 - 80 = 50 km/hr

Percentage Increase $=\frac{50}{80} \times 100$

=62.5 %

(ii) Decreased to 65 km/hr . Find percentage decreased .

Sol :

Decrease = 80 - 65 =15

Percentage Decrease $=\frac{15}{80} \times 100$

=18.75%

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Q12 | Ex-9B | Class 7 | SChand Composite Maths | Percentage and it's application | myhelper

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Question 12

A four meter elastic band is increased by 10%. Find its new length.

Sol :

Increase = 10% of 4

$=\frac{10}{100} \times 4$

=0.4 m

New length = Original + Increase 

=4+0.4=4.4 m

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Q13 | Ex-9B | Class 7 | SChand Composite Maths | Percentage and it's application | myhelper

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Question 13

The length of a rope is decreased by $12\frac{1}{2}$ % . What is the length if its was 150 m ?

Sol :

Decrease = 12.5 % of 150

$=\frac{125}{100} \times \frac{150}{10}$

$=\frac{75}{4}$

=18.75

New length = 150 - 18.75 =131.25

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Q14 | Ex-9B | Class 7 | SChand Composite Maths | Percentage and it's application | myhelper

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Question 14

The monthly expenditure of a family on milk is 700 . If the price of the milk is increased by 8% . Find the increase in the expenditure of the family on milk.

Sol :

Increase = 8% of 700

$=\frac{8}{100} \times 700$

=56

Increased expenditure = 700 + 56

=756

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Q15 | Ex-9B | Class 7 | SChand Composite Maths | Percentage and it's application | myhelper

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Question 15

The population of an Indian State is 8 crore . If it is increased by 2% every year, find the population of the state after one year.

Sol :

Population = 8 crore 

Increase by = 2% of 80000000

$=\frac{2}{100} \times 80000000$

=1600000

New population

=800000000+1600000

=81600000 or 8.16 crore

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Q16 | Ex-9B | Class 7 | SChand Composite Maths | Percentage and it's application | myhelper

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Question 16

A company used 75% of its profits to buy new machinery and raw materials. If this amounted to 85071. Find the total profits of the company.

Sol :

Total Profit = 100%

Invested  75%=85071

$1\%=\frac{85071}{75}$

$100\%=\frac{85071}{75} \times 100$

=113428

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Q17 | Ex-9B | Class 7 | SChand Composite Maths | Percentage and it's application | myhelper

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Question 17

A man left 25% of his money to his mother, 55% to his daughter and the remaining 9000 to his brother. How much money did he leave ? Find the shares of his mother  and daughter.

Sol :

Let the man  have 100% of money 

He gives to mother , daughter=25+55=80%

Remaining gives to his brother=100-80=20%

According to question,

20% = 9000

then $1\%=\frac{9000}{20}$

∴$100\%=\frac{9000}{20} \times 100=45000$

Mother's share = 25% of 45000

$=\frac{55}{100} \times 45000$

=24750

S Chand CLASS 10 Chapter 9 Arithmetic and Geometric Progression Exercise 9B

  Exercise 9B

Question 1 

Ans:(i) First 15 terms of the AP : 2, 5 , 8 , 11....

Here ,  $a=2, d=5-2=3, n=15$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$s_{15}=\frac{15}{2}[2 \times 2+(15-1) \times 3]$

$=\frac{15}{2}[4+42]$

$=\frac{15}{2} \times 46=345$

(ii) First 50 terms of the A.P -27 , -23 , -19....

Here, a = - 27 , d =-23 -(-27)

= -23 +27

= 4 

n= 50

So  $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{50}{2}[2 \times(-27)+(50-1) \times 4]$

$S_{50}=25[-54 \times 49 \times 4]$

$=25[-54+196]$

$=25 \times 142$

$=3550$

(iii) Ans : First 17 terms of the A.P: $=\frac{1}{5}, \frac{-3}{10}, \frac{-4}{5}, \ldots$

Here, $a=\frac{1}{5}, d=\frac{-2}{10}-\frac{1}{5}$

$\frac{-3-2}{10}=\frac{-5}{10}=\frac{-1}{2}$

and $n=17$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$S_{17}=\frac{17}{2}\left[2 \times \frac{1}{5}+(17-1)\left(\frac{-1}{2}\right)\right]$

$=\frac{17}{2}\left[\frac{2}{5}+16\left(\frac{-1}{2}\right)\right]$

$=\frac{17}{2}\left[\frac{2}{5}-8\right]$

$=\frac{17}{2}\left[\frac{2-90}{5}\right)$

$=\frac{17}{2} \times \frac{-38}{5}$

$=\frac{-323}{5}$

$=-64 \frac{3}{2}$

(iv) Ans: First 24 terms of AP : 0 , 6, 1 , 7 , 28 ......

Here a = 0.6 d = 1.7 , 0.6 = 1.1, n = 24 

So , $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$S_{24}=\frac{24}{2}[2 \times 0.6+(24-1)(1.1)]$

$=12[1.2+23 \times 1.1]$

$=12[1.2+25.3] 84$

$=12 \times 26.5$

$=318.0$

Question 2 

Ans: (i) $\begin{aligned}=& 34+32+30+\ldots+2 \\ & Here a=34, d=32-34=-2,1=2\\ & a_{n}=a+(n-1) d \\ \Rightarrow & 2=34+(n-1) \times(-2) \\ \Rightarrow & 2-34=-2(n-1) \\ \Rightarrow & \frac{-32}{-2}=n-1 \end{aligned}$

$\Rightarrow n-1=16$

$n=16+1$

$n=17$

So , $s_{n}=\frac{n}{2}(a+l) .$

$=\frac{17}{2}(34+2)$

$=\frac{17}{2} \times 36=306$

(ii) $7+9-\frac{1}{2}+12+\ldots+67$

Here, $a=7, d=9 \frac{1}{2}-7=2 \frac{1}{2}=\frac{5}{2}, 1=67$

$U=\left(a_{n}\right)=a+(n-1) d$

$\Rightarrow 67=7+(n-1)\left(\frac{5}{2}\right)$

$67-7=\frac{5}{2}(n-1)$

$\Rightarrow \frac{60 \times 2}{5}$

$=n-1$

$\Rightarrow n-1$

$=24$

= 24n = 24 +1 

=25 

So, 

$S_{25}=\frac{n}{2}[a+1]=\frac{25}{2}[7+67]$

=$\frac{25}{2} \times 74$

$=925$

Question 3

Ans: In an AP

$d=-2, a=100,1=-10$

$l=\left(a_{n}\right)=a+(n-1) d$

$-10=100+(n-1)(-2)$

$(n-1)(-2)=-10-100=-110$

$n-1=\frac{-110}{-2}=55$

$\Rightarrow n=55+1$

=56

$s_{56}=\frac{n}{2}[a+1]=\frac{56}{2}[100-10]$

$=28 \times 90$

$=2580$

Question 4

Ans: $\begin{aligned} \Rightarrow & \text { AP is } 54,51,48, \ldots \text { and } S_{n}=513 \\ & \text { Here, } a=54, d=51-54 \\=&-3 \end{aligned}$

$\begin{aligned} & S_{n}=\frac{n}{2}[2 a+(n-1) d] \\ & 513=\frac{n}{2}[2 \times 54+(n-1) \times(-3)] \\ \Rightarrow & 513 \times 2 \\=& n[108-3 n+3] \\ \Rightarrow & 10260 \\=& 108 n-3 n^{2}+3 n \\ \Rightarrow & 3 n^{2}-11 n+1026=0 \\ \Rightarrow & n^{2}-37 n+342=0 \\ \Rightarrow & n^{2}-18 n-19 n+342=0 \end{aligned}$

$\left\{\begin{array}{l}\because 342=-18 \times(-19) \\ -37=-18-19\end{array}\right\}$

$\begin{aligned} & \Rightarrow n(n-18)-19(n-18)=0 \\ \Rightarrow &(n-18)(n-19)=0 \end{aligned}$

Either n - 18 =0 , then n = 18

Or n- 19 =0 , then n = 19

So, Number of terms = 18 or 19 

Question 5

 

Ans: $\Rightarrow S_{9}=72, d=5$

let a be the first term,

n=9

So, $S_{g}=\frac{n}{2}[2 a+(n-1) d]$

$72=\frac{9}{2}[2 a+(9-1) \times 5]$

$\frac{72 \times 2}{9}=2 a+40$

$\Rightarrow 16=2 a+90$

$2 a=16-40=$

$2 a=-24$

$a=\frac{-24}{2}$

$a=-12$

 So a = -12

$a_{10}=a+(n-1) d$

$=-12+(10-1) \times 5$

$=-12+45$

$=-33$

Question 6

Ans: In an AP, 

Sum of first 6 terms 

=42

$a_{10}: a_{30}=1: 3$

Let a be the first term and d be the common difference , then

$S_{6}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{6}{2}[2 a+(6-1) d]$

$42=3(2 a+5 d)=6 a+15 d$

$\Rightarrow 6 a+15 d=42$

$a_{10}: a_{30}=1: 3$

$\frac{a+(10-1) d}{a+(30-1) d}=\frac{1}{3} \Rightarrow \frac{a+9 d}{a+29 d}=\frac{1}{3}$

$3 a+27 d=a+29 d$

$3 a-a=29 d-27 d$

$\Rightarrow 2 a=2 d$

$\Rightarrow a=d$

From (i) 

$6 a+15 a=42 \Rightarrow 21 a=42$

$\Rightarrow a=\frac{42}{21}=2$

$So, a=2, d=2$

So first term = 2

$a_{13}= a+(n-1) d=2+(13-1) \times 2$

$=2+12 \times 2=2+24$

$=26$

Question 7

Ans:  In a AP 

$a_{13}=4 \times a_{3}$

$a_{5}=16$

Let a be the first term and d be the common difference 

 So ,$\operatorname{a}_{5}=a+(n-1) d$

$=a+(5-1) d=a+4 d$

 So , $a+4 d=16$.............(i)

Similarly  

$a_{13}=a+12 d$ and $a_{3}=a+2 d$

$So, a+12 d=4 \times(a+2 d)$

$a+12 d=4 a+8 d$

$12 d-8 d=4 a-a \Rightarrow 3 a=4 d$

$a=\frac{4}{3} d$

From (i)

 $\frac{4}{3} d+4 d=16 \Rightarrow \frac{16}{3} d=16$

$\Rightarrow d=\frac{16 \times 3}{16}=3$

So, d= 3

and a = $\frac{4}{3} d=\frac{4}{3} \times 3=4$

$a=4, d=3$

Now $_{10}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{10}{2}[2 \times 4+(10-1) \times 3]$

$=5(8+27)$

$=5 \times 35=175$

Question 8

Ans:  $\Rightarrow A P$ is $8,10,12, \ldots$.

Were $a=8, d=10-8=2, n=60$

So  $T_{n}=a+(n-1) d$

$\Rightarrow 60=8+(n-1) \times 2$

$\Rightarrow(n-1) \times 2=60-8$

$=52$

$n-1=\frac{52}{2}$

$=26$

$n=26+1$

$=27$

$T_{60}=a+(60-1) d$

$=8+59 \times 2$

$=8+61=69$

Sum of last 10 term $=s_{60}-s_{50}$

$=\frac{60}{2}(2 a+59 d)-\frac{50}{2}(2 a+49 d)$

$=30(2 a+59 d)-25(2 a+49 d)$

$=60 a+1770 d-50 a-1225 d$

$=10 a+545 d=10 \times 8+545 \times 2$

$=80+1090$

$=1170$

Question 9

Ans: $\Rightarrow$ In an $A P$,

$\begin{aligned}&a_{12} 0 \times T_{12}=-13 \\&s_{4}=24\end{aligned}$

Let a be the first term and $b$ e the common difference, then

$a_{12}=a+(n-1) d$

$\Rightarrow-13=a+(12-1) d$

$\Rightarrow a+11 d=-13 \Rightarrow a=-13-11 d$

$s_{4}=\frac{n}{2}[2 a+(n-1) d]$

$24=\frac{4}{2}[2 a+3 d]=2[2 \times(-13-11 d)+3 d]$

$\frac{24}{2}=-26-22 d+3 d \Rightarrow 12=-26-19 d$

$\Rightarrow 12+26=-19 d \Rightarrow-19 d=38$

$d=\frac{38}{-19}=-2$

and $a=-13-11 d=-13+11 \times 2$

$=-13+22=9$

so, $a=9, d=-2$

Now , $s_{10}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{10}{2}[2 \times 9+(10-1)(-2)]$

$\begin{aligned} & 5[18+9(-2)]=5(18-18) \\=& 5 \times 0 . \\=& 0 \end{aligned}$

Question 10

Ans: $\because \Rightarrow$ Natural numbos betuteen 101 and 999 which are divisible by 2&5 both are 110 , $120,130, \ldots, 990$

Here $a,=110$ and $d=10,1=990$

 Now, $l=a_{n}=a+(n-1) d$ 

$990=110+(n-1) \times 10$

$\Rightarrow 990-110=10(n-1) \Rightarrow 10(n-1)=880$

$\Rightarrow n-1=\frac{880}{10} \Rightarrow n-1=88$

So , n = 88+ 1 = 89

Now , $s_{8 9}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{89}{2}[2 \times 110+(89-1) \times 10]$

$=\frac{89}{2}[220+88 \times 10]=\frac{89}{2}[220+880]$

$=\frac{89}{2} \times 110$

$=48950$

Question 11

Ans:  $\Rightarrow 2$ - digit number greater than 50 which are divisible by 7 , leaves a remainder of 4 are:

$53,60,67,74,81,88,95$

Here $a=53$,

$\begin{aligned} d &=7 \\ l &=95 \end{aligned}$

$\begin{aligned} \therefore a_{n}(l) &=a+(h-1) d . \\ 95 &=53+(h-1) \times 7 . \end{aligned}$

$\begin{aligned} 95-53 &=7(n-1) . \\ 42 &=7(n-1) \\ \frac{42}{7} &=(n-1) \\ 6 &=(n-1) \\(n-1) &=6 \\ n &=6+1 \\ n &=7 . \end{aligned}$

Then , 

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{7}{2}[2 \times 53+(7-1) \times 7]$

$=\frac{7}{2}[106+6 \times 7]$

$=\frac{7}{2}[106+42]$

$=\frac{7}{2} \times 48$

$=518$

Question 12

Ans: (i) Integers between 100 and 200 which are divisible by 9 are 108, 117 , 126 ,......198

Here 

a= 108

d =9

and l = 198

$\begin{aligned} \therefore a_{n}(l) &=a+(n-1) d . \\ 198 &=108+(n-1) \times 9 . \\ 198-108 &=9(n-1) \\ 90 &=9(n-1) \\ \frac{90}{9} &=n-1 \\ 10 &=n-1 \\ 10+1 &=n \\ n &=11 \\  & \end{aligned}$

Then 

$\begin{aligned} S_{11} &=\frac{h}{2}[2 a+(h-1) d] . \\ &=\frac{11}{2}[2 \times 108+(11-1) \times 9] \\ &=\frac{11}{2}[216+10 \times 9] \\ &=\frac{11}{2}[216+90] \end{aligned}$

$\frac{11}{2}\times 306$

=1683

(ii) Then sum of integers from 101 to 199

Here, a = 101,

d= 1 

l = 199

and n = 99

$s_{n}=$ $\frac{n}{2}[2 a+(n-1)d]$

$=\frac{99}{2}[2 \times 101+(99-1) \times 1]$

$=\frac{99}{2}[202+98]$

$=\frac{99}{2}\times 300$

=14850

∴ Sum of integers which are not divisible by 9 

$=14850-1683$

$=13167$

Question 13

Ans:  Number between 9 and 95 when divided by 3, 

leaves remainder 1 

$10,13,16,19, \ldots . .99$

Here, 

$\begin{aligned} q &=10, \\ d &=3, \\ \text { and } l &=94 . \end{aligned}$

$a_{n}(l)=a+(n-1) d $

$94=10+(n-1) \times 3$

$94-10=3(n-1) .$

$84=3(n-1)$

$\frac{84}{3}=n-1$

$28=n-1$

$28+1=n$

$29=$n

$n=29$

Then, middle term 

$\begin{aligned} &=\frac{29+1}{2} \text { th } \\ &=\frac{30}{2} \mathrm{th} . \end{aligned}$

=15th term

Sum of first 14 terms 

$=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{74}{2}[2 \times 10+(14-1) \times 3]$

$=7[20+13 \times 3]$

$=7[20+39]$

$=7 \times 59$

$=413$

Middle terms = 10+15\times 3

= 10+45 = 55

After middle term; number of terms = 14 

Whose first terms (a)= 55 , d = 3

and n = 14

$\therefore s_{14}=\frac{19}{2}[2 \times 55+(14-1) \times 3]$

$\begin{aligned} &=7[110+39] \\ &=7 \times 149 \\ &=1043 \end{aligned}$

Question 14

Ans: $S_{n}=$ Sum of first n terms of an A.P.

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$To $ prove $=S_{12}=3\left(S_{8}-S_{4}\right)$

from $R \cdot H \cdot S=3\left[S_{8}-S_{4}\right]$

$=3\left[\frac{8}{2}[2 a+(8-1) d]-\frac{4}{2}[2 a+(4-1) d]\right] .$

$=3\left[\frac{8}{2}(2 a+7 d)-\frac{4}{2}[2 a+3 d)\right]$

$=3[4(2 a+7 d)-2(24+3 d)]$

$=3[84+28 d-4 a-6 d]$

$=3[4 a+22 d]$

$=12 a+66 d .$

$=6[2 a+11 d) .$

$\frac{12}{2}[2 a+(12-1) d]$= $S_{12}$ =L.H.S

Hence proved 

Question 15

Ans: Sum of first n even natural number $=\left(1+\frac{1}{h}\right)$ 

(Sum of first n odd natural numbers)

Sum of first n even natural number (2,4,6,8......2n)

$=\frac{n}{2}[2 a+(n-1) d] .$

$=\frac{n}{2}[2 \times 2+(n-1) \times 2]$

$=\frac{n}{2}[4+2n-2] .$

$=\frac{n}{2}[2+2n] .$

$=\frac{n}{2} \times 2(1+n)$

$=n(1+n) $

and Sum of n odd natural number (1, 3,5 ,.......(2n -1))

$=\frac{n}{2}[2 \times 1+(n-1) 2]$

$=\frac{n}{2}[2+2 n-2]$

$=\frac{n}{2}[2 n]$

$=\frac{n}{2} \times 2[n]$

$=n^{2} .$

Then , 

$n^{2} \times \left(1+\frac{1}{n}\right)$

$=n^{2}\left(\frac{n+1}{n}\right)$

$=n^{2} \times \frac{n+1}{n}$

$=n(n+1)$

Question 16

Ans: Given,

First term $\left(a_{1}\right)=5$

Common difference $\left(d_{1}\right)=36$.

$S_{n}=$ Sum of $2_{n}$ terms of another $A P$ whose first term $\left(a_{2}\right)=36$ and common difference $\left(d_{2}\right)=5$.

 In first $A P$

$\begin{aligned} S_{n} &=\frac{n}{2}[2 a+(n-1) d] . \\ &=\frac{n}{2}[2 \times 5+(n-1) \times 36] \\ &=\frac{n}{2}[10+36 n-36] \\ &=\frac{n}{2}[36 n-26] . \end{aligned}$

$=\frac{n}{2} \times 2[18 n-13]$

$=n[18 n-13]$

In Second AP, 

$S_{2 N}=\frac{2 n}{2}[2 \times 36+(2 n-1) \times 5]$

$=n[72+10{n}-5]$

$=\mathrm{n}[67+10 \mathrm{n}] $

$\therefore S_{n}=S_{2 n}$

18 (18n- 13) = n(67 + 10n)

18 n - 13 =  $\frac{n}{n}(67+10 n)$

$18 n-10 n=67+13$

8n = 80

n = $\frac{80}{8}$

n=10

Question 17

Ans: Sum of first 10 terms of an AP = $4 \times sum$ of first 5 terms 

Let a be the first term and d be the common difference 

$\begin{aligned} \therefore S_{n} &=\frac{n}{2}[2 a+(n-1) d] \\ S_{10} &=\frac{10}{2}[2 a+(10-1) d] \\ &=5[2 a+9 d] \\ &=10a+45d \end{aligned}$

$\begin{aligned} S_{{5}} &=\frac{5}{2}[2 a+(5-1) d] \\ &=\frac{5}{2}[2 a+4 d] \end{aligned}$

$=\frac{5}{2} \times 2[a+2 d]$

$=5 a+10 d$

According to the question,

10a + 45 d =  $4 \times(5 a+10 d)$

$10 a+45 d=20 a+40 d .$

$45 d-40 b=20 a-10 a$

$5 a=109 .$

$\frac{5}{10}=\frac{a}{d}$

$\frac{a}{d}=\frac{5}{10}$

$\frac{a}{d}=\frac{1}{2}$

∴ Ratio in first term to the common difference = 1:2

Question 18

Ans:  Given, 

In the first row, plants are = 37

In second row = 35 

In third row = 33

and in the last row = 5

Here , 

a= 37 

d= 35 - 37 

d= - 2

$\begin{aligned} a_{n}(1)=& a+(n-1) d . \\ 5=& 37+(n-1) \times(-2) \\ 5 &-37=-2 n+2  \\ &-32=-2(n-1) . \\ & \frac{+32}{+2}=n-1 \end{aligned}$

$16=n-1$

$16+1=n$

$17=n$

$n=17$

So, number of rows = 17 

Then, 

Total plants $\left(S_{n}\right)=\frac{n}{2}[24+(n-1) d]$

$=\frac{17}{2}[2 \times 37+(17-1) \times(-2)] .$

$=\frac{17}{2}[74-32]$

=357

Hence , the total plants is 357

Question 19

Ans: Given , 

Total logs = 200 

In the bottom row, number of logs = 20 

and next row above it = 19 

Next row above it = 18

And Soon 

- AP is 20,19,18,17,16....... and $S_{n}=200$

Let the top row be the nth row 

$\therefore a_{n}=a+(n-1)d$

and $S_{n}=200$

So, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$200=\frac{n}{2}[2 \times 20+(n-1) \times(-1)]$

$400=n[40-n+1]$

$400=40 n-n^{2}+h$

$400=41 n-n^{2} $

$n^{2}-41 n+400=0$

$n^{2}-25 n-16 n+400=0$

$n(n-25)-16(n-25)=0$

$(n-25)(n-16)=0$

n -25 =0 or n-16=0

n=25 or n =16

But n = 25 is not possible 

 As number of logs in the first row = 20 

and d = - 1 

Number of row = 16 

and number of log in 16th row 

$=9+(n-1) d$

$=20+(16-1) \times(-1)$

$=20-15$

$=5 logs$

Question 20

Ans: Giver,

Total amount= ₹1590,

Number of cash prizes = 7 

and each prize is Rs 50 less than the proceeding prize 

Let first prize = Rs a 

Then second prize = a - 50 

Third prize = a - 100

And so on 

$\therefore$ first term $=a$,

common difference (d) $=-50$,

$S_{n}=1890$

$n=7 $

$\begin{aligned} S_{n}=& \frac{n}{2}[2 a+(n-1) d] \\ 1890=& \frac{7}{2}[2 a+(7-1) \times(-50)] \\ 3780=& 7[2 a-300] \\ & \frac{3780}{7}=2 a-300 \\ 540=& 2 a-300 \\ 540+300=& 2 a \\ 840=& 2 a \\ \frac{840}{2}=9 \\ 420=a \\ a=420 \end{aligned}$

Hence prizes are $ ₹420, ₹ 370, ₹ 320 ; 270, ₹ 220, ₹ 170$ and $₹ 120$.

RS AGGARWAL CLASS 9 CHAPTER 9 CONGRUENCE OF TRIANGLES AND INEQUALITIES IN A TRIANGLE EXERCISE 9B

 EXERCISE 9B

PAGE NO-296

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Q1 | Ex-9B |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 1:

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.
(i) 5 cm, 4 cm, 9 cm
(ii) 8 cm, 7 cm, 4 cm
(iii) 10 cm, 5 cm, 6 cm
(iv) 2.5 cm, 5 cm, 7 cm
(v) 3 cm, 4 cm, 8 cm

Answer 1:

(i) No, because the sum of two sides of a triangle is not greater than the third side.
5 + 4 = 9

(ii) Yes, because the sum of two sides of a triangle is greater than the third side.
7 + 4 > 8; 8 + 7 > 4; 8 + 4 > 7

(iii) Yes, because the sum of two sides of a triangle is greater than the third side.
5 + 6 > 10; 10 + 6 > 5; 5 + 10 > 6

(iv) Yes, because the sum of two sides of a triangle is greater than the third side.
2.5 + 5 > 7; 5 + 7 > 2.5; 2.5 + 7 > 5

(v) No, because the sum of two sides of a triangle is not greater than the third side.
3 + 4 < 8

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Q2 | Ex-9B |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 2:

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

Answer 2:

Given: In ΔABC, ∠A = 50° and ∠B = 60°

In ΔABC,
∠A + ∠B + ∠C = 180°           (Angle sum property of a triangle)
$\Rightarrow$50° + 60° + ∠C = 180°
$\Rightarrow$110° + ∠C = 180°
$\Rightarrow$∠C = 180° $-$ 110°
$\Rightarrow$∠C = 70°

Hence, the longest side will be opposite to the largest angle (∠C = 70°)  i.e. AB.
And, the shortest side will be opposite to the smallest angle (∠A = 50° ) i.e. BC.

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Q3 | Ex-9B |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 3:

(i) In ∆ABC, ∠A = 90°. Which is its longest side?
(ii) In ∆ABC, ∠A = ∠B = 45°. Which is its longest side?
(iii) In ∆ABC, ∠A = 100° and ∠C = 50°. Which is its shortest side?

Answer 3:

(i) Given: In ∆ABC, ∠A = 90°

So, sum of the other two angles in triangle ∠B + ∠C = 90°

i.e. ∠B, ∠C < 90°

Since, ∠A is the greatest angle.

So, the longest side is BC.

(ii) Given: ∠A = ∠B = 45°

Using angle sum property of triangle,

∠C = 90°

Since, ∠C is the greatest angle.

So, the longest side is AB.

(iii) Given: ∠A = 100° and ∠C = 50°

Using angle sum property of triangle,

∠B = 30°

Since, ∠A is the greatest angle.

So, the shortest side is BC.

PAGE NO-297

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Q4 | Ex-9B |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 4:

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

Answer 4:

In triangle CBA, CBD is an exterior angle.

i.e., ∠CBA+∠CBD=180°

⇒60°+∠CBD=180°

⇒∠CBD=120°


Triangle BCD is isosceles and BC = BD.
Let ∠BCD=∠BDC = x°.
In △CBD, we have:

⇒∠BCD+∠CBD+∠CDB=180°

⇒x+120°+x=180

⇒2x=60°

⇒x=30°

∴∠BCD=∠BDC=30°

In triangle ADC, ∠C=∠ACB + ∠BCD = 50°+30°=80°
∠A=70°and ∠D=30°

∴∠C>∠A

⇒AD>CD ...(1)

Also, ∠C>∠D

⇒AD>AC ...(2)

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Q5 | Ex-9B |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 5:

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

Answer 5:

Given: ∠B < ∠A and ∠C < ∠D

To prove: AD > BC

Proof:

In ΔAOB,∠B<∠A

⇒AO<BO (Side opposite to the greater angle is longer) .....(1)

In ΔCOD,∠C<∠D

⇒OD<OC (Side opposite to the greater angle is longer) .....(2)

Adding (1) and (2), we get

AO+OD<BO+OC

∴AD<BC

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Q6 | Ex-9B |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 6:

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

Answer 6:

Given: In quadrilateral ABCD, AB and CD are respectively the smallest and largest sides.

To prove:
​(i) ∠A > ∠C
(ii) ∠B > ∠D


Construction: Join AC.

Proof:

In ΔABC,∵ BC>AB (Given, AB is the smallest side)

∴ ∠1>∠2 ...(1)

In ΔADC,∵ CD>AD (Given, CD is the largest side)

∴ ∠3>∠4 ...(2)

Adding (1) and (2), we get

∠1+∠3>∠2+∠4

∴ ∠A>∠C


(ii)

Construction: Join BD.

Proof:

In ΔABD,∵ AD>AB (Given, AB is the smallest side.)

∴ ∠5>∠6 ...(3)

In ΔCBD,∵ CD>BC (Given, CD is the greatest side.)

∴ ∠7>∠8 ...(4)

Adding (3) and (4), we get
∠5+∠7>∠6+∠8
∴∠B>∠D

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Q7 | Ex-9B |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 7:

In a quadrilateral ABCD, show that (AB + BC + CD + DA) > (AC + BD).

Answer 7:

Given: Quadrilateral ABCD

To prove: (AB + BC + CD + DA) > (AC + BD)

Proof:

$$\begin{gathered} \mathrm{In} ∆ABC, \\ AB+BC>AC ...\left(\mathrm{i}\right) \\ \mathrm{In} ∆CAD, \\ CD+AD>AC ...\left(\mathrm{ii}\right) \\ \mathrm{In} ∆BAD, \\ AB+AD>BD ...\left(\mathrm{iii}\right) \\ \mathrm{In} ∆BCD, \\ BC+CD>BD ...\left(\mathrm{iv}\right) \end{gathered}$$

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

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

Hence, (AB + BC + CD + DA) < (AC + BD).

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Q8 | Ex-9B |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 8:

In a quadrilateral ABCD, show that
$\left(AB+BC+CD+DA\right)<2\left(BD+AC\right)$

Answer 8:

Given: Quadrilateral ABCD
To prove: (AB + BC + CD + DA) < 2(BD + AC).

Proof:

In ∆AOB, 
$OA+OB>AB .....\left(i\right)$
In ∆BOC, 
$OB+OC>BC .....\left(ii\right)$
In ∆COD, 
$OC+OD>CD .....\left(iii\right)$
In ∆AOD, 
$OD+OA>AD .....\left(iv\right)$
Adding $\left(i\right),\left(ii\right),\left(iii\right) \mathrm{and} \left(iv\right)$, we get
$$\begin{gathered} 2\left(OA+OB+OC+OD\right)>\left(AB+BC+CD+DA\right) \\ \Rightarrow 2\left(OB+OD+OA+OC\right)>\left(AB+BC+CD+DA\right) \\ \Rightarrow 2\left(BD+AC\right)>\left(AB+BC+CD+DA\right) \end{gathered}$$

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Q9 | Ex-9B |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 9:

In ΔABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in X. Arrange AX, BX and CX in descending order.

Answer 9:

Given: In ΔABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in X.

$$\begin{gathered} \mathrm{In} ∆ABX, \\ \because \angle BAX>\angle ABX \\ \therefore BX>AX ...\left(\mathrm{i}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Similarly}, \mathrm{in} ∆ACX, \\ \because \angle ACX>\angle XAC \\ \therefore AX>CX ...\left(\mathrm{ii}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{From} \left(\mathrm{i}\right) \mathrm{and} \left(ii\right), \mathrm{we} \mathrm{get} \\ BX>AX>CX \end{gathered}$$

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Q10 | Ex-9B |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 10:

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

Answer 10:

Since the angle opposite to the longer side is greater, we have:

$$\begin{gathered} PQ>PR \\ \Rightarrow \angle R>\angle Q \\ \Rightarrow \frac{1}{2}\angle R>\frac{1}{2}\angle Q \\ \Rightarrow \angle SRQ>\angle RQS \\ \Rightarrow QS>SR \end{gathered}$$

∴$SQ>SR$

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Q11 | Ex-9B |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 11:

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

Answer 11:

Given: In $∆$ABC, AB = AC

To prove: CD < BD

Proof:

In $∆$ABC,

Since, AB = AC       (Given)

So, $\angle ABC=\angle ACB$      ...(i)

$$\begin{gathered} \mathrm{In} ∆ABC \mathrm{and} ∆DBC, \\ \angle ABC>\angle DBC \\ \Rightarrow \angle ACB>\angle DBC \left[\mathrm{From} \left(\mathrm{i}\right)\right] \\ \Rightarrow BD>CD \left(\mathrm{Side} \mathrm{opposite} \mathrm{to} \mathrm{greater} \mathrm{angle} \mathrm{is} \mathrm{longer}.\right) \\ \therefore CD<BD \end{gathered}$$

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Q12 | Ex-9B |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 12:

Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than $\frac{2}{3}$ of a right angle.

Answer 12:

Given: In $∆$ABC, BC is the longest side.

To prove: $\angle$BAC > $\frac{2}{3}$ of a right angle, i.e., $\angle$BAC > 60$°$

Construct: Mark a point D on side AC such that AD = AB = BD.

Proof:

In $∆$ABD,

$\because$ AD = AB = BD    (By construction)

$\therefore \angle 1=\angle 3=\angle 4=60°$

$$\begin{gathered} \mathrm{Now}, \\ \angle BAC=\angle 1+\angle 2=60°+\angle 2 \\ \mathrm{but} 60° =\frac{2}{3} \mathrm{of} \mathrm{a} \mathrm{right} \mathrm{angle} \\ \mathrm{So}, \angle BAC=\frac{2}{3} \mathrm{o}\mathrm{f} \mathrm{a} \mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t} \mathrm{a}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e} + \angle 2 \end{gathered}$$

Hence, $\angle$BAC > $\frac{2}{3}$ of a right angle.

PAGE NO-298

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Q13 | Ex-9B |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 13:

In the given figure, prove that
(i) CD + DA + AB > BC
(ii) CD + DA + AB + BC > 2AC.

Answer 13:

Given: Quadrilateral ABCD

To prove:
(i) CD + DA + AB > BC
(ii) CD + DA + AB + BC > 2AC

Proof:
(i)
In ΔACD,CD+DA>CA ...(1)

In ΔABC,AB+CA>BC ...(2)

Adding (1) and (2), we get

CD+DA+AB+CA>CA+BC

∴ AB+CD+DA>BC

(ii)
In ΔCDA,CD+DA>CA ...(3)

In ΔBCA,BC + AB > CA ...(4)

Adding (3) and (4), we get

CD+AD+BC+AB>CA+CA

∴ CD+AD+BC+AB>2CA

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Q14 | Ex-9B |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 14:

If O is a point within ∆ABC, show that:
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + OB + OC
(iii) OA+OB+OC>$\frac{1}{2}$(AB+BC+CA)

Answer 14:

Given:

In triangle ABC, O is any interior point.
We know that any segment from a point O inside a triangle to any vertex of the triangle cannot be longer than the two sides adjacent to the vertex.
Thus, OA cannot be longer than both AB and CA (if this is possible, then O is outside the triangle).

(i) OA cannot be longer than both AB and CA.​
AB>OB ...(1)

AC>OC ...(2)

Thus, AB+AC>OB+OC ...[Adding (1) and(2)]

(ii) AB>OA......(3)

BC>OB.....(4)

CA>OC.....(5)

Adding the above three equations, we get:
Thus, AB+BC+CA>OA+OB+OC ...(6)

OA cannot be longer than both AB and CA.​
AB>OB.....(5)

AC>OC.....(6)

AB+AC>OB+OC..........[On adding (5) and (6)]

Thus, the first equation to be proved is shown correct.

(iii) Now, consider the triangles OAC, OBA and OBC.
We have:
OA+OC>AC

OA+OB>AB

OB+OC>BC

Adding the above three 

equations, we get:

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

⇒2OA+OB+OC>AB+AC+BC

Thus, 

OA+OB+OC>$\frac{1}{2}$AB+BC+CA

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Q15 | Ex-9B |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 15:

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

Answer 15:

Given: AD ⊥ BC and CD > BD

To prove: AC > AB

Proof:

$$\begin{gathered} \angle ADB=\angle ADC=90° \left(AD\perp BC\right) ...\left(1\right) \\ \angle BAD<\angle DAC \left(CD>BD\right) ...\left(2\right) \end{gathered}$$

In $∆ABD$,

Using angle sum property of a triangle,

$$\begin{gathered} \angle B=180°-\angle ADB-\angle BAD \\ \angle B=90°-\angle BAD ...\left(3\right) \end{gathered}$$

In $∆ADC$,

Using angle sum property of a triangle,

$\angle ACD=90°-\angle DAC ...\left(4\right)$

From (2), (3) and (4), we get

$\angle B>\angle C$

Therefore, $AC>AB$.

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Q16 | Ex-9B |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 16:

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.

Answer 16:

Given: CD = DE

To prove: AB + AC > BE

Proof:

In $∆ABC$,

$AB+AC>BC ...\left(1\right)$

In $∆BED$,

$$\begin{gathered} BD+CD>BE \\ \Rightarrow BC>BE ...\left(2\right) \end{gathered}$$

From (1) and (2), we get

AB + AC > BE