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Question-1

In ΔPQR, PQ = QR = RP = 7 cm, then find each angle of ΔPQR.
 
  

 

Solution:
In ΔPQR, PQ = PR ...... (Given)
           ... ∠Q = R ...... (angles opposite to equal sides) ... (i)

Again, QR = RP


... ∠Q = P ...... (ii)

From eqns (i) and (ii), we get

... ∠P = Q = R

But in
ΔPQR, we have
        P + Q + R = 180°

... 3 P = 180° P = 60°

Hence
P = Q = R = 60°

Question-2

In the figure AB = AC and ACD = 110° find A.


Solution:
            ACB + ACD = 180° ...... (Linear pair)
          ... ÐACB + 110° = 180°
⇒ ∠ACB = 180° - 110° = 70°

       Since in
ΔABC, AB = AC
                    ÐB = C = 70°

Now, in
ΔABC, we have
           A + B + C = 180° ...... (Sum angles)
      ... ÐA + 70° + 70° = 180°
     ...ÐA = 180° - 140° = 40°

Question-3

In the figure, AP and BQ are perpendiculars to PQ and AP = BQ, prove that R is the mid-point of PQ and AB.


Solution:
In ΔAPR and ΔBQR,

    AP = BQ ...... (Given)

ARP = BRQ ...... (Vertically opposite angles)
APR = BQR ...... (each 90°)
ΔAPR ΔBQR ...... (RHS Criterion)

 PR = RQ ..... (c.p.c.t.)

 
and AR = RB ...... (c.p.c.t.) 

Hence R is the mid-point of AB and PQ.

Question-4

In the figure, l ïï m and M is the mid-point of AB. Prove that M is also the mid-point of CD.  
 

Solution:
Since l ïï m, and AB is the transversal
 ÐCAM = DBM ...... (Alternate interior angles)

Now in
ΔACM and ΔBDM, we have
  CAM = DBM ..... (proved above)

     AM = BM ...... (M is the mid-point)


  ÐAMC = BMD ............(Vertically opposite angles)
  ΔACM ΔBDM

   
 DM = MC .............(CPCT)

Hence, M is the mid-point of DC.

Question-5

ΔABC is an isosceles triangle with AB = AC. Side BA is produced to D such that AB = AD. Prove that BCD = 90°.


Solution:
In ΔABC, AB = AC

            ÐB = C= ∠4                              ......(i) 
   Since, AB = AC ...... (Given)

      and AB = AD ...... (Produced)


  ∴ AD = AC
Now, in ΔACD, AD = AC
∴ D = C = 3                                        ...... (ii) 
Adding eqn (i) and eqn (ii), we get
    B + D = 4 + 3

⇒ ∠B + D = BCD

Now in
ΔBCD, we have
      B + BCD + D = 180°

(B + D) + BCD = 180°

      
⇒ ∠BCD + BCD = 180°

          
2 BCD = 180°
  ...  ÐBCD =   = 90°
                                    

Question-6

In ΔABC, if AB = AC and BE, CF are the bisectors of B and C respectively. Prove that ΔEBC ΔFCB and BE = CF.

Solution:


Since in ΔABC, AB = AC            ABC = ACB ......(i)
Since CF and BE are angle bisectors of
C and B,
      we getABE = EBC ...... (ii)
          and ACF = FCB ...... (iii) 
Now from eqns (i), (ii) and (iii), we get

  1
ABC =  1 ACB
  2               2
    
 ÐEBC = FCB ...... (iv)

Now in
ΔFBC and ΔECB,
 
we have
FBC = ECB   ...... (B = C)

               BC = BC ...... (Common)

            ÐFCB = EBC ...... [From (iv)]
            ΔEBC ≅ ΔFCB
               BE = CF ...... (c.p.c.t.)

Question-7

In the figure AC = BC, DCA = ECB and DBC = EAC. Prove that triangles DBC and EAC are congruent and hence DC = EC.

Solution:
In ΔACE and ΔBCD, we have

               AC = BC ...... (given)

            ÐEAC = DBC ...... (given)
              DCA = ECB   ...... (given)
DCA + DCE = ECB + DCE  ........(Adding DCE  on on both sides)
         ⇒ ∠ACE = BCD
             ΔACE ΔBCD ...... (ASA criterion)

                DC = EC ...... (c.p.c.t.)
 

Question-8

Prove that the medians of an equilateral triangle are equal.

Solution:
Let ABC be an equilateral triangle and let AD, BE and CF be the medians.

Since
ΔABC is an equilateral triangle A = B = C = 60°

Now in
ΔADC and ΔABE,

 we have  
BC = AC
DC = AE

C = A ...... (each 60°)

AC = AB


ΔADC ΔABE

AD = BE ...... (c.p.c.t.)

Similarly, we can prove that BE = CF

AD = BE = CF

Hence medians of equilateral triangle are equal.

Question-9

In the figure, PQRS is a square and SRT is an equilateral triangle. Prove that
(i) PT = QT
(ii)
TQR = 15°


Solution:
Since PQRS is a square

PSR = QRS ...... (each 90°)

Now again in equilateral
ΔSRT, 

we have
TSR = TRS ...... (each 60°)

PSR + TSR = QRS + TRS

       
⇒ ∠TSP = TRQ

Now in
ΔTSP and ΔTRQ, we have

              TS = TR ...... (sides of equilateral triangle)

         
TSP = TRQ ...... (Proved above)

              PS = QR ...... (sides of a square)

          
ΔTSP ΔTRQ

          
∴ PT = QT

Now in ΔTQR, we have

TR = QR ...... (QR = RS = TR)


TQR = QTR and TQR + QTR + TRQ = 180°

             
TQR + QTR + TRS + SRQ = 180°
                            2 TQR + 60° + 90° = 180°...........(TQR = QTR)

                          2 TQR = 180° - 150° = 30°

                                       
TQR =  30 = 15°
                                                     2

Question-10

In the figure x is a point in the interior of square ABCD, AXYZ is also a square. Prove that BX = DZ.

Solution:
Since ABCD and AXYZ both are squares

AZY = AXY

        =
AXB ... (each 90°) and AX = AZ and AB = AD
 
Now in
ΔABX and ΔADZ,

AZD = AXB ...... (each 90°)

   AZ = AX ...... (sides of a square)

   AB = AD ...... (sides of a square)


ΔABX ΔADZ

   BX = DZ ....... (c.p.c.t.)

Question-11

In the figure, the sides AB and BC of square ABCD are produced to P and Q respectively so that BP = CQ. Prove that DP and AQ are perpendicular to each other.


Solution:
Since ABCD is a square, AB = BC

     Also BP = CQ ...... (given)

... AB + BP = BC + CQ

        ... AP = BQ

Now in
ΔAPD and ΔBQA, 

           AP = BQ ...... (proved above)

      
ABQ = DAP ... (each 90°)

     and AB = AD

    ... ΔAPD ΔBQA

   ...
APD = BQA ...... (c.p.c.t.)

 and
ADP = QAP ...... (c.p.c.t.)

Also
DAQ = AQB

   ...
DAO = APO

Now in
ΔAOD and ΔAOP, 

     
ADO = OAP,

and
DAO = APO

        ... 3rd
DOA = 3rd AOP [Since, two angles of DAOD and DAOP are equal, the third angle is also equal]
but DOA + AOP = 180°

           ... 2
DOA = 180°

              ...
DOA = 90°

... DO is perpendicular to AO or DP is perpendicular to AQ.

Question-12

In the figure, B = C and AB = AC. Prove that BD = CE.


Solution:
In ΔABD and ΔACE,

     AB = AC ...... (given)

     B = C ...... (given)
     A = A ...... (common)
... ΔABD ΔACE ...... (By ASA criterion)

   ... BD = CE ...... (c.p.c.t.)

Question-13

In the figure, AD = BE, BC = DF and ABC = EDF. Prove that AC ïï EF and AC = EF.

Solution:
    Since AD = BE

... AD + DB = BE + DB

         AB = DE
 
Now in
ΔABC and ΔEDF

             AB = DE ...... (proved above)

             BC = DF ...... (given)
 
   and
ABC = EDF ...... (given)
      ... ΔABC ΔEDF

         ... AC = EF ...... (c.p.c.t.)

   and BAC = DEF

but these are alternate interior angles of AC and EF with transversal AE

... AC
ïï EF
 

Question-14

Given ABCD is a parallelogram, BC is produced to F and BD is produced to E and AE=CF Prove that:

(i) BE || DF
(ii) BD and EF bisect each other.

 




Solution:

To Prove: BE ïï DF
and EO = OF
and DO = OB

Proof: In ΔABE and ΔCDF

     AB = DC ...... (opposite sides of parallelogram)

      AE = CF ...... (given)

and BE = DF ...... (given)


... ΔABE ΔCDF

... ABE = CDF ...... (i) (c.p.c.t.)

Now since AB
ïï DC and DB is a transversal

... ABD = BDC ...... (ii) (Alternate interior angles)

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


ABE + ABD = CDF + BDC

       
⇒ ∠EBD = BDF BE ïï DF

Now in
ΔOBE and ΔODF, 

               BE = DF ...... (given)

         
EBO = FDO ...... (proved above)

     and
BOE = DOF ...... (Vertically opposite angles)

         ...
ΔOBE ΔODF

           ... OE = OF

         and OB = OD ...... (c.p.c.t.)

Therefore O is the mid-point of EF and DB.

Question-15

O is a point in the interior of a rhombus ABCD. If OA = OC then prove that DOB is a straight line.


Solution:
Given: A rhombus ABCD and a point O in it such that OA = OC

To Prove: DOB is a straight line.

Construction: Join OB and OD

Proof: In
ΔAOD and ΔCOD

      AO = CO ...... (given)

      OD = OD ...... (common)

      AD = CD ...... (sides of a rhombus)


... ΔAOD ≅ ΔCOD

...
AOD = COD ...... (i)

Similarly
ΔAOB ΔCOB

... AOB = COB ...... (ii)

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

                             
AOD + AOB = COD + COB

But
AOD + AOB + COD + COB = 360° ...... (Angles at a point)

  
... AOD + AOB + AOD + AOB = 360°
                     2 (AOD + AOB) = 360°

                          
⇒ ∠AOD + AOB = 180° but it is a linear pair

... OD and OB are in a line

Hence DOB is a straight line.

Question-16

In the figure PQ > PR, QM and RM are the bisectors of Q and R respectively. Prove that QM > RM.


 

Solution:
Since QM and RM are angle bisectors of Q and R.

 1 = 2 and 3 = 4

Now in
ΔPQR, PQ > PR
 
            ... PRQ > PQR ...... (Angles opposite to larger side)

...
PRM + MRQ > PQM + MQR

         ...
4 + 3 > 1 + 2

             
 Þ 2 3 > 2 2

                
⇒ ∠3 > 2

                ... QM > RM.

Question-17

In the figure, PQ = PR, show that PS > PQ.

 

Solution:
In ΔPQR, PQ = PR

     
PQR = PRQ ..(i)

Now in
ΔPSQ, PQR is the exterior angle
      PQR = PSQ + SPQ

     
PQR > PSQ

     
PRQ > PSQ ...... (PQR = PRQ)

         
PS > PR ...... (side opposite to greater angle)

         ... PS > PQ ...... (PR = PQ)
 

Question-18

In ΔABC, AC > AB and AD is the bisector of A show that y > x.

Solution:
In ΔABC, since AC > AB

 B > C ...... (i)

Also
1 = 2 ...... [given]

Now, in
ΔABD, we have

1 + B + x = 180° ...... (ii)

And in
ΔADC, we have

2 + C + y = 180° ...... (iii)
 
Comparing eqns (i) and (ii), we get


1 + B + x = 2 + C + y

      
B + x = C + y

Now since
B > C ...... [proved above]

           ...
x < y

            or
y > x

Question-19

In ΔPQR, PS QR and SR > SQ, show that PR > PQ.

 

Solution:
PR > PQ

Construction: Draw PT such that SQ = ST

Proof: In
ΔPQS and ΔPTS,

       PS = PS         ...... [common]

      SQ = ST          ......[By construction]

  
PSQ = PST       ...... [each 90°]

...
ΔPQS ΔPTS   ...... [SAS criterion]

...
1 = 2                 ...... [c.p.c.t.]

Now in
ΔPRT, we have the exterior angle 2

...
2 = 3 + TPR

or
2 > 3

...
1 > 3 ...... (1 = 2)

Now in
ΔPQR, since

   
1 > 3

... PR > PQ ...... [side opposite to greater angle]

Question-20

Prove that the difference of any two sides of a triangle is less than the third side.

Solution:
To Prove: AC - AB < BC,

               BC - AB < AC

and          AC - BC < AB

Construction: Mark a point D on AC such that

AB = AD and join BD.

Proof
In
ΔABD, AB = AD

         
1 = 2 ...... (i)

Now in
ΔABD, AD is produced to C

        
3 = 1 + A ...... (Exterior angle)

        
3 > 1 ...... (ii)

Now in
ΔBCD, CD is produced to A

       
2 = 4 + C

       
2 > 4 ...... (iii)

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

          
3 > 4

        ... BC > CD

       or BC > AC - AD

        or BC > AC - AB

Similarly we can prove other two inequalities

     BC - AB < AC and  AC - BC < AB.

Question-21

Show that the sum of the three altitudes of a triangle is less than the sum of the three sides of the triangle.

Solution:
ΔABC, in which altitudes AQ, BR and CP meet at M.

To Prove: 
AB + BC + CA > BR + CP + AQ

I Proof:
In
ΔABQ, we have AQB = 90°
... ABQ will be a acute angle

...
AQB > ABQ

    ...AB > AQ ...... (i)

Similarly, we can prove that:

      BC > BR ...... (ii)

and CA > CP ...... (iii)

Adding three inequalities, we get

AB + BC + CA > AQ + BR + CP

II Proof:
AB > AQ, AC > AQ ...... [ ... Perpendicular is shortest]

  ... AB + AC > 2AQ ...... (i)

Similarly BC + CA > 2PC ...... (ii)

       and AB + BC > 2BR

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

2AB + 2BC + 2CA > 2AQ + 2BR + 2CP

  or AB + BC + CA > AQ + BR + CP

Question-22

In a quadrilateral PQRS, the diagonals PR and QS intersect each other at O. Show that
(i) PQ + QR + RS + SP > PR + QS.
(ii) PQ + QR + RS + SP < 2 (PR + QS). 



Solution:
We know that in a triangle the sum of two sides of a triangle is greater than the third side. Therefore in ΔPSR, we have   
     
PS + SR > PR   ...... (i)

In
ΔQSR, QR + RS > QS ...... (ii)

In
ΔPQS, PQ + SP > QS   ...... (iii)

In
ΔPQR,PQ + QR > PR    ..... (iv)

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

2 (SP + PQ + QR + RS) > 2 (QS + PR)  PQ + QR + RS + SP > PR + QS

Similarly in
ΔPQO, ΔQRO, ΔRSO and ΔSPO, 

we have
                  QO + PO > PQ ...... (v)

                  QO + RO > QR ...... (vi)

                  RO + SO > RS ...... (vii)

                  SO + PO > SP ...... (viii)

Adding (v), (vi), (vii) and (viii), 

we get
2 (QO + SO + PO + OR) > PQ + QR + RS +SP 2 (QS + PR) > PQ + QR + RS + SP

   or PQ + QR + RS + SP < 2 (QS + PR).

 




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