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1.(v) Find the remainder whenx^ 3 + 3x^ 2 + 3x + 1 is divided by
   (v)   5+2x

When we divide    by  .  By long division method, we will get  Therefore, the remainder is   

6. In Fig. 6.44, the side QR of \Delta PQR is produced to a point S. If the bisectors of \angle PQR and \angle PRS meet at point T, then prove that  \angle QTR = \frac{1}{2}   \angleQPR.


We have,  PQR is produced to a point S and  bisectors of PQR and PRS meet at point T, By exterior angle sum property, PRS = P + PQR Now,  ................(i) Since QT and QR are the bisectors of  PQR and PRS respectively. Now, in QRT, ..............(ii) From eq (i) and eq (ii),  we get Hence proved

5. In Fig. 6.43, if PQ \bot PS, PQ ||SR, \angle SQR = 28° and \angle QRT = 65°, then find the values of x and y.


We have,  PQ PS, PQ || SR, SQR = 28° and QRT = 65° Now, In  QRS, the side SR produced to T and PQ || RS therefore, QRT =  =   So,  Also, QRT = RSQ + SQR (By exterior angle property of a triangle) Therefore, RSQ = QRT - SQR                     Now, in  PQS, P + PQS + PSQ =   

4. In Fig. 6.42, if lines PQ and RS intersect at point T, such that \angle PRT = 40°, \angle RPT = 95° and \angle TSQ = 75°, find \angle SQT.


We have, lines PQ and RS intersect at point T, such that PRT = 40°, RPT = 95° and TSQ = 75° In PRT, by using angle sum property PRT + PTR + TPR =  So, PTR  =     Since lines, PQ and RS intersect at point T therefore, PTR = QTS (Vertically opposite angles)                 QTS =  Now, in QTS, By using angle sum property TSQ + STQ + SQT =  So, SQT = 

3.   In Fig. 6.41, if AB || DE, \angle BAC = 35° and \angle CDE = 53°, find \angle DCE.


We have,  AB || DE,  BAC = 35° and  CDE = 53° AE is a transversal so,  BAC =  AED =  Now, In  CDE, CDE + DEC + ECD =  (By angle sum property) Therefore, ECD =                   

2. In Fig. 6.40, \angle X = 62°, \angle XYZ = 54°. If YO and ZO are the bisectors of \angle XYZ and \angle XZY respectively of \Delta XYZ, find \angle OZY and \angle YOZ.


We have  X = , XYZ =  YO and ZO bisects the XYZ and XZY Now, In XYZ, by using angle sum property XYZ + YZX + ZXY =  So, YZX =        YZX =  and, OYZ =  also, OZY =  Now, in OYZ  Y + O + Z =   [Y =  and Z = ] So, YOZ = 

1. In Fig. 6.39, sides QP and RQ of \DeltaPQR are produced to points S and T respectively. If \angle SPR = 135° and \angle PQT = 110°, find \angle PRQ.


Given, PQR is a triangle, SPR =,  PQT =  Now, TQP + PQR =  (Linear pair) So, PQR =  Since the side of QP of the triangle, PQR is produced to S So, PQR + PRQ =  (Exterior angle property of triangle)

6.  In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.



Draw  a ray BL PQ and CM  RS Since PQ || RS (Given) So, BL || CM and BC is a transversal  LBC =  MCB (Alternate interior angles).............(i) It is known that, angle of incidence  = angle of reflection So, ABL = LBC and MCB =  MCD ..................(ii) Adding eq (i) and eq (ii), we get ABC = DCB Both the interior angles are equal Hence AB || CD

5.  In Fig. 6.32, if AB || CD, \angle APQ = 50° and \angle PRD = 127°, find x and y.


Given, AB || CD, APQ =  and PRD =  PQ is a transversal. So,  APQ = PQR=  (alternate interior angles) Again, PR is a transversal. So,  y + =  (Alternate interior angles)

4.  In Fig. 6.31, if PQ || ST, \angle PQR = 110° and \angle RST = 130°, find \angle QRS.
            [Hint : Draw a line parallel to ST through point R.]


Draw a line EF parallel to the ST through R. Since PQ || ST and ST || EF   EF || PQ PQR = QRF =   (Alternate interior angles) QRF = QRS + SRF .............(i) Again, RST + SRF =  (Interior angles of two parallels ST and RF)   (RST = , given) Thus, QRS = 

3.  In Fig. 6.30, if AB || CD, EF \bot CD and \angle GED = 126°, find \angle AGE, \angle GEF and \angle FGE.


Given AB || CD, EFCD and GED =  In the above figure,  GE is transversal. So, that AGE = GED =   [Alternate interior angles] Also, GEF = GED - FED                       =            Since AB is a straight line  Therefore, AGE  + FGE =   So, FGE = 

2. In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.


Given AB || CD and CD || EF and  therefore, AB || EF and  (alternate interior angles)..............(i) Again, CD || AB  .............(ii) Put the value of  in equation (ii), we get Then  By equation (i), we get the value of   

1. In Fig. 6.28, find the values of x and y and then show that AB || CD.


Given that, In the figure, CD and PQ intersect at  F Therefore,   (vertically opposite angles) PQ is a straight line. So,  Hence AB || CD (since  and are  alternate interior angles)

6. It is given that \angle XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects \angle ZYP, find \angle XYQ and reflex \angle QYP.

Given that, XYZ =  and XY produced to point P and Ray YQ bisects ZYP   Now, XYP is a straight line So, XYZ + ZYQ + QYP =  Thus reflex of QYP =  Since XYQ = XYZ + ZYQ  [      = 

5. In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that    \angle \textup{ROS} = \frac{1}{2}(\angle \textup{QOS} - \angle \textup{POS})


Given that, POQ is a line, OR  PQ and  ROQ is a right angle. Now,  POS + ROS +  ROQ =   [since POQ is a straight line]  .............(i) and,  ROS +  ROQ =  QOS        ..............(ii) Add the eq (i ) and eq (ii),  we get hence proved

4. In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.


Given that,  ..............(i) It is known that, the sum of all the angles at a point =    ..............(ii) From eq (i) and eq (ii), we get Hence proved AOB is a line.  

3.  In Fig. 6.15, \angle PQR = \angle PRQ, then prove that \angle PQS = \angle PRT.


Given that, ABC is a triangle such that  PQR =  PRQ  and ST is a straight line. Now,  PQR +  PQS =      {Linear pair}............(i) Similarly,  PRQ +  PRT = ..................(ii) equating the eq (i) and eq (ii), we get    {but  PQR =  PRQ } Therefore,  PQS = PRT Hence proved.

2. In Fig. 6.14, lines XY and MN intersect at O. If \angle POY = 90^o and a : b = 2 : 3, find c.


Given that, Line XY and MN intersect at O and POY =  also ..............(i)  Since XY is a straight line Therefore, ...........(ii) Thus, from eq (i) and eq (ii), we get So,   Since MOY = c [vertically opposite angles]           a + POY = c            

1. In Fig. 6.13, lines AB and CD intersect at O. If \angle AOC + \angle BOE = 70° and \angle BOD = 40°, find \angleBOE and reflex \angleCOE.


Given that, AB is a straight line. Lines AB and CD intersect at O.  and  BOD =  Since AB is a straight line  AOC + COE + EOB =   [since ] So, reflex COE =  It is given that AB and CD intersect at O Therefore, AOC  = BOD  [vertically opposite angle]  [ GIven  BOD = ] Also,  So,  BOE = 

15.12)  For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?

(i) (a) All the points vibrate with the same frequency of 60 Hz. (b) They all have the same phase as it depends upon time. (c) At different points, the amplitude is different and is equal to A(x) given by   (ii)