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Question-1

Lines AB and CD intersect at O. If + = 700 and = 400, find and reflex

Solution:
 Lines AB and CD intersect at O
AOC = BOD                                                  | Vertically Opposite Angles

But BOD = 40
° …(1)                                             | Given

AOC = 40° ….(2)

Now, AOC + BOE = 70°
40° + BOE = 70°                                                |Using (2)        ⇒ ∠ BOE = 70° - 40°

        ⇒ ∠ BOE = 30°
Again, Reflex COE = COD + BOD + BOE

                              = COD + 40
° + 30° ........(3)     |Using (1) and (2)

                              = 180
° + 40° + 30°                       Since, AOC + AOD = 180°   COD = 180°    (Linear Pair Axiom)
                             
                              = 270
°  

Question-2

In the figure, lines XY and MN intersect at O. If  POY =  90° and a : b = 2 : 3 find c.

Solution:
Ray OP stands on line XY
POX + POY = 180
°                |Linear Pair Axiom    POX + 90°  = 180°               |Since, POY = 90°  (Given)             POX = 180° - 90°
             POX =  90° 
   POM +  XOM = 90°  
              a + b = 90°   ……..(1)


     a : b = 2 : 3 = Let = = k

    a = 2k, 
b = 3k         

Substituting the values of a and b in (1), we get
     2k + 3k =
90°            5k = 90°   
           ⇒ k =          k = 18° 

a = 2k = 2(18°) = 36°
   b = 3k = 3(18°) = 54° …...........(2)

Ray OX stands on line MN               | Linear Pair Axiom
              b + c = 180°
            54° + c = 180°
              c = 180° - 54°           | Using (2)

                    c = 180° - 54°
                    c = 126°

Question-3

 


Solution:

New Page 1 Ray QP stands on line ST PQS + PQR = 180° …..(1)         | Linear Pair Axiom Ray RP stands on line ST

PRQ + PRT = 180° …(2)           | Linear pair Axiom

From (1) and (2), we obtain

  PQS + PQR = PRQ + PRT

            PQS = PRT                  | Since, PQR = PRQ(Given)


Question-4

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

Solution:
x + y = w + z …..(1)                       | Given The sum of all the angles round a point is equal to 360°
x + y + w + z = 360°

  Þ x + y+ x + y = 360°            Using (1)
        2(x + y) = 360°
        x + y = 180° 
AOB is a line.

Question-5

In the figure POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ROS = (∠QOS - POS).

Solution:
Given, Ray OR is perpendicular to line PQ QOR = POR = 90°               ...(1)
   QOS = QOR + ROS           ...(2)

   POS = POR - ROS             ...(3)

From (2) and (3), QOS - POS =  QOR +  ROS -  POR +  ROS
                         = ( QOR - POR) + 2 ROS
                         = 2 ROS                    | Using (1)
             ROS = ( QOS -  POS).

Question-6

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

Solution:
 

Ray YZ stands on line PX

XYZ + ZYP = 180°        | Linear Pair Axiom     ⇒ 64° + ZYP = 180°        |Since, XYZ = 64° (Given)             ZYP = 180° - 64°

             ZYP = 116° …….(1) Given Ray YQ bisects ZYP

  PYQ = ZYQ = ZYP = (116
° = 58° …..(2)              | Using (1)
 

Reflex QYP = 360° - 58° = 302°

  Again, XYQ = XYZ + ZYQ

                      = 64° + 58°              | XYZ = 64° (Given) and ZYQ = 58° [From (2)]

                      = 122°.

Question-7

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

Solution:
Ray AE stands on line GH
AEG + AEH = 180°          | Linear Pair Axiom             50°+ x = 180°   x = 180° - 50°= 130° ...(1)

                       y = 130° ...(2)
     | Vertically Opposite Angles

From (1) and (2), we conclude that

x = y

But these are alternate interior angles and they are equal.

So, we can say that AB || CD.

Question-8

In the figure, if AB || CD, CD || EF and y: z = 3 : 7, find x.

Solution:
 

Given, AB || CD and CD || EF

 Þ AB || EF           | Lines parallel to the same line are parallel to each other
              \ x = z    ..............(1)

x + y = 180° ...(2)    |Consecutive interior angles on the same side of the transversal GH to parallel lines AB and CD

From (1) and (2),

z + y = 180°

 y : z = 3 : 7

Sum of the ratios = 3 + 7 = 10   y = = 540 and z = 1260   x = z = 1260

Question-9

In the figure, if AB || CD, EF CD and GED = 126°, find AGE, GEF and FGE.

Solution:
(i) AGE = GED = 126°                   | Alternate Interior Angles
(ii)           GED = 126°

              ⇒ GEF + ∠ FED = 126°             GEF+ 90°= 126°                     |Since, EF CD and FED = 90°             GEF = 126° - 90° = 36°

(iii) GEC + GEF + FED = 180°      | CD is a line            GEC + 36° +  90° = 180°                     GEC + 126°= 180°          GEC = 180° - 126° = 54°

             Now, FGE = GEC = 54°           

Question-10

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

Solution:
Construction: Draw a line RU parallel to ST through point R.

RST + SRU = 180°                |Sum of the consecutive interior angles on the same side
                                                |of the transversal is 180°

130° + SRU = 180°
SRU = 180° - 130° = 50°             ...... (1)
    QRU = PQR = 110°           | Alternate Interior Angles
 ⇒ ∠ QRS + SRU = 110°
    QRS + 50° = 110°           | Using (1)
             QRS = 110° - 50° = 60°

Question-11

In the figure, if AB || CD, APQ = 50° and PRD = 127°, find x and y.

Solution:
x = APQ = 50°                   | Alternate Interior Angles
 PRD = x + y = 127°              | Sum of the two Interior opposite Angles
50° + y = 127°
y= 127° - 50°
y= 77°

Question-12

In the figure, 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.

Or

In the figure, m and n are two plane mirrors parallel to each other. Prove that incident ray CA is parallel to reflected ray BD.


[Hint: Draw perpendiculars (normals) at A and B to the two plane mirrors. Recall that angle of incidence is equal to angle of reflection.]

 

Solution:
Construction: Draw ray BL PQ and ray CM RS.

Proof: BL PQ, CM RS and PQ || RS

BL || CM

LBC = MCB ...(1)           | Alternate Interior Angles ABL = LBC ...(2)            | Angle of incidence = Angle of reflection

MCB = MCD ...(3)           | Angle of incidence = Angle of reflection

From (1), (2) and (3), we get ABL = MCD ...(4)

Adding (1) and (4), we get

  LBC + ABL = MCB + MCD
         ABC = BCD


But these are alternate interior angles and they are equal.

So, AB || CD.

Question-13

In the figure, sides QP and RQ of Δ PQR are produced to points S and T respectively. If SPR =135° and PQT = 110°, find PRQ.

Solution:
TR is a line PQT +  PQR = 180°

  110° + PQR = 180°              PQR = 180° - 110° = 70°   ......... (1) QS is a line

         ∴ ∠ SPR + QPR = 180°          135° + QPR =180° QPR = 180° - 135° = 45°                ......... (2)

In Δ PQR,  PQR + QPR + PRQ = 180°       | The sum of all the angles of a triangle is 180°     70° + 45° + PRQ = 180°           | Using (1) and (2)            115° + PRQ =180°   PRQ =180° - 115° = 65°.

Question-14

In the figure, X = 62°, XYZ = 54°. If YO and ZO are the bisectors of XYZ and XZY respectively of Δ XYZ, find OZY and YOZ.

 

Solution:
In XYZ, XYZ + YZX + ZXY = 180°                                   | The sum of all the angles of a triangle is 180°     54° + YZX + 62° = 180°            116° + YZX = 180°   YZX = 180° - 116° = 64°  ...(1)

YO is the bisector of XYZ XYO = OYZ = XYZ = (54°) = 27° ...(2) ZO is the bisector of YZX
XZO = OZY = YZX = (64°) = 32° ...(3)         | Using (1)

In triangle OYZ, OYZ + OZY + YOZ = 180°                                  | The sum of all the angles of a triangle is 180°     27° + 32° + YOZ = 180°                                  | Using (2) and (3)               59° +  YOZ = 180°     YOZ = 180° - 59° = 121°.

Question-15

In the figure, if AB ½½ DE, BAC= 35° and CDE = 53°, find DCE.

Solution:
DEC = BAC = 35° …(1)            | Alternate Interior Angles
CDE = 53° …(2)                         | Given

In Δ CDE, CDE + DEC + DCE = 180°    | The sum of all the angles of a triangle is 180°.
  53° + 35° = CDE + DEC | Using (1) and (2)
  88° + DCE = 180°
  DCE = 180° - 88° = 92°.

Question-16

In the figure, if lines PQ and RS intersect at point T, such that PRT = 40°, RPT = 95°,  TSQ = 75°, find SQT.

Solution:
In Δ PRT, PTR + PRT + RPT = 180°  | The sum of all the angles of a triangle is 180° 
   PTR + 40° + 95° = 180°           PTR + 135° = 180°                      PTR = 45°    | Vertically Opposite Angles          QTS = PTR = 45°

In Δ TSQ, QTS + TSQ + SQT = 180° | The sum of all the angles of a triangle is 180
°
   45° + 75° + SQT = 180°           120° + SQT = 180°   SQT = 180° - 120° = 60°.

Question-17

In the figure, if PQ PS, PQ || SR, SQR = 28° and QRT = 65°, then find the values of x and y.

Solution:
QRT = RQS + QSR                 |The exterior angle is equal to the sum of the two
                                                   |interior opposite angles

65° = 28° + QSR
QSR = 65° - 28° = 37° PQ SP

⇒ ∠ QPS = 90° PQ || SR

∴∠ QPS + PSR = 180°                 | The sum of consecutive interior angles on the same side
                                                          of  the transversal is 180° 90° + PSR = 180° PSR = 180° - 90° = 90°
PSQ + QSR = 90°
  y + 37° = 90° y = 90° - 37° = 53°

In Δ PQS, PQS + QSP + QPS = 180°      | The sum of all the angles of a triangle is 180° x + y + 90° = 180°

  x + 53° + 90° =  180°
x + 143° = 180°
  x = 180°- 143° = 37°.

Question-18

In the figure, the side QR of Δ PQR is produced to a point S. If the bisectors of PQR and PRS meet at point T, then prove that QTR = QPR.
                                                                     

Solution:
TRS is an exterior angle of ΔTQR
TRS = TQR + QTR ...(1)                   | Since, the exterior angle is equal to the sum of the two
                                                             |interior opposite angles. 
∠ PRS is an exterior angle of ΔPQR
PRS = PQR + QPR ...(2)                   | Since, the exterior angle is equal to the sum of the
                                                                 |two interior opposite angles 2 TRS = 2 TQR + QPR                      |QT is the bisector of PQR and RT is the bisector of PRS
2 TRS
                                          
  2 ( TRS - TQR) = QPR ...(3)

From (1),          TRS - TQR = QTR ...(4)

From (3) and (4), we obtain

                   2 QTR = QPR
                 QTR = QPR.




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