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Question-1

If B lies between A and C and AC = 10 cm, BC = 5 cm, what is AB2.


Solution:
Here B lies between A and C, and A, B, C are collinear. AC = 10 cm
          
    AC = AB + BC

    10 = AB + 5


AB = 10 - 5 = 5 cm

   AB2 = (5)2 = 25 cm.

Question-2

In the figure, POQ is a line, POR = 4x and QOR = 2x. Find the value of x.
                              

Solution:
POR + QOR = 180° (Linear pair axiom)

 

But POR = 4x and QOR = 2x

4x+2x = 180°

     6x = 180°

       Þ x = 30°

The value of x = 30° .

Question-3

Given three collinear points A, B, C, name all the lines-segments they determine.


Solution:
A, B and C are collinear
Therefore, line segments will be AB, AC and BC.

Question-4

In the figure, OA and OB are opposite rays.
 (i) If ∠ BOC= 75°, find ∠ AOC
 (ii) If ∠ AOC= 110°, find ∠ BOC.

                        

Solution:
(i) OA and OB are opposite rays, then ∠ AOC and ∠ BOC form a linear pair.
 
∴ ∠ AOC + ∠ BOC = 180°
If ∠ BOC = 75°, then
    ∠ AOC = 180° -75°
              = 105°

(ii) If ∠ AOC = 110°, then
       ∠ BOC = 180° - 110°
                 = 70°

Question-5

In the figure, lines AB, CD and EF intersect at O. Find the measures of AOE, EOD, BOD and COF.


Solution:

Since line AB and EF intersect at O

AOE = BOF = 35°

Again, since line AB and CD intersect at O,


DOB = AOC = 30°

Now
AOC + COF + BOF = 180°

30° + COF + 35° = 180°

     ...
COF = 180° - 30° - 35° = 115°

Again
EOD = COF = 115°
 

Question-6

In the figure, POR  and QOR form a linear pair. If a – b = 80°, find the values of a and b.
              

Solution:
POR and QOR form a linear pair. (given)

...
a + b  = 180°      (Two adjacent angles form a linear if and only if they are
                                 supplementary) --------------- (1) 
    
a - b =  80°                        ------------------- (2)

Adding (1) and (2)
    2
a = 180o +80° = 260o
   
a = 130°

Substituting
a = 130° in (1) 
     130° +
b = 180°
..
b = 180° - 130° = 50°.

Question-7

In the figure, a is greater than b by one third of a right angle. Find the values of a and b.
            

Solution:
Given that a > b by of a right angle.
   a = b + × 90° a = b + 30°
   But a + b = 180° (Linear pair axiom)
b + 30 + b = 180°
2b + 30° = 180°
       2b = 180° -30°
       2b = 150°.
        b = 75°
         a = 180° -75°
              = 105°
The measure of a = 105° and b = 75°.

Question-8

In the figure, two straight lines AB and CD intersect at O. If COT = 60°, find a, b, c.


Solution:
Since AB and CD intersect at O

BOD = AOC

     
a = 4b …. (i)

Also a + b + 60° = 180°

          
 Þ a + b = 120° …. (ii)

          or 4b + b = 120°

              
5b = 120° 
                 b = 24°

a = 4b = 4 x 24 = 96°

Since
AOD = COB …. (Vertically opposite angles)

         
2c = 60° + b

         
2c = 60° + 24° = 84°

           ... c = 42°

Hence a = 96°, b = 24°, c = 42°

Question-9

In the figure, if AOC + BOD = 70°, find COD.
           

Solution:
AOC + BOD =  70° (Given) ……………….(i)

AOC + COD + BOD = 180° (Linear pair axiom)

                       \ COD = 180° - ( AOC +  BOD)

                                    = 180° - 70° [From (i)]

                                    = 110°

The measure of COD = 110°.

Question-10

In the figure, find the value of y
.

Solution:
Since line EF and AB intersect at O, hence vertically opposite angles EOB and AOF will be equal. 

EOB = AOF = 5y …. (i)

Now two rays EO and BO stands on line CD


COE + EOB + BOD = 180°

             
4y + 5y + 3y = 180°

                           ... 12y = 180° 

                               y = 15°.

Question-11

In the figure, find the value of y.
       

Solution:
By linear pair axiom,

5y + 3y + 2y = 180°

      10y = 180°

          y = 18°.

Question-12

In figure, m||n and the 1 = 62°. Find 6 and 7.

Solution:
         1 = 3 …. (Vertically opposite angles)

Since
1 = 62°

      ..
3 = 62°

  Now
3 = 7 …. (Corresponding angles)

     ...
7 = 62°

        Now
6 + ∠7 = 180°
            .. Ð6 + 62° = 180°

   ...  Ð6  = 180° - 62° = 118°

Thus ∠6 = 118° and ∠7 = 62°

Question-13

If ray OC stands on a line AB such that ∠ AOC = ∠ BOC (Fig), then show that ∠ AOC = 90°.
             

Solution:
OC stands on the line AB.
⇒ ∠ AOC + ∠ BOC = 180°
⇒ ∠ AOC = 180° - ∠ BOC
But ∠ AOC = ∠ BOC (Given)
   2 ∠ AOC = 180°   (Since ∠ AOC = ∠ BOC)
∴ ∠ AOC = = 90°. ∴ The measure of ∠ AOC = 90°.

Question-14

In figure 1 = 110° and 8 = 60°. Is m || n.


Solution:
         1 = 3 …. (Vertically opposite angles)

Since
1 = 110°

    
3 = 110°

Again
8 =6 …. (Vertically opposite angles) 

  and
8 = 60° …. (given)

    
6 = 60°

Now
3 + 6 = 110° + 60° = 170° 180°

... m is not parallel to n

Question-15

In the figure, if AOB is a line, OP bisects ∠ BOC and OQ bisects ∠ AOC, show that
∠ POQ is a right angle.

Solution:
Ray OP bisects ∠ BOC
... ∠ COP = ∠ POB
... ∠ COP =  1/2 ∠ BOC           ------------(1)

Ray OQ bisects ∠ AOC
... ∠ AOQ = ∠ QOC
       ∠ QOC = 1/2 ∠ AOC             -----------(2)

Adding (1) and (2) 

... ∠COP + ∠ QOC = 1/2 [ ∠BOC + ∠ AOC ]    (∠AOC and ∠ BOC form a linear pair.)

                         = 1/2 x 180°                      (Linear Pair Axiom)

                 ∠QOP =  90°

             ... ∠POQ is a right angle.

Question-16

In quadrilateral ABCD, AB || CD and AD || BC. Prove that ABC = ADC.


Solution:
Since AB || DC and AD is the transversal

BAD + ADC = 180°     …. (Consecutive interior angles) …(i)

Again AD || BC and AB is the transversal


⇒ ∠DAB + ABC = 180° ….(ii)

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


⇒ ∠BAD + ADC = DAB + ABC

           ...
ADC = ABC

Question-17

Ray OE bisects ∠AOB and OF is the ray opposite OE. Show that FOB =FOA.

Solution:
Ray OE bisects ∠BOA

...
BOE = AOE          ----------------------(1)

FOB  and BOE  form a linear pair.

FOB + BOE = 180° (Linear Pair Axiom)

         
FOB  = 180° - BOE  ----------------- (2)

Similarly,
FOA and AOE form a linear pair.

FOA + AOE = 180°   

          
FOA = 180° - AOE  ----------------- (3)

From (1) 
BOE = AOE

    180° -
BOE =180° - AOE  

        ..
FOB  = FOA   (From 2 and 3) .

Question-18

In the figure, show that AB || EF.


Solution:
     ∠ABC = 70° 

    ∠BCD = ∠BCE + ∠ECD = 30° + 40° = 70°

 =>   ∠ABC = ∠BCD = 70°

Hence, interior alternate angles are equal

 =>   AB || CD

Now ∠FEC + ∠ECD = 140° + 40° = 180°

But these angles are consecutive interior angles

   =>  CD || EF

Also CD || AB
   ... AB || EF

Question-19

Find the value of x in the following figures if
(a) AB || CD
(b) AB || CD and BC || DE
(c) AB || CD and BC || ED
(d) BA || CD

    

(a)                              (b)



(c)                          (d)


 

Solution:
(a) Draw EF || AB



Since AB || EF and AE is a transversal

      ... ∠A + ∠1 = 180° 
⇒ 120° + ∠1 = 180° Since AB || CD and AB || EF

 ∠1 = 180 - 120 = 60° ….(i)

Since AB || CD and AB || EF

... CD || EF. Also EC is a transversal
 
       ⇒ ∠2 + ∠C = 180° ….(Consecutive interior angles)
     ∠2 + 125° = 180°
... ∠2 = 180 - 125 = 55° ….(ii)

Adding equations (i) and (ii), we get ..∠1 + ∠2 = 60° + 55° = 115°

            ... x =  ∠1 + ∠2 = 115°


(b)

Since AB || CD and BC is a transversal
     ..∠B = ∠C = 40° ….(Alternate angles)

Now BC || DE and CD is transversal
⇒ ∠C + ∠CDE = 180° ….(Consecutive interior angles)

⇒ 40° + ∠CDE = 180°     ... ∠CDE = x = 180° - 40° = 140°

(c)

Since AB || CD and BC is the transversal     ⇒ ∠B = ∠C .....(Alternate angles)

but ∠B = 75°,

  .. ∠C = 75°

Again we have BC || ED with transversal CD.

   ...  ∠C + ∠D = 180° …. (Consecutive interior angles) ⇒ 75° + ∠D = 180°      ... ∠D = x = 180° - 75° = 105°


(d)

Through O draw OE parallel to BA || CD

          ⇒ ∠1 + ∠B = 180°          ⇒ ∠1 + 55° = 180° ... ∠1 = 180° - 55° = 125° ….(i)

     Again ∠2 + ∠C = 180° ….(Consecutive interior angles)           ... ∠2 + 50° = 180°    ... ∠2 = 180 - 50 = 130° ….(ii)

Adding (i) and (ii), we get
... ∠1 + ∠2 = x = 125° + 130° = 255°

Hence x = 255°

Question-20

Rays OA, OB, OC, OD and OE have the common end point O. Show that AOB + BOC+ COD + DOE + EOA = 360°.

Solution:
Given: OA, OB, OC, OD and OE have common end point O.

To prove: AOB + BOC+ COD + DOE + EOA = 360°.

Construction: Draw the ray opposite to OA.

Proof: Adjacent
s, COF and COA form a linear pair.

COF + COA = 180°     (Linear Pair Axiom) --------------(1)

Similarly, adjacent
s, ÐFOD and DOA form a linear pair.

FOD + DOA = 180°                              --------------- (2) 

Adding (1) and (2)

                            
COF + COA + FOD + DOA  = 180° + 180°

(
COF + FOD) + COB + BOA + (EOA + DOE)  = 360°

             AOB + BOC+ COD + DOE + EOA = 360°.

Question-21

m and n are two plane mirrors placed parallel to each other as shown in the figure. An incident ray AB to the first mirror is reflected twice in the direction CD, Prove that AB || CD.

Solution:
Draw CN n and BM m. Since BM m, CN n and m || n. 

Therefore CN
|| BM. Now according to laws of reflection, we know that angle of incidence is equal to angle of reflection

     ... At B, 1 = 2 ….(i)

and at C,
3 = 4 ….(ii)

Since CN
|| BN with transversal BC,     2 = 3 ….(Alternate interior angles)       Þ 2 2 = 2 3

        Þ 2 +2 = 3 + 3

        Þ 1 + 2 = 3 + 4
     ABC = BCD

... AB || CD

Question-22

Given that each of the angles AOC and AOB is a right angle. Show that BOC is a line.
         

Solution:
Given: BOA = 90° and COA = 90°

To prove: BOC is a line.

Proof: BOA and COA are adjacent angles.

                        BOA = 90°    (given)
                        COA = 90°      (given)
BOA + COA = 90°+ 90° = 180°         
  BOA and COA form a linear and it have a common arm OA,
... BOC is a line.

Question-23

A transversal intersects two given lines in such a way that the interior angles on the same side of the transversal are equal. Show by means of a figure that the lines need not be parallel. State the condition under which the two lines will be parallel.

Solution:
Here 1 = 2 ….(Given)

AB will be parallel to CD only when

    1 + 2 = 180° (1 and 2 should be supplementary)       2 1 = 180° 
        1 = 90° ⇒ ∠1 = 2 = 90°

There AB will be parallel to CD when
1 = 2 = 90°, but AB will not be parallel to CD if two angle (1 and 2) are either acute or obtuse.

Question-24

OP bisects AOC, OQ bisects ∠COB  and OP ⊥ OQ. Show that A, O, B are collinear.
     
   

Solution:
Given: OP bisects ∠AOC
        OQ bisects ∠COB
        OP ⊥ OQ

To prove: A, O, B are collinear.

Proof: OP bisects ∠AOC
         AOP =  POC ⇒ COP = 1/2 COA --------------(1)

OQ bisects ∠BOC
        COQ =  QOB
⇒ ∠COQ = 1/2  BOC                ------------------(2)

     ∠POQ = COP + COQ =  90°  -------------------(3)

Substitute (1) and (2) in (3)
     1/2 ∠COA + 1/2  BOC =  90°
        1/2 [ COA +  BOC] = 90°
   AOC + COB  = 90° x 2 = 180°
=>  AOC , COB  form a linear pair.
...  A, O, B are collinear (Linear Pair Axiom).

Question-25

What value of x would make AOB a line in the figure, if ∠ AOC=4x and ∠ BOC=6x+30° ?
         

Solution:

 ∠ AOC = 4x (given)

∠ BOC = 6x+30° (Given)

∠ AOC + ∠ BOC = 4x + (6x+30° ) = 180° (Linear pair axiom)

⇒ 10x + 30° = 180°

        ⇒ 10x = 180° - 30°

                  = 150°

             \ x = 15°

∴ The value of x = 15°.

Question-26

If two lines are intersected by a transversal in such a way that the bisectors of a pair of corresponding angles are parallel, show that two lines are parallel.

Solution:

Given: Two lines AB and CD are intersected by the transversal EF and GP is the angle bisector of EGB and HQ is the bisector of GHD. Also GP || HQ

To Prove: AB || CD

Proof: Since GP || HQ and EF is the transversal
 
  
  Þ 1 = 3

  Þ 2 1 = 2 3    
     
...
EGB = GHD

But these are corresponding angles of AB and CD with the transversal EF. 

... AB || CD

Question-27

In the figure, AB and CD are parallel lines. The bisectors of the interior angles on the same side of the transversal EF intersect at P. Prove that ∠GPH = 90°.


Solution:
  ∠BGH + ∠DHG = 180°  ….(Consecutive interior angles of AB || CD)….(i)

Now ray GP bisects ∠BGH

... ∠1 = ∠2 =  ∠BGH ….(ii)

Similarly ∠3 = ∠4 = ∠DHG ….(iii)

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

... ∠1 + ∠3 =   ∠BGH +   ∠DHG

... ∠1 +∠3 = (∠BGH + ∠DHG) ….(iv) 

From equations (i) and (iv), we get

... ∠1 + ∠3 = x 180° = 90°

Now in ΔPGH, we have

... ∠1 + ∠3 + ∠GPH = 180° ….(sum of the angles of Δ)

        ⇒ 90° + ∠GPH = 180°

⇒ ∠GPH = 180° - 90° = 90°

Question-28

In the figure, AOF and FOG form a linear pair. EOB = FOC = 90° and DOC =  FOG =  AOB = 30°.
          

 Find the measure of FOE, COB and DOE.
        
      

Solution:
AOF + FOG = 180° [Linear pair axiom]
AOG = 180°
AOB + EOB + FOE + FOG = 180°
30° +90° + FOE+ FOG = 180°
FOE + FOG = 180° - (30° + 90° )
                                 =  180° - 120°
                                 =  60°

FOE + 30° = 60° [ FOG =30° ]
FOE = 30°
AOB + EOB + FOE + FOG = 180°
AOB + BOC + COF + FOG = 180°
30° + BOC + 90° + 30° = 180° [ AOB = 30° , COF = 90° and FOG = 30° ]
BOC = 180° - (30° + 90° + 30° )
                = 180° - 150°
                = 30° ∴ ∠ COB = 30° ……………….(i)

EOB = 90° [Given]
BOC + COD + DOE = 90°
30° + 30° + DOE = 90° [ BOC = 30° from (i) and COD = 30° ,Given]
DOE = 90° - (30° + 30° )
                = 90° - 60°
                = 30°

DOE = 30° ……………(ii)
FOE = 30° , COB = 30° and DOE = 30° .

Question-29

The sum of two angles of a triangle is 120° and their difference is 20°. Find all the three angles of the triangle.

Solution:
Let three angles of the triangle be A, B and C.

Let A + B = 120° ….(i)

 and A - B = 20 ….(ii)

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


A = 70° and B = 50°

Now the sum of three angles of a triangle is 180°

...
A + B + C = 180°

 70° + 50° +
C = 180°

...
C = 60°

Hence, three angles of DABC are 70°, 50° and 60°.

Question-30

In figure,  lines p and r intersect at O. If x = 45°, find y, z and u.
          

Solution:
Given: Lines p and r intersect and forming x, y, z and u.
  x = 45°

To find: y, z and u


Solution: In the figure x, z are vertically opposite angles.

x = z (Vertically opposite angles are equal) 
x = 45° (given)
z = 45°

Adjacent angles x and y form a linear pair.
x + y = 180° (Linear Pair Axiom)
            y = 180° - 45°
                 = 135° ---------------- (1)

y and u are vertically opposite angles.
     Ð y = u
     y = 135° (from 1)

u = 135°
y = 135°  
z = 45°.

Question-31

In figure,  AB, CD and PQ are three lines concurrent at O. If ∠ AOP = 5y, ∠ QOD = 2y and ∠ BOC = 5y, find the value of y.

         

Solution:
Lines AB, CD and PQ intersect at O.
∠ DOQ = 2y°, ∠ AOP = 5y° and ∠ COB = 5y°.

AB and CD intersect at O.
∠ COB = ∠ AOD (Vertically opposite angles)

∴ ∠ AOD = 5y (Since ∠ COB = 5y)
 ∠ AOP + ∠ AOQ = 180°
⇒ ∠ AOP + ∠ AOD + ∠ DOQ = 180°
⇒ 5y + 5y +  2y = 180°
⇒ 12y = 180°
⇒ y = = 15°
∴ The value of y = 15°.

Question-32

In the figure, the bisectors of DABC and DACB intersect each other at O. 
Prove that: 
BOC = 90° + A

Solution:
In DABC, Since BO is the bisector of ABC

        ...
1 = 2

Similarly
3 = 4

and
A + B + C = 180° ….(i)

Now in DBOC, we have


2 + BOC + 3 = 180° ….(ii)

Multiplying equation (ii) by 2, we get

2
2 + 2BOC + 23 = 2 x 180 = 360°

B + 2BOC + C = 360°

               
2BOC = 360° - (B +C)

               
2BOC = 360° - (180° - A)

                              = 360° - 180° +
A

               
2BOC = 180° + A

                  
⇒ ∠BOC = 90° + A

Question-33

In figure,  three lines p, q and r are concurrent at O. If a = 50° and b = 90°, find c, d, e and f.
        

Solution:
Given: Three coplanar lines AB, CD and PQ meet at O.
       BOQ = a = 50° and BOC = b = 90°
 
To find: c, d, e and f.
Proof: Adjacent angles, BOQ, BOP form a linear pair.
     .. BOQ + BOP = 180° (Linear Pair Axiom)
 a + BOC + COP = 180°
                   a + b + c = 180°
                 50° + 90° + c = 180° (given)
                                    c = 180°- (90° + 50°)
                                       = 180° - 140°
                                       = 40°
 d = a = 50° (Vertically opp. angles)
 e = b = 90° (Vertically opp. angles)
 f = c = 40° (Vertically opp. angles).

Question-34

In the figure, the sides AB and AC of a triangle ABC are produced to P and Q respectively. The bisectors of ∠PBC and ∠QCB intersect at O. Prove that ∠BOC = 90° - ∠A.


Solution:
Since BO and CO are angle bisectors of PBC and QCB respectively

... ∠1 = ∠2 and ∠3 = ∠4

Side AB and AC of DABC are produced to P and Q respectively .

Hence Exterior ∠PBC = ∠A + ∠C ….(i)

   and Exterior ∠QCB = ∠A + ∠B ….(ii)

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

∠PBC + ∠QCB = 2∠A + ∠B + ∠C

 => 2∠2 + 2∠3 = ∠A +(∠A + ∠B + ∠C)
 => 2∠2 + 2∠3 = ∠A + 180°  

    ⇒ ∠2 + ∠3 = 90° + ∠A ….(iii)

Now in ΔBOC, ∠2 + ∠BOC + ∠3 = 180° ….(iv) 

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

∠BOC + 90° + ∠A = 180°
\ ∠BOC = 90° - ∠A

Question-35

In the figure, PS is the bisector of QPR and PT ^ QR. Show that TPS = (Q - R)


Solution:
Since PS in the bisector of QPR

...
QPS = SPR

...
QPS = P

Also in the right triangle PTQ, we have


Q +TPQ = 90° (... PTQ = 90°)

    ...
TPQ = 90° - Q

  Now
TPS = QPS - QPT

                =
QPR - (90° - Q)

                =
QPR - 90° + Q

                = [180° - (
Q + R)] - 90° + Q

                = 90° -
Q - R - 90° + Q

                = (Q - R)

Question-36

In figure, AB and CD are straight lines and OP and OQ are respectively the bisectors of a BOD and AOC.  Show that the rays OP and OQ are opposite rays.
         

Solution:
Given: AB and CD are straight lines and OP and OQ are respectively the bisectors of angles BOD
and AOC. 
 To prove: The rays OP and OQ are in the same line.
 Proof:  OQ is the bisector of AOC

... AOQ = 1/2 AOC
Similarly, OP is the bisector of DOB,
 
 ... DOP = 1/2 DOB
AOQ + AOD + DOP = 1/2 AOC + 1/2 DOB + AOD
2 ( AOQ + AOD + DOP) = AOC + DOB + 2 AOD
                                             = AOC + DOB + AOD + COB           (  AOD =  COB vertically opp. angles )                            
                                             = 360° (Angles around a point 360°)

  AOQ + AOD + DOP = 360° /2 = 180°

.. Points Q, O, P are collinear.
or OP and OQ are opposite rays.

Question-37

One of the four angles formed by two intersecting lines is a right angle. Show that the other three angles will also be right angles.

Solution:


Let AB and CD be the intersecting lines intersecting at O.

Given that one of the angles formed say AOC = 90° .

AOC and AOD form a linear pair.

AOC + AOD = 180°

     90° + AOD = 180°

              AOD = 180° -90°

                            = 90°
AOC = DOB (Vertically opposite angles)

AOC = DOB = 90°

AOD= BOC (Vertically opposite angles)

AOD = BOC = 90°

One of the four angles formed by two intersecting lines is a right angle, then the other three angles will also be right angles.

Question-38

If two parallel lines are intersected by a transversal, prove that the bisectors of the two pairs of interior angles enclose a rectangle.

Solution:



Given: Two lines AB and CD are parallel with transversal EF intersecting AB at M and CD at O. Bisectors of consecutive interior angles meet at N and P respectively.

To Prove: Quadrilateral MNOP is a rectangle.
 
Proof: Since MP, OP, MN and ON are angle bisectors

∴   ∠1 = ∠2

    ∠3 = ∠4

    ∠5 = ∠6

    ∠7 = ∠8

Again AB || CD and EF is a transversal

      ∠BMO = ∠MOC

and ∠AMO = ∠DOM (alternate angles)

   ∠AMO =   ∠DOM

           ∠3 = ∠7

But these are alternate angles of sides MN and OP.

  MN || OP

Similarly, MP || ON

Hence MNOP is a parallelogram.

                       Now ∠ BMO + ∠ MOD = 180° (Consececutive interior angles)

     ⇒   ∠ BMO + ∠ MOD = x 180° = 90°
But in ∠ MOP, we have ∠2 + ∠7 + ∠ P = 180°.

              ⇒ ∠ BMO + ∠ MOD + ∠ P = 180°
                                        90° + ∠ P = 180°

                           or ∠ P = 180° - 90° = 90°

Now since MNOP is a || gm and one angle is 90°.

Hence MNOP is a rectangle.

Question-39

In the figure, if l || m, find the value of x. 


Solution:
Since l || m and DC is the transversal

...
D + 1 = 180°        ............................(i)

60° + 1 = 180° 1 = 120°  ..............(ii)

      Now
2 = 1 = 120°    (Vertically opposite angles)...................... (iii)

In
Δ ABC, we have A + B + C = 180°

                  
25° + x° + 120° = 180°

           
x = 180° - 120° - 25° = 35°

Question-40

AB, CD and EF are three concurrent lines passing through the point O such that OF bisects BOD. If BOF=35° , find BOC and AOD.

Solution:
 
 AB, CD and EF are three concurrent lines passing through the point O.
OF bisects BOD.
   BOF = FOD = 35°
But BOD = BOF + FOD
               = 35° +35°
               = 70°
BOD and AOD form a linear pair.
BOD + AOD = 180°
             AOD = 180° - BOD
                           = 180° -70°
                           = 110°
BOC = AOD (Vertically opposite angles)
But AOD = 110°
   BOC = 110°

Question-41

Sides BC, BA and CA of ABC are produced to D, F and E respectively. If ACD = 120° and EAF = 45°, find three angles of ABC.


Solution:
Since lines CE and BF intersect at A.
    BAC = EAF = 45° ….(Vertically opposite angles)
 
Now
ACD = BAC + ABC

    
120° = 45° + ABC

  
  ABC = 120° - 45° = 75°

Now in
ABC, we have

            
A + B + C = 180°

      
 Þ 45° + 75° + C = 180°

...
C = 180° - 45° - 75° = 60°

Hence three angles of
ΔABC are 45°, 75° and 60°.

Question-42

Which of the following statements are true (T) and which are false (F)? Give reasons.

         (i) Angles forming linear pair are supplementary.
         (ii) If two adjacent angles are equal, then each angle measures 90°.
         (iii) Angles forming a linear pair can both be acute angles.
         (iv) Two distinct lines in a plane can have two points in common.
         (v) If angles forming a linear pair are equal, then each of these angles is of measure 90°.
        (vi) If two lines intersect and if one pair of vertically opposite angles is formed by acute angles, then the other pair of vertically opposite angles will be formed by obtuse angles.
        (vii) If two lines intersect and one of the angles so formed is a right angle, then the other three angles will not be right angles.

Solution:

(i) T. All angles forming linear pair must be supplementary. 

(ii) F. Two adjacent angles will be equal as 90° only,when the lines are perpendicular to each other. 

(iii) F. Both angles forming linear pair need not be acute angles, can be one obtuse and one acute angle or both equal to 90
°

(iv) F. Two distinct lines cannot have two points in common.

(v) T. If angles forming a linear pair are equal, the measure of each angle formed will be equal to 90° since the lines will be perpendicular to each other.

(vi) T. If two lines intersect, and if one pair of vertically opposite angles formed are acute, then the other pair will be obtuse since the sum of the four angles will be 360°.

(vii) F. If two lines intersect and one of the angles so formed is a right angle, then the two lines will be perpendicular to each other. So all the angles formed will right angles. 

Question-43

Fill in the blanks so as to make the following statements true:
(i) Two distinct points in a plane determine a …………… line. (ii) Two distinct……….. in a plane cannot have more than one point in common. (iii) Given a line and a point, not on the line, there is one and only …………. line which passes through the given point and       is …….. to the given line.
(iv) A line separates a plane into ………….. parts namely the two ……… and the …………. itself.
(v) If one angle of a linear pair is acute, then its other angle will be ………..
(vi) If a ray stands on a line, then the sum of the two adjacent angles so formed is ………..
(vii) If the sum of two adjacent angles is 180°, then the ……….. arms of the two angles are opposite rays.
(viii) If two lines intersect, then vertically opposite angles are …………

Solution:
(i) unique
(ii) lines
(iii) one, parallel or perpendicular
(iv) three, half planes, line
(v) obtuse
(vi) 180°
(vii) non-common arms
(viii) equal

Question-44

In the figure AB || CD and ECD = 100° and ABE = 110°, find the value of x.


Solution:
Since AB || CD

     ... AF || CD ….(AB was produced to F) and CF is transversal

  ...
DCF = BFC = 100°

Now
BFC + BFE = 180° ….(Linear pair)

    ... 100° +
BFE = 180°

               ...
BFE = 180° - 100° = 80°

Now in
ΔBFE, we have

  
ABE = BFE +BEF

110° = 80° + x°
     Þ x = 110° - 80° = 30°

Question-45

In the given figure, p is a transversal to lines m and n, 2 = 120° and 5 = 60° . Prove that m || n.
               

Solution:
m and n are two lines cut by a transversal l.     2 = 120°

   2 = 4 (Vertically opposite angles are equal).

4 = 120°

Adjacent angles, 3, 4 form a linear pair.

3 + 4 = 180° (Linear Pair Axiom)

        3 = 180° - 4

               = 180° - 120° ( from (1) )

               = 60°

   But, 5 = 60° (given)

       3 = 5 (Alternate interior angles are equal)

m || n.

Question-46

The side BC of a triangle ABC is produced, such that D is on ray BC. The bisector of A meets BC at E. Show that ABC +ACD = 2 AEC.


Solution:
1 = 2

Since
ACD is the exterior angle of triangle ABC

...
ACD = A + B ….(i)

Now in
ΔABE, we have

    
AEC = 1 + 6

2AEC = 21 + 26

2AEC = A + 2B

2AEC = (ACD - B) + 2B

2AEC = ACD + B

             =
ACD +ABC

Question-47

In figure, if p is a transversal to lines q and r, q || r and 1 and 2  are in the ratio 3 : 2, find all angles from 1 to 8.

                  

Solution:
Adjacent angles 1 and 2 form a linear pair.
  1 + 2 = 180° (Linear Pair Axiom)
      1 :  2 = 3 : 2
           ... 1 = 3/5   x 180°   
                      = 108°
  2 = (2/5) x 180 = 72° ---------- (1)

Lines p and q intersect.
  2 =  4 (Vertically opposite angles)
   .. 4 = 72° (from 1)
Since 3 and 4  form a linear pair.
.. 3 = 180° - 72o = 108° (same as (2))

    4 = 6 = 72°  (Alternate interior angles)
    3 =  5 = 108° (Alternate interior angles)
Lines p and r intersect
5 = 7 = 108° (Vertically opposite angles)
6 = 8  = 72°  (Vertically opposite angles) 

.. 1 = 108°, 5 = 108°, 2 = 72°,  6 = 72°, 3 = 108°, 7 = 108°, 4 = 72°  
and 8 = 72°  

Question-48

The sides AB, BC, CA of a ABC are produced in order, forming exterior angles CBF, ACD and BAE show that ACD + BAE + CBF = 360°


Solution:
In ΔABC

exterior
1 = 3 + 5 ….(i)

exterior
4 = 2 + 5 ….(ii)

exterior
6 = 3 +2 ….(iii)

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

               
1 + 4 +6 = 23 + 25 + 22
 

ACD + BAE + CBF = 2(3 + 5 +2)

      
But 3 + 5 +2 = 180° ….(Sum of three angles of triangle)

...
ACD + BAE + CBF  =  2 x 180° = 360°

Question-49

In figure, show that AB || EF.
           

Solution:
Given: AB and OF are two lines cut by the transversal BO. CE is joined. ABO = 65° and FEC = 145°.

Construction: Produce FE to cut BC at O.

Proof:  In Δ COE, FEC = 145° is the exterior angle.
... FEC = EOC + OCE  (In a Δ the exterior angle is equal to sum of interior opposite angles).
     145° = EOC + 30°
   EOC = 145° - 30° = 115° --------------- (1)
BOE and EOC form a linear pair.
     ... BOE + EOC = 180° (Linear Pair Axiom).
                    BOE = 180° - 115° 
                            = 65°
AB and OF are two lines cut by the transversal BO.
... ABO and BOE are alternate angle, 
    and BOE = ABO = 65°
Since alternate interior angles are equal.
     ... AB || EF.

Question-50

In the figure, find the value of x. 


Solution:
Produce AE to meet BC at D.

Now in
ΔABD, we have exterior ADC = A + B

  ...
ADC = 25° + 55° = 80°

Again in
DCE, we have
 
exterior
AEC = EDC + ECD

  
 Þ ÐAEC = ADC + ECD

           x° = 80° + 60° = 140°

Question-51

In the figure l || m, A = 100°, B = 30° find x.

Solution:
Since l || m and DA is the transversal, 

A = CDB = 100° ….(i)

Now in
ΔBCD, we have

ext
x = CBD +CDB

       x = 30° + 100° = 130°

   ... x = 130°

Question-52

In figure, p is a transversal to lines m and n, 1 = 60° and Р2 =  of a right angle. Prove that m || n.
            

Solution:
1 = 60° (Given)
1 =3(Vertically opposite angles)
2 = of a right angle (Given)
     = × 90° = 60° ∴ 2 =3
Since 2 and 3 are a pair of corresponding angles which are equal, m||n.

Question-53

In figure, n||m and p||q. If 1=75° , prove that 2= 1+of a right angle.
         

Solution:
2 = 105° =75° + 30°

      = 1+30° (Since 1 = 75° )

      = 1+× 90°

The measure of 2 = 1+of a right angle.

Question-54

In the figure, AM BC and AN is the bisector of BAC. If B = 70° and C = 35°, find  MAN.


Solution:
In the ΔABC, AN is the angle bisector of BAC.

...
BAN = NAC   ........    (i)

       Also
A + B + C = 180°

        
⇒ ∠A + 70° + 35° = 180°

BAN + NAC + 105° = 180°

2 BAN = 180° - 105° = 75°

                       ...
 ÐBAN = 37.5° 

In Δ ABM, we have B + Δ BAM +Δ AMB = 180°

... 70° +
BAM + 90° = 180°

                  ...
BAM = 20°  .......  (ii)

               Now
MAN = BAN - BAM = 37.5° - 20° = 17.5°

Question-55

In the given figure lines AB and CD intersect at O. If Ð BOC Ð BOE = 700 and Ð BOD = 1300, find ÐBOC and reflex Ð AOC

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

But BOD = 130
° …(1)                    | Given
AOC = 130° ….(2)
Now, AOC + COB = 180°                     | Linear Pair Axiom

 1300 +  COB = 1800                   |Using (2)         Þ   COB   = 500     

Again, Reflex  AOC = COB + BOA
                            = 
500  +180° ........(3)     Since,  BOA = BOD + AOD = 180°   (Linear Pair Axiom)
                            = 230°                                                                  
 
 

Question-56

Which pair of lines in Fig. are parallel? Give reason.
        

Solution:
AB || DC and BC || AD.
BAD + ADC = 115° +65°
                       = 180°

BAD and ADC are supplementary angles.
AB and CD are two lines and AD is a transversal. One pair of interior angles on the same side of the transversal are supplementary.

AB and CD are parallel lines.
|||y CBA + BAD = 65° +115°
                             =180°

CBA and BAD are supplementary angles.
BC and AD are two lines and BA is a transversal. One pair of interior angles on the same side of the transversal are supplementary angles.

Question-57

In figure, if x = y and a = b, prove that r || n.
        

Solution:
Given that r and m are two lines cut by a transversal p and x = y.
x and y are two corresponding angles.

r and m are parallel.
Given that m and n be two lines cut by a transversal q and a = b.
a and b are corresponding angles.

m and n are parallel.
r || m proved and m || n proved.

r || n.

Question-58

If p, m, n are three lines such that p || m and n p, prove that n m.

Solution:


p || m (Given) and n is a transversal cutting p and m.
Let a and b be two angles made by p and m when cut by a transversal.

  a = b = 90° (Since it is given that n p
Since n is intersecting m and b is 90°

 n ^ m.
Hence proved.

Question-59

In figure, EF is a transversal to two parallel lines AB and CD. GM ad HL are the bisectors of the corresponding angles EGB and EHD. Prove that GM||HL.
                   

Solution:
Given: AB || CD and EF is a transversal cutting AB at G and CD at H.

GM is the bisector of EGB and HL bisector of GHD.

To prove: MGB = LHD.

Proof: Since AB || CD and EF is a transversal.

EGB = GHD (Corresponding angles are equal). --------------- (1)

GM is the bisector of EGB


EGM = MGB

or MGB = EGB------------- (2)

Similarly, HL is the bisector of GHD


LHD = GHD ---------------(3)

From (1)

    Ð EGB = GHD

  EGB = GHD


MGB = LHD [From (2) and (3)]

GM and HL are two rays cut by a transversal EF and MGB = LHD.


GM || HL

If a transversal intersects two lines in such a way a pair of corresponding angles are equal then the two lines are parallel.

Question-60

If two parallel lines are intersected by a transversal, then prove that the bisectors of any two alternate angles are parallel.
                           

Solution:

Given: AB and CD are 2 parallel intersected by a transversal EF.

PO and QR are the bisectors of the alternate angles APQ and PQD respectively.

To prove: OP || QR.

Proof: Since AB || CD they cut by a transversal EF.

APQ = PQD (Alternate angles are equal)----------- (1)
 

OP is the bisector of APQ

 \ APO = OPQ

or, APO = APQ -------------- (2)
 

Similarly, QR is the bisector of PQD

PQR = RQD = PQD ………….. (3)
 

From (1)

  APQ = PQD

APQ = PQD (From 2 and 3)

  Ð OPQ = PQR


Since, OP and QR are two lines cut by transversal EF such OPQ = PQR.

OP || QR (If a transversal intersects two lines in such a way that a pair of alternate

angles are equal then the two lines are parallel.)

The bisectors of any pair of alternate interior angles are parallel.

Question-61

In figure, bisectors GM and HL of alternate angles AGH and DHG respectively are parallel to each other. Prove that AB || CD.
                   

Solution:
GM and HL are parallel lines.
EGHF is a transversal.
      MGH =GHL …………………….(i)

2MGH = 2GHL
MGH +AGM =GHL +LHD

AGH =DHG

AB and CD are two lines and EGHF is a transversal, if one pair of alternate angles are equal, the lines AB and CD are parallel.

Question-62

If two lines are intersected by a transversal in such a way that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel.

                   

Solution:

Given: AB and CD are two lines cut by transversal EF.
      QP is the bisector of the angle
EQB. SR is the bisector of the corresponding QSD.
     PQ || SR.

To prove: AB || CD

Proof: QP || SR and are cut by the transversal EF.

    
 ÐEQP = QSR        (Corresponding angles are equal.)------ (1)

PQ is the bisector of
EQB

...
EQP = PQB = (1/2) EQB     ------------------(2)

Similarly, SR is the bisector of
QSD

   
QSR = RSD = (1/2) QSD ------------------(3)

But, from (1)

   
EQP = QSR

  (1/2) EQB = (1/2) QSD (From 2 and 3)

...
EQB = QSD.

AB and CD are two lines cut by a transversal EF such that corresponding angles.

   
EQB = QSD

... AB || CD (If a transversal intersects two lines such that a pair of corresponding angles are equal then the lines are parallel.)

Question-63

A transversal intersects two given lines in such a way that the interior angles on the same side of the transversal are equal. Is it always true that the given lines are parallel? If not, state the condition (s) under which the two lines will be parallel.

Solution:

Given:
AB and CD are two lines intersected by a transversal EF.
   
BPQ = PQD.

To prove: AB and CD need not be parallel. When will AB and CD be parallel?

Proof: If the two lines are parallel. Then,

  
BPQ + PQD = 180° (Sum of the interior angles is 180°)------------ (1) 

      ... If
BPQ = PQD, then (1) need not be true.

... Conditions under which the two lines are parallel is
BPQ + PQD = 180°

     Since
BPQ = PQD

                     BPQ = PQD = = 90°

... The two lines should be perpendicular to the transversal.

For the two lines to be parallel, any pair of corresponding angles made by the transversal should be equal, or any pair of adjacent angles on the same side of the transversal will be supplementary.

Two lines that are respectively perpendicular to two parallel lines are parallel to each other.

Question-64

In figure, arms BA and BC of ABC are respectively parallel to arms ED and EF of DEF. Prove that ABC = DEF.


Solution:


Given: BA || ED and BC || EF.

To Prove: ABC = DEF

Construction: Produce ED to meet BC at P.

Proof:

AB || ED and BPC is a transversal.

   1 = 2 (a pair of alternate angles)    …………..(i)

BC and EP’ are two lines intersecting at P.


  2 = 3 (Vertically opposite angles)  …………..(ii)

and 4 = 5(Vertically opposite angles) …………..(iii)

From (i) and (ii),

1 = 3………………(iv)

BC || EF (Given)

3 = 6 (Corresponding angles on the same side of the transversal)

But 3 = 2 and 2 = 1


   1 = 6

But 1 = ABC and 6 = DEF


ABC = DEF.

Question-65

In figure, arms BA and BC of ABC are respectively parallel to arms ED and EF of DEF. Prove that ABC +  DEF = 180°.


Solution:

Given: BA || ED and BC|| EF.
To Prove: ABC +DEF = 180° .
Construction: Produce FE to meet AB at E’.
Proof: BA || ED and FEE’ is the transversal.
BE’E =DEF……………..(i)

BC || E’EF and AB is the transversal.
BE’E = E’BC………………(ii)

BC || E’EF and AB is the transversal.
E’BC +BE’E = 180° [ Angles on the same side of the transversal are supplementary]
ABC +BE’E = 180° [E’BC =ABC]
ABC + DEF = 180° [From (i) BE’E =DEF]
  \ ABC + DEF = 180°

Question-66

Prove that two lines that are respectively perpendicular to two parallel lines are parallel to each other.

Solution:

Given: l and m are parallel. p and q are to l and m.

To prove: p and q are parallel to each other.

Proof:

Parallel lines l and m are cut by a transversal p, 1 = 3 (Corresponding angles)
Given that p is to l and m.
1 = 3 = 90° ……………………….(i)

Similarly parallel lines l and m are cut by a transversal q, 2 = 4(Corresponding angles)
Given that q is to l and m.
2 = 4 = 90° ………………………..(ii)

From (i) and (ii), 1= 2 = 90° and 3 = 4 = 90°
One pair of corresponding angles are equal, then the two lines are parallel.
p and q are parallel to each other.

Question-67

Prove that through a given point we can draw only one perpendicular to a given line.

Solution:


Given: AB is a line and P is any point outside AB. PM AB and PN is another line drawn from P to AB.
To prove: We can draw only one perpendicular from a given point P to a given line AB.

Proof: Let PM and PN be two perpendiculars drawn from a point P to the line AB. 

Then 1 = 2 = 90°
But these two angles form a pair of corresponding angles.
PM || PN
But two parallel lines never intersect.
So our assumption is wrong.
It is proved that we can draw only one perpendicular to a given line through a given point.

Question-68

Prove that two lines that are respectively perpendicular to two intersecting lines intersect each other.

Solution:


Given: Let n and p be two intersecting lines. Let l n and m p.
To Prove: l and m will intersect.
Proof:
Let n and p be two intersecting lines.
Let us assume that l and m do not intersect.
Now, l n [Given]
l || m [By assumption]
m n …………………..(i)

Also, m p, ……………(ii)

From (i) and (ii), we have n || p which is wrong since n and p intersect. [Given]
Our assumption is wrong.
l and m intersect.

Question-69

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

Solution:
    

Given: m and n are parallel to each other.

To prove: ray BD || ray CA.

Proof:

BQ is the normal to the reflecting surface.

BQ n.

Similarly, AP is the normal to the reflecting surface.

AP m.

Since m || n.

AP || BQ.

  QBA = BAP (Alternate angles are equal)

2QBA = 2BAP

     DBA = BAC (The angle of incidence = The angle of reflection)

ray BD || ray CA ( Alternate angles are equal ).

Question-70

(v) If two parallel lines are intersected by a transversal, then the interior angles on the same side of the transversal are equal.

 


Solution:
(i) False. If two parallel lines are cut by a transversal, only then the corresponding angles will be equal.
(ii) True. If two parallel lines are cut by a transversal, then the alternate angles will be equal.
(iii) False. Two lines perpendicular to the same line are parallel to each other.
(iv) True.
(v) False. If two parallel lines are intersected by a transversal, then the interior angles on the same side of the transversal are supplementary.

Question-71

Fill in blanks in each of the following to make the statement true:
(i) If two parallel lines are intersected by a transversal, then each pair of corresponding angles are …………..
(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are ……………
(iii) Two lines perpendicular to the same line are ………….. to each other.
(iv) If a transversal intersects a pair of lines in such a way that a pair of alternate angles are equal, then the lines are ………….
(v) Two lines parallel to the same line are ……………to each other.
(vi) If a transversal intersects a pair of lines in such a way that the sum of interior angles on the same side of the transversal is 180°,
then the lines are ………….

Solution:
(i) equal
(ii) supplementary
(iii) parallel
(iv) parallel
(v) parallel
(vi) parallel

Question-72

One of the angles of a triangle is 65°. Find the remaining two angles, if their difference is 25°.

Solution:
In Δ ABC, let A = 65° ………….(i)

        B - C = 25° ……………………….(ii)

A + B + C = 180° [ Sum of the angles of a Δ is 180° ]
65° + B + C = 180°
      B + C = 180° - 65°
      B + C = 115° …………………….(iii)

Adding (ii) and (iii),
 2 B = 140°
B = = 70°

Substituting in (iii), 70° + C = 115°
C = 115 - 70° = 45°
B = 70° and C = 45°.

Question-73

If the angles of a triangle are in the ratio 2 : 3 : 4, find the three angles.

Solution:
Let the three angles be 2x, 3x and 4x.

2x + 3x + 4x = 180° (The sum of three angles a triangle is 180° )

                 9x = 180°

                   x = = 20°

The three angles are 2 × 20°, 3 × 20°, 4 × 20°

i.e. The three angles are 40°, 60° and 80°.

Question-74

Prove that if one angle of a triangle is equal to the sum of the other two angles, the triangle is a right angled.

Solution:
In Δ ABC, let A = B + C.
But A + B + C = 180° [Sum of the angles of a triangle is 180° ]
A + A = 180°
      2 A = 180°
       A = 90°
Hence the Δ ABC is a right angled triangle.

Question-75

An exterior angle of a triangle is 115° and one of the interior opposite angles is 35°. Find the other two angles.

Solution:

In Δ ABC, let A = 35° and exterior ACD = 115°.
Exterior ACD = A + B
115° = 35° + B
B = 115° - 35° = 80°
In Δ ABC, A + B + C = 180° [Sum of the three angles of a triangle is 180° ]
35° + 80° + C = 180°
C = 180° - (80° +35° )
          = 180° - 115°
          = 65°.
The other two angles are B = 80° and C = 65°.

Question-76

Can a triangle have
     (i) Two right angles?
     (ii) Two obtuse angles?
     (iii) Two acute angles
     (iv) All angles more than 60° ?
     (v) All angles less than 60° ?
     (vi) All angles equal to 60° ?

Solution:
(i) A triangle cannot have two right angles as the sum of the three angles of a triangle is 180°.
If two angles are right angles then their sum will be 180° which contradicts the above
statement.
A triangle cannot have two right angles.

(ii) If two angles are obtuse angles then their sum will be > 180°. But sum of three angles of
a triangle is 180°.
A triangle cannot have two obtuse angles.

(iii) If two angles are acute angles then their sum will be <180o. Since sum of three angles of a triangle is 180° .

A triangle can have two acute angles.

(iv) If all angles are > 60° then sum of three angles will be > 3 × 60° = 180o .Since sum of three angles of a triangle is 180°.

All angles cannot be more than 60°.

(v) If all the angles are < 60° then sum of three angles will be < 3 × 60° = 180o.Since sum of three angles of a triangle is 180°.

All angles cannot be less than 60°. 


(vi) If each angle = 60° then sum of three angles will be = 3 × 60° = 180°.

Since sum of three angles of a triangle is 180°.

All angles can be equal to 60°.

Question-77

Three angles of a quadrilateral are respectively equal to 110°, 40° and 50°. Find the fourth angle.

Solution:
Sum of the four angles of a quadrilateral = (2n - 4) right angles where n = 4
                                                          = (2 × 4 - 4) right angles
                                                          =  4 right angles
                                                          = 4 × 90°
                                                          = 360°
Sum of the three angles = 110° + 50° + 50° = 210°
The fourth angle = 360° - 210° = 150°

Question-78

In figure, the bisectors of ABC and BCA intersect each other at O.Prove that BOC = 90°+ A. 

         

Solution:
Given: OB is the bisector of ABC and OC is the bisector of ACB of Δ ABC.

To prove: 90° +
A
Proof: In a triangle the sum of three angles is 180°  ----------- (1)

In Δ ABC, 
...
A + B + C = 180°
Dividing by 2,
 
A + B + C = = 90°
[
B + C ] = 90° -   A    --------------  (2)
In Δ BOC, 

BOC + OBC + OCB = 180°   (same as (1)
                          
BOC = 180° - OBC - OCB                                  
                      
  ... BOC = 180o( B+ B) ………..……….. (3)
(Since OB, OC are bisector of
ABC, ACB).
Substitute (2) in (3)
                       ...
BOC = 180° - [90° - A ]
                                     = 180° - 90° +
A
                                     = 90° +
A

Question-79

In figure, the bisectors of the exterior angles at B and C formed by producing sides AB and AC of Δ ABC intersect each other at the point O. Prove that BOC=90° - A .

Solution:
Given: ABC is a triangle . Let AB be produced to P and AC be produced to Q. Let the bisectors of  PBC and  QCB meet at O.
To prove: BOC = 90° -
Proof: In Δ ABC,
A + B + C = 180° (Sum of the three angles of a triangle is 180° )
A + B + C = × 180°
= 90° - ……………………….(1)
= 90° -
In Δ BOC,
BOC + CBO + BCO = 180°
BOC = 180° - ( CBO + BCO) ………………(2)
PBC = 180° - ABC (Since ABC and CBP form a linear pair)
Dividing by 2,
PBC = (180° - ABC)
CBO = 90° …………………..(3)
   Ð QCB = 180° - ACB (Since QCB and ACB form a linear pair)
QCB = (180° -  Ð ACB)
BCO = 90° - ………………………(4)
Substituting (3) and (4) in (2),
BOC = 180° - ( CBO + BCO)
           = 180° - (90° - + 90° - )
           = 180° - (180° - - )
           =+…………………………….(5)
But from (1), + = 90° -
Substituting in (5),
BOC = 90° -

Question-80

Side BC of a Δ ABC is produced to both directions. Prove that the sum of the two exterior angles so formed is greater than 180° .

Solution:


Given: ABC is a Δ and side BC is produced in both directions. Let CB be produced to P and BC be produced to Q.
To prove: ABP + ACQ > 180°
Proof:
When CB is produced to P,
Ext. ABP = A + C …………………(1)
(Ext. angle so formed will be equal to sum of the interior opposite angles)
When BC is produced to Q,
Ext. ACQ = A + B ……………….(2)
(Ext. angle so formed will be equal to sum of the interior opposite angles)
Adding (1) and (2), ABP+ ACQ = A+ C+ A+ B
                                               = A + ( A+ B+ C)
                                               = A + 180° [ A + B + C = 180° , since A, B and C are the three angles of a triangle .
∴ ∠ ABP + ACQ > 180°

Question-81

In figure, AP and DP are the bisectors of two adjacent angles A and D of a quadrilateral ABCD. Prove that 2APD = B + C.
       

Solution:
Given: ABCD is a quadrilateral. AP and DP are the bisectors of A and D intersecting at P.
To prove: 2APD =B +C
Proof:
From quadrilateral ABCD, A +B +C +D = 360° [Sum of the four angles of a quadrilateral is 360° ]
        ÐA +D = 360° - ( B + C) ……………………………..(1)
From Δ APD, APD + PAD + ADP = 180° [Sum of the three angles of a triangle is 180° ]
APD ++ = 180° [Since +PAD = , ADP = ]
APD = 180° --
             = 180° -(A +D)
             =
2APD = 360° -[360° - (B +C)]   [from (1)]
              = 360° -360° +B +C
              = B +C.
2APD = B +C.

Question-82

In figure, prove that p||m.
        

Solution:
In Δ BCD, B + C + D = 180° [Sum of the angles of a triangle is 180° ]
 
B + 45° + 35° = 180°
                 B = 180° - (45° +35° )
                           = 180° - 80°
              DBC = 100° ………………………..(i)
      DBE + DBC = 180° [Linear pair axiom]
     Þ DBE + 100° = 180° [From (i)]
               Þ DBE = 180° - 100° = 80°
               DBE = FAB
But these angles form a pair of corresponding angles when p and m are cut by a transversal n, which are equal.
p || m.

Question-83

In figure, side QR of Δ PQR has been produced to S. If P :  Q :  R = 3 : 2 : 1 and RT PR, find TRS.
     

Solution:
Given: In Δ PQR, P :  Q :  R = 3 : 2 : 1.
RT PR, then PRT = 90°
To find: TRS
Proof:
In Δ PQR,
P + Q + R = 180° [Sum of the three angles of a Δ is 180° ]
Ratio of the angles of the Δ PQR = P : Q :  R = 3 : 2 : 1
                                     R =× 180° = 30°
Now, PRQ + PRT + TRS = 180°   [Linear pair axiom]
            30° + 90° + TRS = 180°
                             TRS = 180° - (30° + 90° )
                                          = 180° - 120°
                                          = 60°

Question-84

If two parallel lines are intersected by a transversal, then prove that the bisectors of the interior angles form a rectangle.

Solution:
          

Given:
AB and CD are two parallel lines intersected by a transversal EF at G and H respectively.
GM is the bisector of AGH, GL is the bisector of BGH, HL is the bisector of DHG, HM is the bisector of CHG.
To Prove: GMHL is a rectangle.
Proof:
AB || CD and EF intersects them.
      AGH = GHD [Alternate angles]
AGH = GHD
      1 = 2 …………………(i)
But these angles form a pair of equal alternate angles for lines GM and HL and transversal GH.
GM || HL
Similarly HM || GL.
GLHM is a parallelogram.
AB || CD and EF is the transversal.
BGH + GHD = 180° [ The sum of the interior angles on the same side of the transversal is 180° ]
BGH + GHD = = 90°
              3 + 2 = 90° ……………………….(ii)
In Δ GHL, 3 + 2 + GLH = 180° [Sum of the three angles of a triangle is 180° ]
            90° + GLH = 180°
   GLH = 180° - 90° = 90°
One angle of the parallelogram is a right angle.
Parallelogram GLHM is a rectangle.

Question-85

The side EF, FD and DE of a triangle DEF are produced in order forming three exterior angles DFP, EDQ and FER respectively. Prove that
DFP+ EDQ+ FER=360°

Solution:
        

Given: DEF is a triangle . The sides EF, FD and DE are produced to P, Q, and R respectively.
To prove: DFP + EDQ + FER = 360° Proof:
DFP = D + E………………….(i) [Exterior angle is equal to the sum of the interior opposite angles]
EDQ = E + F…….…………(ii) [Exterior angle is equal to the sum of the interior opposite angles]
FER = F + D ………………….(iii) [Exterior angle is equal to the sum of the interior opposite angles]
Adding (i), (ii) and (iii),
DFP + EDQ + FER = ( D + E) + ( E + F) + ( F + D)
                                  = 2( D + E + F)
                                  = 2 × 180° [Sum of the three angles of a triangle is 180° .]
                                   = 360°
∴∠ DFP + EDQ + FER = 360°

Question-86

Prove that if the arms of an angle are respectively perpendicular to the arms of another angle, then the angles are either equal or supplementary.
(i) If the arms of an angle are respectively perpendicular to the arms of another angle, then the angles will be equal:
(ii) If the arms of an angle are respectively perpendicular to the arms of another angle, then the angles will be supplementary:

Solution:
(i) 


Given: AB and BC are the arms of an ABC. EQ and QD are the arms of another angle.
QEO = 90° , ODB = 90° .
To Prove: OBD = EQO

Proof:

In Δ BOD, OBD + BOD + ODB = 180° [Sum of the three angles of a triangle is 180° ]
OBD + BOD + 90° = 180°
OBD + BOD = 90° …………………(i)
In Δ EQO, EQO + QOE + OEQ = 180° [Sum of the three angles of a triangle is 180° ]
EQO + QOE + 90° = 180°
EQO + QOE = 90° …………………(ii)
From (i) and (ii),
OBD + BOD = EQO + QOE
But, BOD = QOE [Vertically opposite angles]
OBD = EQO
(ii)

Given:
ABC and PQR are two angles whose arms are respectively perpendicular to each other respectively.
AB QR and BC QP.
To Prove: DBE and EQD are supplementary.
Construction: Join BQ.
Proof:
In Δ BDQ, DBQ + BQD + QDB = 180° [Sum of the three angles of a triangle is 180° ]
DBQ + BQD + 90° = 180°
DBQ + BQD = 90° ………………………..(i)
In Δ BQE, EBQ + BQE + BEQ = 180° [Sum of the three angles of a triangle is 180° ]
EBQ + BQE + 90° = 180°
EBQ + BQE = 90° ………………………….(ii)
Adding (i) and (ii),
( DBQ + BQD) + ( EBQ + BQE) = 90° + 90°= 180°                                                      
DBE + EQD = 180°
DBE and EQD are supplementary.

Question-87

In fig, Q > R and M is a point on QR such that PM is the bisector of QPR. If the perpendicular from P on QR meets QR at N, then prove that
MPN = ( Q -  R).
           

Solution:
Given: Δ PQR, PM is the bisector of QPR such that QPM = MPR and PN  ^  QR.
To prove: NPM = ( Q - R)
Proof: In Δ NPM, PNM = 90°
∴ ∠ NPM = 90° - PMN [Sum of the angles of a triangle is 180° ]……….(i)
In Δ PMR, PMN = MPR + PRM (Ex. Angle of a triangle is sum of the interior opposite angles]
But MPR = QPM = [MP is the bisector of QPR]
∴ ∠ PMN = + PRM ………………….(ii)
In Δ PQR, P + Q + R = 180° {Sum of the three angles of a triangle is 180° ]
+ + = = 90°
                        = 90° - ( +)………………..(iii)
Substituting (iii) in (ii),
PMN = 90° - ( + ) + R
          = 90° - - + R
          = 90° - + ……………………..(iv)
Substituting (iv) in (i),
NPM = 90° - [90° -   + ]
          = 90° - 90° + -
          = -
          = ( Q - R)
NPM = ( Q - R)

Question-88

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

               

Solution:
Given: m and n are two perpendicular mirrors. CA is the incident ray and BD is the reflected ray.
To prove: CA || BD.
Construction: Draw normals at A and B, m and n respectively to meet at P.
Proof: AP m and BP n.
But m n (Given)
 Þ AP   BP  
APB = 90° ………………(i)


At A the angle made by the incident ray and normal is CAP will be equal to the angle made by the normal and the reflected ray ie, PAB.
CAP = PAB …………………..(ii)

Similarly ABP = PBD…………………...(iii)

In Δ APB, APB = 90°
  Ð ABP + PAB = 90° …………………………….(iv)

( In a rt. Angled triangle , the sum of the other two angles will be 90° )
CAP + PBD = 90° [from (ii) and (iii)]
\CA || BD.

Question-89

Prove that the sum of the angles of a hexagon is 720° .

Solution:


Given: ABCDEF is a hexagon.
To prove: A + B + C + D + E + F = 720°
Construction: Join AC, AD and AE.
Proof:
In Δ ABC, 1 + B + 5 = 180° [Sum of the three angles of a triangle is 180° ] …….(i)

In Δ ACD, 2 + 6 + 7 = 180° [Sum of the three angles of a triangle is 180° ] …..(ii)

In Δ ADE, 3 + 8 + 9 = 180° [Sum of the three angles of a triangle is 180° ] …….(iii)

In Δ AEF, 4 + 10 + F = 180° [Sum of the three angles of a triangle is 180° ]……..(iv)
Adding (i), (ii), (iii) and (iv), we get,
( 1 + B + 5) + ( 2 + 6 + 7) + ( 3 + 8 + 9) + ( 4 + 10 + F)  = 720°
( 1 + 2 + 3 + 4) + B + ( 5 + 6) + ( 7 + 8) + 9 + 10) + F = 720°
A + B + C + D + E + F = 720°

Question-90

In a ΔABC, B = 45°, C = 55° and bisector of A meets BC at a point D. Find ADB and ADC.

Solution:

Given: In Δ ABC, B = 45°, C = 55°. AD is the bisector of A meets BC at D

To find: ADB and ADC.

Proof: In Δ ABC, A + B + C = 180°
  Þ A + 45° + 55° =180°
A = 180° - 100° = 80°
BAD = CAD = A [AD is the bisector of A]
                         = × 80° = 40°

In Δ ABD, ADB + DBA + BAD = 180° [Sum of the three angles of a triangle is 180° ]
ADB + 45° + 40° =180°
         ADB + 85° = 180°
РADB = 180° - 85° = 95°

Now take angles РADB and ADC. They form a linear pair of angles.
           \ 95° + ADC = 180°
ADC = 180° - 95° = 85°

Question-91

ABCDE is a regular pentagon and bisector of BAE meets CD at M. If bisector of BCD meets AM at P, find CPM.

Solution:


Given: ABCDE is a regular pentagon. The bisector of BAE meets CD at M and bisector of BCD meets AM at P.
To find: The measure of CPM.
Proof: Each angle of a pentagon =

=

=

=

=

= 108°
In quadrilateral ABCM,
BAM + B + C + CMP = (2n - 4) right angles
= (2 × 4 - 4) right angles
= 4 right angles
= 4 × 90°
= 360°

+108° +108° + CMP = 360°
                  270° + CMP = 360°
                           CMP = 360° -270°
                           CMP = 90°

PCM = BCM = × 108° = 54°
In Δ CPM, CPM + PCM + CMP = 180° [Sum of the three angles of a triangle is 180° ]
           CPM + 54° + 90° = 180°
                  CPM + 144° = 180°
                             Þ CPM = 180° - 144° = 36°

Question-92

In figure, bisectors of ∠ B and ∠ D of a quadrilateral ABCD meet CD and AB produced at P and Q respectively. Prove that ∠ P +∠ Q=(∠ ABC + ∠ ADC).
           

Solution:
Given: ABCD is a quadrilateral. The bisectors of ∠ B and ∠ D meet CD and AB produced at P and Q.
To Prove: ∠ P + ∠ Q = (∠ ABC + ∠ ADC).
Proof:
In Δ PBC,
         ∠ P = 180° - (∠ C + ∠ PBC) [ Angle sum property of a triangle]
But ∠ PBC = (Given)
      ∴ ∠ P = 180° - (∠ C+) …………………(i)

|||y, in Δ ADQ,
∠ Q = 180° - (∠ A + )…………………..(ii)

Adding (i) and (ii),
∠ P + ∠ Q = 180° - (∠ C + ) + 180° - (∠ A + )

Adding and subtracting ( + )

∠ P + ∠ Q = 360° - (∠ A + ∠ B + ∠ C + ∠ D) ++

=360 - 360° + +

= +

= +

= (∠ ABC + ∠ ADC)

∴ ∠ P + ∠ Q = (∠ ABC + ∠ ADC)

Question-93

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

(i)An exterior angle of a triangle is less than either of its interior opposite angles.

(ii) Sum of the three angles of triangle is 180° .

(iii) Sum of the four angles of a quadrilateral is three right angles.

(iv) A triangle can have to right angles.

(v) A triangle can have two acute angles.

(vi) A triangle can have two obtuse angles.

(vii) An exterior angle of a triangle is equal to the sum of the two interior opposite angles.


Solution:
(i) F

(ii) T

(iii) F

(iv) F

(v) T

(vi) F

(vii) T

Question-94

Fill in the blanks to make the following statements true:

         (i) Sum of the three angles of a triangle is …………..

        (ii) An exterior angle of a triangle is equal to the two ……….. opposite angles.

        (iii) An exterior angle of a triangle is always …….. than either of the angles.

        (iv) A triangle cannot have more than ………. right angles.

       (v) A triangle cannot have more than ………… obtuse angles.

       (vi) Sum of the angles of a quadrilateral is ……………


Solution:
(i) 180°

(ii) interior

(iii) greater

(iv) one

(v) one

(vi) 360°  





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