Saturday, May 2, 2020

LINES AND ANGLES (LECTURE 7)

LINES AND ANGLES

LESSON  LINES AND ANGLES
DEAR STUDENTS

WELCOME TO THE CLASS!!




FEW REMINDERS:

The text in Red, is to be written in your register

·        
                                                                                                           


1.     Take your  Mathematics Register

2.      Our handwriting reflects a lot about us. It will be awesome if you use good presentation and cursive hand writing
3.      Make a column on the right hand side, if you need to do any rough work
4.      Leave two lines where you finished yesterday’s work and draw a horizontal line
5.      Write today's date on the line after that.
6.      Pending Queries from yesterday (if any).......






7.      Please write the learning outcomes as mentioned below



Today's learning outcomes are  


I will be able to:
1 RECALL THE PROPERTIES OF PARALLEL LINES AND ANGLES FORMED.
2. APPLY THE PROPERTIES LEARNT IN SOLVING QUESTIONS.

Let's revise all the properties done in the previous classes










let's do solved eg 4
Example 4 : In Fig. 6.24, if PQ || RS, ∠ MXQ = 135° and ∠ MYR = 40°, find ∠ XMY .




After the construction the figure will look like 




 Now, AB || PQ and PQ || RS.
so, can we say that AB //RS?
Yes, because  lines parallel to the same line are parallel to one another.

Now, ∠ QXM + ∠  XMB = 180° (AB || PQ, Interior angles on the same side of the transversal XM)



But ∠ QXM = 135°   (given)



So, 135° + ∠ XMB = 180°



Therefore, ∠ XMB = 45° (1)



Now, ∠ BMY = ∠ MYR (AB || RS, Alternate angles)
Therefore, ∠ BMY = 40° (2)


Adding (1) and (2), you get

∠ XMB + ∠ BMY = 45° + 40°
That is, ∠ XMY = 85°


⭐⭐Important
Example 5 : If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel


Given:  A  transversal AD intersects two lines PQ and RS at points B and C respectively. 
 BE is the bisector of ∠ ABQ and  CG is the bisector of ∠ BCS
 and BE || CG.

 To prove that PQ || RS

Look at the figure☟




We are given, BE is the bisector of ∠ ABQ
so, ∠ ABE = ∠ EBQ (SAY ∠1)

Also, CG is the bisector of ∠ BCS


so, ∠ BCG =∠ GCS (SAY ∠ 2)

, BE || CG and AD is the transversal.

Therefore, ∠ ABE = ∠ BCG     (Corresponding angles axiom)  


so,  ∠1  =  ∠2
 therefore 2 ∠1  = 2 ∠2  ( remember the Euclid's Axiom,DOUBLE OF EQUALS ARE EQUAL))


That is, ∠ ABQ = ∠ BCS  


But, they are the corresponding angles formed by transversal AD with PQ and RS; and are equal.





Therefore, PQ || RS
(Converse of corresponding angles axiom)
LET'S START WITH EX 6.2

Q1 In Fig. 6.28, find the values of x and y and then show that AB || CD




REMEMBER !!

WHEN WE HAVE TO PROVE LINES PARALLEL 
TWO PROPERTIES ARE VERY USEFUL
LINEAR PAIR AXIOM & VERTICALLY OPPOSITE ANGLES




We are given, two lines AB and CD intersected by a transversal
To prove: AB // CD

IN THE FIG.  x +50°  =180°  (why?)

🔴 check the important point mentioned in the beginning of solution)
            =>x  =180°  - 50°
            =>x = 130°


also, y =130° (why?)
🔴 check the important point mentioned in the beginning of solution)
which shows x =y

but, x and y form alternate interior angles

so, we get  AB // CD


Q2. In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.




DONT FORGET TO MARK !!
∠1 IN THE FIG.









since y :z = 3 :7
so, let y =3a and z = 7a where a is any constant
since AB//CD,CD//EF

so, AB//EF

therefore,  ∠ x = ∠ z (alternate interior angles)

so, ∠ x  = 7a

since AB//CD  
we get  ∠ x + ∠ y = 180°   (why?)


3a +7a = 180°
10a= 180°
 a =18°

so, x  = 7a = 7x18° = 126°






GO THROUGH SOLVED EG 6 ON PAGE 103   AND NOTE DOWN IN YOUR NOTEBOOK
REVISE YOUR CW


WE END OUR CLASS HERE!!
SEE YOU FOR THE NEXT CLASS!!

Friday, May 1, 2020

LINES AND ANGLES (LECTURE 6)

Ex 6.2 (Q3 to Q6) lines and angles


LINES AND ANGLES
EX.6.2 (Q3 TO 6)

LETS   RECALL  YESTERDAYS LEARNING OUTCOMES .



 CHECK  THE ANSWERS.WHICH ARE GIVEN BELOW . REMEMBER  ALL THESE PROPERTIES OF PARALLEL LINES , WHEN  IT IS INTERSECTED BY  A TRANSVERSAL OR TRANSVERSALS



TODAYS LEARNING OUTCOMES 
I will be able to recall 
  • The properties of a transversal intersecting 2 or more 
      parallels lines 
  • To   prove and  evaluate the unknown angle .
LETS SOLVE AN   IMPORTANT  QUESTION  👇( DO IT IN REGISTER)





LETS SOLVE  ONE MORE   IMPORTANT  QUESTION  👇



practice the  above 👆 questions again ( HOMEWORK)


                                     

Lets solve Q3 of ev.6.2






I hope you understood  Lets try Q4


solution 4
click the link  for solution 6  

you  must have understood .Now write the solution in your 
register. (cw)





ALL SOLUTIONS TO BE DONE IN THE REGISTER




FEW  QUESTIONS FOR HOMEWORK 




EXERCISE  6.2  Q5

QUESTION 1






QUESTION 2 
                      P           D                 C
QUESTION 3

   
Please join the google meet classroom by clicking on the following link. I will be available here from 11:50 - 12:40 am to sort your queries.
Rules:
Keep your mic on mute. You will switch it on only to ask questions or answer the questions when called on to do so. Enter meeting with your own names.


Keep your videos off.

MEETING ID
meet.google.com/nng-szza-rvv


          
THAT S ALL FOR THE DAY 


HAVE A GREAT DAY







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