LINES AND ANGLES (LECTURE 5)

30TH APRIL
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!!