6.8 SUMMARY

If a transversal intersects two parallel lines then

( i ) corresponding angles are equal

( ii ) alternate interior angles are equal

( iii ) each pair of interior angles on the same side of the transversal is supplementary

6 ) PQ ∥ RS ∥ MN. Also MP ⊥ PQ

If ∠ QMN = 60*. Find the values of x, y, z

Solution:

R

i ) Consider RS ∥ MN P x M

Here MS becomes the transversal.

∴ y + 60* = 180* S 60*

( because interior angles on the same side of the y

transversal are supplementary )

∴ y = 180 – 60

∴y = 120*

ii ) Now consider PQ ∥ RS and BD is the transversal z = y

( corresponding angles are equal )

∴z = 120*

iii ) Since PQ ∥ RS and RS ∥ MN we have PQ ∥ MN

MP is the transversal

∠ MPQ + ∠ NMP = 180*

( Interior angles on the same side of the transversal

are equal )

90* + x + 60* = 180*

150* + x = 180* x = 180* – 150*

∴x = 300*

7 ) Prove that the sum of the angles of a triangle is 1800.

Solution:

Given:

PQR is a triangle ( 4 ) x ( 5 ) x y z

To prove: ∠ 1 + ∠ 2 + ∠ 3 = 1800 ( 1 )

Const:Draw line XPY

Proof:

Now since XPY is a line we get

∠ 4 + ∠ 1 + ∠ 5 = 1800 – – – – – – – – – – ( 1 )

( Angles on a straight line is 1800 ) ( 2 ) ( 3 )

Line XPY ∥ QR Q R

And PQ is the transversal

∠ 4 = ∠ 2 ( alternate interior angles ) – – – – – – – – – – ( 2 )

( Alternate interior anglesare Z shaped )

Again XPY ∥ QR

And PR is the transversal

∠ 3 = ∠ 5 ( alternate interior angles ) – – – – – – – – – ( 3 )

Substituting the value of ( 2 ) and ( 3 ) in ( 1 )

We get

∠ 4 + ∠ 1 + ∠ 5 = 180*

∠ 2 + ∠ 1 + ∠ 3 = 180* ( ∵ ∠ 4 = ∠ 2 , ∠ 5 = ∠ 3 )

∴ ∠ 1 + ∠ 2 + ∠ 3 = 180*

8 ) In Fig. PQ ∥ ST

∠ QPR = 50* and ∠ RST = 70*, find ∠ SRT.

Solution:

Proof:PQ∥ST and PT is the transversal

∴ ∠ QPR = ∠ RTS ( alternate interior angles )

∠ RTS = 50* ( ∵ ∠ QPR = 50* given )

PQ

Consider ⊿RST50*

∠ RST + ∠ RTS + ∠ SRT = 180*

70* + 50* + ∠ SRT = 180*R

120* + ∠ SRT = 180*

∴ ∠ SRT = 180* – 1200 = 60* 70*

ST

AB

( 9 ) In the fig. if AB⊥AD and AB ∥DE

∠ DBC = 400 and

∠ BCE = 70* find the

values of x and y

Solution:

y 70*

Since AB ∥ DE and BC is the transversalDCE

We get ∠ ABC = ∠ BCE ( alternate interior angles )

x + 40* = 70* ( since ∠ ABC = x + 400 and ∠ BCE = 700 given )

∴ x = 70* – 40*

∴x = 30*

Now consider ∆ ADB

∠ ADB + ∠ DBA + ∠ DAB = 180* ( Angles of a triangle is 1800 )

y + x + ∠ DAB = 1800 – – – – – – – – – – – – ( 1 )

( But x = 30* proved above )

∠ DAB = 900 since AB ⊥AD

Putting these values in ( 1 ) we get

y + 30* + 90* = 180*

y + 120* = 180*

y = 180* – 120*

y = 60*

∴ The values are x = 30*, y = 60* proved