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