Lines and Angles ⇒⇒ Exercise 6.2

Question 1

In the figure, if AB || CD, CD || EF and y : z = 3 : 7, Find x

Solution :

It is given that, y:z = 3:7

Given, AB || CD

and CD || EF

$ Thus, AB || EF $

( ∵ the angles x and z are alternate interior angles and thus are equal )

∴ $ \angle $ x = $ \angle $ $ z .....(i) $

Now, Again AB || CD

( ∵If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary )

∴ $ \angle $ x + $ \angle $ $ y = 180°$

⇒ $ \angle $ z + $ \angle $ $ y = 180° .... (ii)$

( From equation (i) )

Since $ \angle $ y and $ \angle $ z are in the ratio 3 : 7

∴ Let y = 3m and z = 7m

( Substituting value of y and z in equation (ii) )

$⇒ 3m + 7m = 180°$

$⇒ 10m = 180°$

$⇒ m = 18°$

Now, since y = 3m , Substituting value of m

$⇒ y = 3 × 18° $

$⇒ y = 54° $

Now, since z = 7m , Substituting value of m

$⇒ z = 7 × 18° $

$⇒ z = 126° $

$ \angle $ x = $ \angle $ z

(Because $ \angle $ x and $ \angle $ z are alternate angles and thus are equal. From equation (i) )

∴ $ \angle $ x = $ \angle $ z = 126°

Answer, x = 126° .

Question 2

In the figure, If AB||CD, EF ⊥ CD and $ \angle $ GED = 126°, find $ \angle $ AGE, $ \angle $ GEF and $ \angle $ FGE.

Solution :

Given, $ \angle $ GED = 126°

AB || CD

EF perpendicular CD

Then, $ \angle $ AGE =? , $ \angle $ GEF =? , $ \angle $ FGE =?

Given here, AB || CD and cut by a transversal,

( The $ \angle $ GED and $ \angle $ AGE are alternate interior angles and thus are equal )

∴ $ \angle $ GED = $ \angle $ AGE

( Given, $ \angle $ GED = 126° )

⇒ $ \angle $ AGE = 126°

Now, we can see that $ \angle $ FED = 90°

( Given, EF perpendicular CD )

∴ $ \angle $ GEF = $ \angle $ GED - $ \angle $ FED

⇒ $ \angle $ GEF = 126° - 90°

⇒ $ \angle $ GEF = 36°

Now,Since ray GE stands in a straight line AF

( Sum of linear pair is always 180° )

∴ $ \angle $ AGE + $ \angle $ FGE = 180°

( ∵ Proved above, $ \angle $ AGE = 126° )

⇒ 126° + $ \angle $ FGE = 180°

⇒ $ \angle $ FGE = 180° - 126°

⇒ $ \angle $ FGE = 54°

Therefore, $ \angle $ AGE = 126° , $ \angle $ GEF = 36° and $ \angle $ FGE = 54°

Question 3

In the figure, if PQ || ST, $ \angle $ PQR = 110° and $ \angle $ RST = 130°, find $ \angle $ QRS.

Solution :

Given, $ \angle $ PQR = 110° and $ \angle $ RST = 130°

PQ || ST

To prove : $ \angle $ QRS = ?

Construction :Draw a line XY parallel to ST through point R.

Since XY || ST and PQ || ST. So, PQ || XY.

Given here, PQ || XY and cut by a transversal,

( The $ \angle $ PQR and $ \angle $ QRX are internal angles on the one side of transversal are complementary angles )

∴ $ \angle PQR + \angle QRX = 180°$

( Given, $ \angle $ PQR = 110° )

$ ⇒ 110° + \angle QRX = 180° $

$ ⇒ \angle QRX = 180° - 110° $

$ ⇒ \angle QRX = a = 70° $

Now, Given here, XY || ST and cut by a transversal,

( The $ \angle $ RST and $ \angle $ SRY are internal angles on the one side of transversal are complementary angles )

∴ $ \angle RST + \angle SRY = 180° $

( Given, $ \angle $ RST = 130° )

$ ⇒ 130° + \angle SRY = 180° $

$ ⇒ \angle SRY = 180° - 130° $

$ ⇒ \angle SRY = c = 50° $

Now, XRY is a straight line.

∴ $ \angle XRY = 180°. $

∴ $ \angle QRX + \angle QRS + \angle SRY = 180° $

( The Angles made on a straight line )

$ OR ( a + b + c = 180°) $

$ ⇒ 70° + \angle QRS + 50° = 180°$

$ ⇒ \angle QRS = 180° - 70° - 50°$

$ ⇒ \angle QRS = 180° - 120°$

$ ⇒ \angle QRS = b = 60°$

Therefore $ \angle $ QRS = 60°

Question 4

In the figure, if AB||CD, $ \angle $ APQ = 50° and $ \angle $ PRD = 127°, find x and y

Solution :

Given, $ \angle $ APQ = 50° and $ \angle $ PRD = 127°

AB || CD

To prove : $ \angle $ x = ? and $ \angle $ y = ?

Given here, AB || CD and cut by a transversal PQ,

( The $ \angle $ APQ and $ \angle $ PQR are alternate interior angles and thus are equal )

∴ $ \angle $ APQ = $ \angle $ PQR

( Given, $ \angle $ APQ = 50° )

⇒ $ \angle $ PQR = x = 50°

Now, Given here, AB || CD and cut by a transversal PR,

( The $ \angle $ APR and $ \angle $ PRD are alternate interior angles and thus are equal )

∴ $ \angle $ APR = $ \angle $ PRD

⇒ $ \angle $ APQ + $ \angle $ QPR = $ \angle $ PRD

( Given, $ \angle $ PRD = 127° and $ \angle $ APQ = 50° )

⇒ 50° + y = 127°

⇒ y = 127° - 50°

⇒ y = 77°

Thus, x = 50° and y = 77°

Question 5

In figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Proved that AB||CD.

Solution :

Given, PQ and RS are two mirrors placed parallel to each other.

Construction :

Draw perpendicular line BE to PQ and CF perpendicular to RS .

Since PQ and RS parallel to each other, perpendiculars drawn are parallel, BE || CF.

According to laws of reflection,

We know that angle of incidence is equal to angle of reflection.

$ \angle $ 1 = $ \angle $ 2 and $ \angle $ 3 = $ \angle $ 4 ...... (i)

Now, here, BE || CF and cut by a transversal BC,

( The $ \angle $ EBC and $ \angle $ BCF are alternate interior angles and thus are equal )

∴ $ \angle $ EBC = $ \angle $ BCF

$ \angle $ 2 = $ \angle $ 3 ........(ii)

Now, To Prove, AB || CD ,

We need to prove that $ \angle $ ABC = $ \angle $ DCB

( $ \angle $ ABC = $ \angle $ 1 + $ \angle $ 2 and $ \angle $ DCB = $ \angle $ 3 + $ \angle $ 4 )

$ \angle $ 1 + $ \angle $ 2 = $ \angle $ 3 + $ \angle $ 4

From Eq.(i) , we get

$ \angle $ 2 + $ \angle $ 2 = $ \angle $ 3 + $ \angle $ 3

2$ \angle $ 2 = 2$ \angle $ 3

From Eq.(ii), we get

2$ \angle $ 2 = 2$ \angle $ 2

$ \angle $ ABC = $ \angle $ DCB

We know that, if a transversal line intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel.Therefore we can say

AB || CD

Hence Proved.