# Class 9 NCERT Solutions – Chapter 6 Lines And Angles – Exercise 6.2

### Question 1. In given figure, find the values of x and y and then show that AB || CD.

Solution:

After given names to the remaining vertices we get,

Now, Given âˆ AEP = 50Â°, âˆ CFQ = 130Â°

=> âˆ EFD = âˆ CFQ  [vertically opposite angles are equal]

=> y = 130Â°   [Given âˆ CFQ = 130Â°]

=> y = 130Â°  —eq(i)

Now, PQ is taking as straight line so, sum of all angles made on it is 180Â°

=> âˆ AEP  + âˆ AEQ = 180Â°

=> 50Â° + x = 180Â°

=> x = 180Â° – 50Â° = 130Â°

=> x = 130Â° –eq(ii)

Now, from eq(i) and eq(ii) We conclude that x = y

As they are pair of alternate interior angles

So, AB || CD

Hence proved!!!

### Question 2. In given figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Solution:

Given AB || CD and CD || EF  y : z = 3 : 7

=> AB || CD || EF

=> AB || EF

So, x=z [alternate interior angles] –eq(i)

Again AB || CD

=> x + y = 180Â° [Co-interior angles]

=> z + y = 180Â°  –eq(ii) [from eq(i)]

But given that y : z = 3 : 7

=> z = (7/3) y = (7/3)(180Â° – z) [from eq(ii)]

=> 10z = 7 * 180Â°

=> z = (7 * 180Â°)/10 =126Â°

=> z  = 126Â° –eq(iii)

from eq(i) and eq(iii) we have

=> x = 126Â°

### Question 3. In given figure, if AB || CD, EF âŠ¥ CD and âˆ GED = 126Â°, find âˆ AGE, âˆ GEF and âˆ FGE.

Solution:

Given AB || CD, EF âŠ¥ CD, âˆ GED = 126Â° and âˆ FED =90Â°

=> âˆ GED =  âˆ GEF + âˆ FED

=> 126Â° = âˆ GEF + 90Â°  [Given]

=> âˆ GEF = 36Â°

As, AB || CD and GE is a transversal

So, âˆ FGE + âˆ GED = 180Â° [ sum of Co- interior angles is 180 ]

=>  âˆ FGE + 126Â° = 180Â° [ Given ]

=>  âˆ FGE = 54Â°

As, AB || CD and GE is a transversal

So, âˆ AGE = âˆ GED [ alternate angles are equal ]

=> âˆ AGE = 126Â°

### Question 4. In given figure, if PQ || ST, âˆ PQR = 110Â° and âˆ RST = 130Â°, find âˆ QRS.

[Hint: Draw a line parallel to ST through point R.]

Solution:

Firstly we have drawn a line EF parallel to ST (EF || ST)

Since, PQ || ST [Given] and EF || ST [ Construction ]

So, PQ || EF and QR is a transversal

=> âˆ PQR = âˆ QRF [ Alternate interior angles ]

=> âˆ QRF = 110Â°   [Given âˆ PQR =110Â° ]

=> âˆ QRF = âˆ QRS + âˆ SRF

=> âˆ QRS + âˆ SRF = 110Â°  –eq(i)

Again ST || EF and RS is a transversal

=>âˆ RST + âˆ SRF = 180Â° [ sum of  Co-interior angles is 180Â° ]

=>130Â° + âˆ SRF =180Â° [Given]

=>âˆ SRF =50Â°

Now , from eq(i)

=> âˆ QRS + âˆ SRF = 110Â°

=> âˆ QRS +50 = 110Â°

=> Thus, âˆ QRS = 60Â°

### Question 5. In Fig. 6.32, if AB || CD, âˆ  APQ = 50Â° and âˆ  PRD = 127Â°, find x and y.

Solution:

Given  AB || CD and PQ is a transversal

=> âˆ APQ = âˆ PQR [ Alternate interior angles ]

=> x= 50Â° [ Given âˆ APQ = 50Â° ]

=> x = 50Â°

Again, AB || CD and PR is a transversal

=>âˆ APR = âˆ PRD [ Alternate interior angles ]

=> âˆ APR = 127Â° [ Given âˆ PRD = 127Â° ]

=> 50Â° + y =127Â° [Given âˆ APQ = 50Â° ]

=> y =127Â° – 50Â° = 77Â°

=>  y = 77Â°

Thus,  x = 50Â° and y = 77Â°

### Question 6. In given 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. Prove that AB || CD.

Solution:

Draw ray BL âŠ¥ PQ and CM âŠ¥ RS

Since, PQ || RS => BL || CM

=>[ So, BL || PQ and CM || RS ]

Now, BL || CM and BC is a transversal

=> âˆ LBC = âˆ MCB –eq(i) [ Alternate interior angles ]

Since, angle of incidence = angle of reflection

=>  âˆ ABL = âˆ LBC  and  âˆ MCB = âˆ MCD

=> âˆ ABL = âˆ MCD –eq(ii) [By eq(i)]

=>Adding eq(i) and eq(ii) we get

=>  âˆ LBC + âˆ ABL = âˆ MCB + âˆ MCD

=> âˆ ABC = âˆ BCD

i.e, a pair of alternate angles are equal

Thus, AB || CD

Hence, Proved !!!

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