NCERT Solutions for Class 9 Maths Chapter 6 - Lines And Angles Exercise 6.2

Exercise: 6.2

1. In the given figure, find the values of x and y and then show that AB $\parallel $CD.

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Ans:

Proof:

It is observed that,

$x + {50^o} = {180^o}$ (Linear pair)

$x = {180^o} - {50^o}$

$x = {130^o}$   ---  (1)

Also, $y = {130^o}$ (vertically opposite angles)

As x and y are alternate interior angles for lines AB and CD and also measures of these angles are equal to each other, therefore, line AB $\parallel $CD.

2. In the given figure, if AB $\parallel $CD, CD $\parallel $ EF and $y:z = 3:7$, find $x$.

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Ans:

Given:

AB $\parallel $ CD, CD $\parallel $ EF and $y:z = 3:7$

Proof:

It is given that AB$\parallel $CD and CD$\parallel $EF.$$

i.e, AB$\parallel $CD$\parallel $EF

Lines parallel to the same line are parallel to each other.

It is observed that

$x = z$(Alternate interior angles)  ---(1)

It is given that

$y:z = 3:7$

Let the common ratio between y and $z$.

$\therefore y = 3a$ and $z = 7a$

Also,

$x + y = {180^o}$

Co-interior angles on the same side of the transversal.

$z + y = {180^o}$

$ \Rightarrow 7a + 3a = {180^o}$

$ \Rightarrow 10a = {180^o}$

$ \Rightarrow a = {\dfrac{{180}}{{10}}^o}$

$ \Rightarrow a = {18^o}$

To find x

$ \Rightarrow x = 7a$

$ \Rightarrow x = 7 \times {18^o}$

$ \Rightarrow x = {126^o}$

The value of $x = {126^o}$.

3. In the given figure, If AB $\parallel $CD, EF $ \bot $CD and $\angle GED = {126^o}$, find $\angle AGE$, $\angle GEF$ and $\angle FGE$.

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Ans:

Given:

AB parallel to CD.

EF perpendicular to CD.

$\angle GED = {126^o}$.

To find:

$\angle AGE$, $\angle GEF$ and $\angle FGE$.

Proof:

It is given that,

AB $\parallel $CD, EF $ \bot $CD. 

$\angle GED = {126^o}$

$\angle GEF + \angle FED = {126^o}$

$\angle GEF + {90^o} = {126^o}$

$\angle GEF = {126^o} - {90^o}$

$\angle GEF = {36^o}$$$

As $\angle AGE$ and $\angle GED$ are alternate interior angles.

$\angle AGE = \angle GED = {126^o}$

$\angle AGE + \angle FGE = {180^o}$

${126^o} + \angle FGE = {180^o}$

$\angle FGE = {180^o} - {126^o}$

$\angle FGE = {54^o}$

Hence $\angle GEF = {36^o}$,$\angle AGE = {126^o}$and $\angle FGE = {54^o}$.

4. In the given figure, if PQ $\parallel $ST, $\angle PQR = {110^o}$ and $\angle RST = {130^o}$, find $\angle QRS$.

(Hint: Draw a line parallel to ST through point R.)

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Ans:

Given:

$\angle PQR = {110^o}$

$\angle RST = {130^o}$

Proof:

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Let us draw a line XY parallel to ST and passing through point R.

$\angle PQR + \angle QRX = {180^o}$.

(Co-interior angles on the same side of transversal QR)

${110^o} + \angle QRX = {180^o}$

$\angle QRX = {180^o} - {110^o}$

$\angle QRX = {70^o}$

Also,

$\angle RST + \angle SRY = {180^o}$

(Co-interior angles on the same side of transversal SR)

${130^o} + \angle SRY = {180^o}$

$\angle SRY = {180^o} - {130^o}$

$\angle SRY = {50^o}$

XY is a straight line, RQ and RS stand on it.

$\angle QRX + \angle QRS + \angle SRY = {180^o}$.

${70^o} + \angle QRS + {50^o} = {180^o}$

$\angle QRS = {180^o} - {70^o} - {50^o}$

$\angle QRS = {180^o} - {120^o}$

$\angle QRS = {60^o}$

The value of $\angle QRS = {60^o}$.

5. In the given figure, if AB$\parallel $CD, $\angle APQ = {50^o}$ and $\angle PRD = {127^o}$,find x and y.

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Ans:

Given:

AB parallel to CD

$\angle APQ = {50^o}$

$\angle PRD = {127^o}$

To find:

x and y

Proof: 

$\angle APR = \angle PRD$(Alternate interior angles)

${50^o} + y = {127^o}$

$y = {127^o} - {50^o}$

$y = {77^o}$

Also,

$\angle APQ = \angle PQR$ (Alternate interior angles)$$

$x = {50^o}$

Hence $x = {50^o}$and $y = {77^o}$.

6. In the 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.

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Ans:

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Proof:

Let us draw BM $ \bot $PQ and CN $ \bot $RS.

PQ is parallel to RS

$\therefore $BM parallel to CN

BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.

$\angle MBC = \angle BCN$(Alternate interior angles)

$\angle 2 = \angle 3$

However,

$\angle 1 = \angle 2$ and $\angle 3 = \angle 4$(By laws of reflection)

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

Also,

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

$\angle ABC = \angle DCB$

However, these are alternate interior angles.

Hence AB parallel to CD

$\therefore AB\parallel CD$.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (Ex 6.2) Exercise 6.2

Opting for the NCERT solutions for Ex 6.2 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 6.2 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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