Lines and Angles: 9 Math

NCERT Exercise 6.2: 9th math

Lines and Angles Class ninth math NCERT Exercise 6.2 Question (1) In figure, find the values of x and y and then show that AB||CD.

Solution

Given,

∠ANM = 50^o

And, ∠CYP = 130^o

Thus, ∠x and ∠y = ?

And to prove that AB || CD.

Since, ∠ANM and ∠x are linear pair, thus their sum will be 180^o

i.e. ∠ANM + ∠x = 180^o

⇒ 50^o + ∠x = 180^o

⇒ ∠x = 180^o – 50^o

⇒ ∠x = 130^o

Now, since ∠COP and ∠NOD are vertically opposite angles, thus they are equal.

i.e. ∠NOD = ∠COP

⇒ ∠y = 130^o

Now, here ∠x = ∠y = 130^o

i.e. Alternate interior angles x and y are equal.

Thus, by the corresponding angles theorm, which says that if a ransversal intersects two lines such that a pair of alternate interior angles is equal, the the two lines are parallel.

Thus, AB || CD Proved

And, ∠x = ∠y = 130^o Answer

Lines and Angles Class ninth math NCERT Exercise 6.2 Question (2) In figure, if AB||CD, CD||EF and y : z = 3 : 7, find x.

Solution

Given, AB || CD and CD||EF

And, y : z = 3 : 7

Thus, x = ?

Here, x and z are alternate internal angles and hence are equal.

i.e. x = z - - - - - (i)

As given y : z = 3 : 7

Thus, let y = 3m and z = 7m

Now, according to theorem based on parallel lines, which says that If a transversal intersects two parallel lines, the each pair of interior angles of the same side of the transversal is supplementary.

Here, x and y are the pair of interior angles of the same side of the transversal and hence are supplementary.

Thus, ∠x + ∠y = 180^o

⇒ ∠z + ∠y = 180^o

[Because ∠x and ∠z are alternate angles and thus are equal. From equation (i)]

⇒ 7m + 3m = 180^o

⇒ 10m = 180^o

`=> m = 180^o/10`

⇒ m = 18^o

Now, z = 7m

Thus after substituting the value of m in above expression we get

z = 7 × 18^o

⇒ y = 126^o = x

Thus, x = 126^o Answer

Lines and Angles Class ninth math NCERT Exercise 6.2 Question (3) In figure, if AB||CD, EF `_|_` CD and ∠GED = 126^o, find ∠AGE, ∠GEF and ∠FGE.

Solution

Given, AB||CD, EF `_|_` CD

And ∠GED = 126^o

Then, ∠AGE, ∠GEF and ∠FGE = ?

Here, ∠GED and ∠AGE are alternate interior angles and hence are equal.

i.e. ∠AGE = ∠GED

⇒ ∠AGE = 126^o

Now, ∠AGE and ∠FGE are linear pair of angles and hence = 180^o

i.e. ∠AGE + ∠FGE = 180^o

⇒ 126^o + ∠FGE = 180^o

⇒ ∠FGE = 180^o – 126^o

⇒ ∠FGE = 54^o

Now, ∠GEF + ∠FED = 126^o

⇒ ∠GEF + 90^o = 126^o

[Because EF `_|_` CD and as given in question ∠GED = 126^o ]

⇒ ∠GEF = 126^o – 90^o

⇒ ∠GEF = 36^o

Thus, ∠AGE = 126^o, ∠GEF = 36^o and ∠FGE = 54^o Answer

Lines and Angles Class ninth math NCERT Exercise 6.2 Question (4) In figure, if PQ||ST, ∠PQR = 110^o and ∠RST = 130^o, find ∠QRS.

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

Solution

Given, PQ||ST

And, ∠PQR = 110^o

And, ∠RST = 130^o

Thus, ∠QRS = ?

Construction A line parallel to ST is drawn through point R.

Since, AB||ST

Thus, ∠PQR and ∠QRB are alternate angles and hence are equal.

i.e. ∠PQR = ∠QRB = 110^o

Similarly, ∠RST and ∠SRA are alternate angles and hence are equal.

i.e. ∠RST = ∠SRA = 130^o

Let ∠QRA = a, ∠QRS = b and ∠SRB = c

Now, a + b + c = 180^o - - - - - - (i)

[Because these angles together form a straight angle]

And, ∠SRA + ∠QRB = 130^o + 110^o

⇒ (a + b) + (b + c) = 240^o

⇒ a + b + b + c = 240^o

⇒ a + 2b + c = 240^o - - - - - (ii)

Now, after subtracting equation (i) from equation (ii) we get

Thus, b = ∠ QRS =60^o

Thus, ∠QRS = 60^o Answer

Alternate Method

Given, PQ||ST

And, ∠PQR = 110^o

And, ∠RST = 130^o

Thus, ∠QRS = ?

Construction A line parallel to ST is drawn through point R.

According to theorem related to parallel line If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.

Here, ∠RST and angle c are interior angle of the same side of transversal, and hence are supplementary.

i.e. ∠RST + ∠c = 180^o

⇒ 130^o + ∠c = 180^o

⇒ ∠c = 180^o – 130^o

⇒ ∠c = 50^o - - - - - (i)

And angle PQR and angle QRB are a pair of alternate angles and hence are equal.

Because one of the theorem related to parallel lines says that if a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.

i.e. ∠PQR = ∠QRB = 110^o

Now, ∠QRS = ∠QRB – ∠c

⇒ ∠QRS = 110^o – 50^o

[From equation (i) ∠c = 50^o]

⇒ ∠QRS = 60^o Answer

Lines and Angles Class ninth math NCERT Exercise 6.2 Question (5) In figure, if AB||CD, ∠APQ = 50^o and ∠PRD = 127^o, find x and y.

Solution

Given, AB||CD

And, ∠APQ = 50^o

And ∠PRD = 127^o

Then, x and y = ?

One of the theorem of parallel lines says that If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.

∠PQR and ∠APQ are pair of alternate interior angles, and hence are equal.

i.e. ∠PQR = ∠APQ = 50^o

⇒ x = 50^o

Considering QR as a transversal

∠APR and ∠PRD are pair of alternate angles, and hence are equal.

i.e. ∠APR = ∠PRD = 127^o

Now, ∠y = ∠APR – ∠APQ

= 127^o – 50^o

⇒ y = 77^o

Thus, x = 50^o and y = 77^o Answer

Lines and Angles Class ninth math NCERT Exercise 6.2 Question (6) 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.

AB is a incident ray and BC is reflected ray.

And when BC is incident ray, then CD is reflected ray.

Thus, to prove AB||CD.

We know that, in the case of a plane mirror, incident ray = reflected ray

Here, `/_ABP =i` is incident ray.

And, `/_CBQ =r` is reflected ray.

Thus, `/_ABP = i=/_CBQ=r`

And when BC is incident to mirror RS,

`/_RCB = i` is incident ray

And, `/_DCS = r` is reflected ray.

Thus, `/_RCB=i=/_DCS=r`

Now, ∠ABP, ∠ABC and ∠CBQ together form a straight angle

Thus, ∠ABP + ∠ABC + ∠CBQ = 180^o

⇒ ∠ABC = 180^o – (∠ABP + ∠CBQ) - - - - - (i)

Similarly, ∠RCB + ∠BCD + ∠DCS = 180^o

[Because these three angles together form a straight angle]

⇒ ∠ABC + ∠BCD + ∠CBQ = 180^o

[Because = ∠RCB = ∠ABC and ∠DCS = ∠CBQ]

⇒ ∠BCD = 180^o Ó (∠ABC + ∠CBQ) - - - - - (ii)

Thus, from equation (i) and equation (ii)

∠ABC = ∠BCD

Now, one of the theorem related to parallel lines says that If a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel.

Here AB and CD are two lines and BC is a transversal.

Now, since ∠ABC and ∠BCD are alternate interior angles and are equal

Thus, AB||CD Proved

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