Mathematics Part II Solutions Solutions for Class 9 Maths Chapter 2 Parallel Lines are provided here with simple step-by-step explanations. These solutions for Parallel Lines are extremely popular among Class 9 students for Maths Parallel Lines Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part II Solutions Book of Class 9 Maths Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part II Solutions Solutions. All Mathematics Part II Solutions Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 17:

#### Question 1:

^{° }

#### Answer:

(i) Since, RP is a straight line and ray HD stands on it, then

∠RHD + ∠DHP = 180^{∘} (angles in a linear pair)

⇒∠RHD + 85^{∘}^{ }= 180^{∘}

⇒∠RHD = 180^{∘} − 85^{∘}^{ }= 95^{∘}^{ }

(ii) Since the straight lines RP and DK intersect at H, then

∠PHG = ∠RHD (Vertically opposite angles)

⇒∠PHG = 95^{∘}

(iii) Since, the line RP || line MS and line DK is their transversal intersecting them at H and G, then

∠HGS = ∠DHP (Corresponding angles)

⇒∠HGS = 85^{∘}

(iv) Since, the straight lines MS and DK intersect at G, then

∠MGK = ∠HGS (Vertically opposite angles)

⇒∠MGK = 85^{∘}

#### Page No 17:

#### Question 2:

*p*|| line

*q*and line

*l*and line

*m*are transversals. Measures of some angles are shown.

Hence find the measures of $\angle a,\angle b,\angle c,\angle d.$

#### Answer:

Let us mark the points R and S on line *p*, T and U on line *q*, A and B on line *l* and C and D on line *m*.

Suppose the lines *p* and *q* intersect the line *l* at K and L respectively and line *m* at N and M respectively.

Since, AB is a straight line and ray KN stands on it, then

∠NKL + ∠NKA = 180^{∘} (Angles in a linear pair)

⇒110^{∘} + *a* = 180^{∘}

⇒*a* = 180^{∘} − 110^{∘} = 70^{∘}

Since, line *p* || line* q ** *and line *l* is a transversal intersecting them at K and L, then

∠TLB = ∠NKA (Alternate exterior angles)

⇒*b* = *a*

⇒*b* = 70^{∘}

Since, line *p* || line* q ** *and line *m* is a transversal intersecting them at N and M, then

∠CNK = ∠NML (Corresponding angles)

⇒*c* = 115^{∘}

Since, CD is a straight line and ray ML stands on it, then

∠NML + ∠LMD = 180^{∘} (Angles in a linear pair)

⇒115^{∘ }+ *d* = 180^{∘}

⇒*d *= 180^{∘} − 115^{∘} = 65^{∘}

#### Page No 17:

#### Question 3:

In the given figure, line *l* || line *m *and line *n *|| line* p*.

Find $\angle a,\angle b,\angle c$ from the given measure of an angle.

#### Answer:

Let us mark the points R and S on line *n*, T and U on line *p*, A and B on line *l* and C and D on line *m*.

Suppose the lines *n* and *p* intersect the line *l* at K and L respectively and line *m* at M and N respectively.

Since, the line *l* and line *p* intersect at L, then

∠KLN = ∠BLT (Vertically opposite angles)

⇒∠KLN = 45^{∘}

Since, line *l* || line *m* and line *p* is a transversal intersecting them at L and N, then

∠KLN + ∠MNL = 180^{∘ }(Pair of interior angles on the same side of transversal is supplementary)

⇒45^{∘} + ∠*a* = 180^{∘}

⇒∠*a* = 180^{∘} − 45^{∘} = 135^{∘}

Since, the line *m* and line *p* intersect at N, then

∠DNU= ∠MNL (Vertically opposite angles)

⇒∠*b* = ∠*a* = 135^{∘}

Since, line *n* || line *p* and line *m* is a transversal intersecting them at M and N, then

∠NMS = ∠DNU (Corresponding angles)

⇒∠*c* = ∠*b* = 45^{∘}

#### Page No 17:

#### Question 4:

In the given figure, sides of $\angle $PQR and $\angle $XYZ are parallel to each other.

Prove that, $\angle $ PQR $\cong $ $\angle $ XYZ

#### Answer:

We have, XY || PQ and YZ || QR.

Let us produce rays YZ and YX backwards to point Z' and X' respectively such that ZYZ' and XYX' are the straight lines.

Similarly, produce the rays QP and QR to points P' and R' respectively such that PQP' and RQR' are straight lines.

Suppose line ZYZ' intersects the line PQP' at L and line RQR' intersects the line XYX' at K.

Now, line XYX' || line PQP' and line ZYZ' || line RQR'.

Since, line ZYZ' || line RQR' and line XYX' is a transversal intersecting them at Y and K, then

*m*∠XYZ = *m*∠XKR (Corresponding angles) ....(1)

Since, line XYX' || line PQP' and line RQR' is a transversal intersecting them at Q and K, then

*m*∠PQR = *m*∠XKR (Corresponding angles) ....(2)

From (1) and (2), we get

*m*∠XYZ = *m*∠PQR

⇒∠XYZ ≅ ∠PQR

#### Page No 18:

#### Question 5:

(i) ∠ART

(iv) ∠PRB

#### Answer:

(i) Since AB is a straight line and ray RT stands on it, then

∠ART + ∠BRT = 180^{∘} (Angles in a linear pair)

⇒∠ART + 105^{∘} = 180^{∘}

⇒∠ART = 180^{∘} − 105^{∘} = 75^{∘}

(ii) Since line AB || line CD and line PQ is a transversal intersecting them at R and T, then

∠ART = ∠CTQ (corresponding angles)

⇒75^{∘} = ∠CTQ

⇒∠CTQ = 75^{∘}

(iii) Since line AB || line CD and line PQ is a transversal intersecting them at R and T, then

∠BRT = ∠DTQ (corresponding angles)

⇒105^{∘} = ∠DTQ

⇒∠DTQ = 105^{∘}

(iv) Since line AB and line PQ intersect at point R, then

∠PRB = ∠ART (vertically opposite angles)

⇒∠PRB = 75^{∘} (∵ ∠ART = 75^{∘})

#### Page No 21:

#### Question 1:

In the given figure, *y *= 108^{° }and *x* = 71^{° }Are the lines* m* and* n *parallel ? Justify ?

#### Answer:

We have, *x* = 71^{°}and *y *= 108^{°}

Now, *x* + *y* = 71^{∘} + 108^{∘} = 179^{∘}

So, *x* + *y *≠ 180^{∘}

But *x* and *y* are the interior angles formed by a transversal *l* of two lines *m* and *n*.

It is known that, if the sum of the interior angles formed by a transversal of two distinct lines is 180^{∘}, then the two lines are parallel.

∴ line *m* is not parallel to line *n*.

#### Page No 21:

#### Question 2:

In the given figure, if $\angle $a $\cong \angle $b then prove that line *l *|| line *m*.

#### Answer:

Let us mark the points A and B on line *l*, C and D on line *m* and P and Q on line *n*.

Suppose the line *n* intersect line *l* at K and line *m* at L.

Since PQ is a straight line and ray KA stands on it, then

*m*∠AKP + *m*∠AKL = 180^{∘} (Angles in a linear pair)

⇒*m*∠*a* + *m*∠AKL = 180^{∘}

⇒*m*∠*a *= 180^{∘} − *m*∠AKL ....(1)

Since PQ is a straight line and ray LD stands on it, then

*m*∠DLQ + *m*∠DLA = 180^{∘} (Angles in a linear pair)

⇒*m*∠*b* + *m*∠DLA = 180^{∘}

⇒*m*∠*b *= 180^{∘} − *m*∠DLA ....(2)

Since, ∠*a *≅ ∠*b*, then *m*∠*a = m*∠*b*.

∴ from (1) and (2), we get

180^{∘} − *m*∠AKL = 180^{∘} − *m*∠DLA

⇒*m*∠AKL = *m*∠DLA

⇒∠AKL ≅ ∠DLA

It is known that, if a pair of alternate interior angles formed by a transversal of two lines is congruent, then the two lines are parallel.

∴ AB || CD or line *l* || line *m*.

#### Page No 21:

#### Question 3:

In the given figure, if $\angle $*a* $\cong $ $\angle $*b *and $\angle $*x* $\cong $ $\angle $*y*

then prove that line *l* || line *n*.

#### Answer:

We have,

∠*a* ≅ ∠*b*

⇒*m*∠*a* = *m*∠*b*

But ∠*a* and ∠*b *are corresponding angles formed by a transversal *k* of line *m* and line *l*.

∴ line *m* || line *l* (corresponding angles test)

Also we have,

∠*x* ≅ ∠*y*

⇒*m*∠*x* = *m*∠*y*

But ∠*x* and ∠*y *are alternate interior angles formed by a transversal *k* of line *m* and line *n*.

∴ line *m* || line *n* (alternate angles test)

So, we have line *m* || line *l *and line *m* || line *n.*

It is known that, if two lines in a plane are paralle to a third line in the plane, then those two lines are paralle to each other.

∴ line *l * || line *n*

#### Page No 21:

#### Question 4:

^{°}and $\angle $D = 100

^{°}. Find the measure of $\angle $ABC.

#### Answer:

Draw a line XY passing through point C and parallel to AB.

We have, AB || DE and AB || XY.

It is known that, if two lines in a plane are parallel to a third line in the plane, then those two lines are parallel to each other.

∴ DE || XY.

Since DE || XY and DC is a transversal intersecting them at D and C, then

∠EDC + ∠YCD = 180^{∘} (Pairs of interior angles on the same side of transversal are supplementary)

⇒100^{∘} + ∠YCD = 180^{∘}

∠YCD = 180^{∘} − 100^{∘} = 80^{∘}

Since, sum of all the angles on a straight line at a point is 180^{∘}, then

∠XCB + ∠BCD + ∠YCD = 180^{∘}

⇒∠XCB + 50^{∘} + 80^{∘} = 180^{∘}

⇒∠XCB + 130^{∘} = 180^{∘}

⇒∠XCB = 180^{∘} − 130^{∘} = 50^{∘}

Since, AB || XY and BC is a transversal intersecting them at B and C, then

∠ABC + ∠XCB = 180^{∘} (Pairs of interior angles on the same side of transversal are supplementary)

⇒∠ABC + 50^{∘} = 180^{∘}

⇒∠ABC = 180^{∘} − 50^{∘} = 130^{∘}

#### Page No 22:

#### Question 5:

In the given figure, ray AE || ray BD, ray AF is the bisector of $\angle $EAB and ray BC is the bisector of $\angle $ABD.

Prove that line AF || line BC.

#### Answer:

Since, ray AF bisects ∠EAB and ray BC bisects ∠ABD, then

∠EAF = ∠FAB = ∠*x* = $\frac{1}{2}$∠EAB and ∠CBA = ∠DBC = ∠*y* = $\frac{1}{2}$∠ABD

∴ ∠*x* = $\frac{1}{2}$∠EAB and ∠*y* = $\frac{1}{2}$∠ABD ....(1)

Since, ray AE || ray BD and segment AB is a transversal intersecting them at A and B, then

∠EAB = ∠ABD (Alternate interior angles)

On multiplying both sides by $\frac{1}{2}$, we get

$\frac{1}{2}$∠EAB = $\frac{1}{2}$∠ABD

Now, using (1), we get

∠*x* = ∠*y *

But ∠*x* and ∠*y *are alternate interior angles formed by a transversal AB of ray AF and ray BC.

∴ ray AF || ray BC (Alternate angles test)

#### Page No 22:

#### Question 6:

A transversal EF of line AB and line CD intersects the lines at point P and Q respectively. Ray PR and ray QS are parallel and bisectors

$\angle $BPQ and $\angle $PQC respectively.

Prove that line AB || line CD.

#### Answer:

Since ray PR bisects ∠BPQ and ray QS bisects ∠PQC, then

∠RPQ = ∠RPB = $\frac{1}{2}$∠BPQ and ∠SQP = ∠SQC = $\frac{1}{2}$∠PQC

∴ ∠BPQ = 2∠RPQ and ∠PQC = 2∠SQP ....(1)

Since PR || QS and PQ is a transversal intersecting them at P and Q, then

∠RPQ = ∠SQP (Alternate interior angles)

On multiplying both sides by '2', we get

2∠RPQ = 2∠SQP

Now, using (1), we get

∠BPQ = ∠PQC

But ∠BPQ and ∠PQC are alternate interior angles formed by a transversal EF of line AB and line CD.

∴ line AB || line CD (Alternate angles test)

#### Page No 22:

#### Question 1:

(i) If a transversal intersects two parallel lines then the sum of interior angles on the same side of the transversal is ............

^{° }(B) 90

^{° }(C) 180

^{° }(D) 360

^{°}

(iii) A transversal intersects two parallel lines. If the measure of one of the angles is 40

^{∘}then the measure of its corresponding angle is .............

^{° }(B) 140

^{° }(C) 50

^{° }(D) 180

^{°}

^{°},$\angle $B = 48

^{° },$\therefore $ C = ..............

^{°}then the measure of the other angle is .............

^{° }(B) 15

^{° }(C) 75

^{° }(D) 45

^{°}

#### Answer:

(i) It is known that, when two parallel lines are intersected by a transversal, then pairs of interior angles on the same side of transversal are supplementary i.e., sum of the interior angles on the same side of transversal is 180^{∘}.

Hence, the correct answer is option (C).

(ii)

In the above figure, line *n* is a transversal of line *l* and line *m*. The angles formed by the transversal *n* of two lines *l* and *m *are :

*a*, *b*, *c*, *d*, *e* , *f* , *g* and *h*.

∴ the number of angles formed by a transversal of two lines is 8.

Hence, the correct answer is option (C).

(iii)

Suppose line RP || line MS and line DK is the transversal intersecting them at H and G respectively.

Let ∠DHP = 40^{∘}

Now, ∠HGS is the corresponding angle of ∠DHP.

It is known that, corresponding angles formed by a transversal of two parallel lines are of equal measure.

∴ ∠HGS = 40^{∘}

Hence, the correct answer is option (A).

(iv) In ∆ABC,

∠A + ∠B + ∠C = 180^{∘} (Angle sum property of triangle)

⇒76^{∘} + 48^{∘} + ∠C = 180^{∘}

⇒124^{∘} + ∠C = 180^{∘}

⇒∠C = 180^{∘} − 124^{∘} = 56^{∘}

Hence, the correct answer is option (B).

(v)

Suppose line RP || line MS and line DK is the transversal intersecting them at H and G respectively.

Let ∠PHG = 75^{∘}

Now, ∠MGH is the alternate angle to ∠PHG.

It is known that, alternate interior angles formed by a transversal of two parallel lines are of equal measure.

∴ ∠MGH = 75^{∘}

Hence, the correct answer is option (C).

#### Page No 22:

#### Question 2:

(i) A pair of complementary angles (ii) A pair of supplementary angles. (iii) A pair of congruent angles.

#### Answer:

Since, ray PQ ⊥ ray PR, then *m*∠QPR = 90^{∘} and ray PA ⊥ ray PB, then *m*∠APB = 90^{∘}.

(i) Two angles, sum of whose measures is 90^{∘}, are called complementary angles.

Here, *m*∠QPR = 90^{∘}

⇒*m*∠BPQ + *m*∠BPR = 90^{∘}

∴ ∠BPQ and ∠BPR are complemenatry angles.

(ii) Two angles, sum of whose measures is 180^{∘}, are called supplementary angles.

Here, *m*∠APB + *m*∠QPR = 90^{∘} + 90^{∘} = 180^{∘}

∴ ∠APB and ∠QPR are supplemenatry angles.

(iii) Angles with exactly the same measures are called congruent angles.

Here, *m*∠QPR = *m*∠APB = 90^{∘}

∴ ∠APB and ∠QPR are congruent angles.

#### Page No 23:

#### Question 3:

Prove that, if a line is perpendicular to one of the two parallel lines, then it is perpendicular to the other line also.

#### Answer:

Suppose *m* and *n* are two parallel lines and line *l* is their transversal. Suppose line *l* ⊥ line *m*.

Let us mark the points A and B on line *m*, C and D on line *n* and P and Q on line *l*.

Suppose the line *l* intersects line *m* and line *n* at K and L respectively.

Since, line *l* ⊥ line *m*, then ∠PKB = 90^{∘}.

Since, line AB || line CD and transversal PQ intersects them at K and L respectively, then

∠KLD = ∠PKB (Corresponding angles)

⇒∠KLD = 90^{∘}

∴ line *l* ⊥ line *n*.

#### Page No 23:

#### Question 4:

In the given figure, measures of some angles are shown.

Using the measures find the measures of $\angle $x and $\angle $y and hence show that line *l *|| line* m*.

#### Answer:

Suppose *n* is a transversal of the given lines *l* and *m*.

Let us mark the points A and B on line *l*, C and D on line *m* and P and Q on line *n*.

Suppose the line *n* intersects line *l* and line *m* at K and L respectively.

Since PQ is a straight line and ray KA stands on it, then

∠AKL + ∠AKP = 180^{∘} (angles in a linear pair)

⇒∠*x* + 130^{∘} = 180^{∘}

⇒∠*x* = 180^{∘} − 130^{∘} = 50^{∘}

Since CD is a straight line and ray LK stands on it, then

∠KLC + ∠KLD = 180^{∘} (angles in a linear pair)

⇒∠*y* + 50^{∘} = 180^{∘}

⇒∠*y* = 180^{∘} − 50^{∘} = 130^{∘}

Now, ∠*x + *∠*y = *50^{∘} + 130^{∘} = 180^{∘}

But ∠*x *and* *∠*y *are interior angles formed by a transversal *n* of line *l* and line *m*.

It is known that, if the sum of the interior angles formed by a transversal of two distinct lines is 180^{∘}, then the lines are parallel.

∴ line *l* || line *m*.

#### Page No 23:

#### Question 5:

Line AB || line CD || line EF and line QP is their transversal. If *y *: *z* = 3 : 7 then find the measure of $\angle $*x.*

#### Answer:

Suppose the transversal QP intersects line AB, line CD and line EF at K, L and M respectively.

Since line CD and line QP intersect at L, then

∠DLM = ∠CLK (Vertically opposite angles)

⇒∠DLM = ∠*y*

Since line CD || line EF and QP is a transversal intersecting them at L and M, then

∠DLM + ∠FML = 180^{∘} (Pair of interior angles on the same side of transversal is supplementary)

⇒∠*y* + ∠*z* = 180^{∘} ....(1)

We have, *y* : *z* = 3 : 7. Let *y* = 3*k* and *z* = 7*k*

Now, from (1), we get

3*k + *7*k = *180^{∘}

⇒10*k* = 180^{∘}

⇒*k* = 18^{∘}

∴ ∠*y* = 3*k* = 3 × 18^{∘} = 54^{∘} and ∠*z* = 7*k* = 7 × 18^{∘} = 126^{∘}

Since line AB || line CD and QP is a transversal intersecting them at K and L, then

∠AKL + ∠CLK = 180^{∘} (Pair of interior angles on the same side of transversal is supplementary)

⇒∠*x* + ∠*y* = 180^{∘}

⇒∠*x* + 54^{∘} = 180^{∘}

⇒∠*x* = 180^{∘} − 54^{∘} = 126^{∘}

So, measure of ∠*x *is 126^{∘}.

#### Page No 23:

#### Question 6:

In the given figure, if line *q* || line* r*, line *p* is their transversal and if* a *= 80^{°} find the values of *f *and *g*.

#### Answer:

Since line *q* || line *r* and line *p* is their transversal, then

∠*g* = ∠*a* (Alternate exterior angles)

⇒∠*g *= 80^{∘}

Since ∠*g *and ∠*f* are forming a linear pair, then

∠*f + *∠*g = *180^{∘} (Angles in a linear pair are supplementary)

⇒∠*f * = 180^{∘} − ∠*g*

⇒∠*f * = 180^{∘} − 80^{∘} = 100^{∘}

Hence, the values of *f* and *g* are 100^{∘} and 80^{∘} respectively.

#### Page No 23:

#### Question 7:

In the given figure, if line AB || line CF and line BC || line ED

then prove that $\angle $ABC = $\angle $FDE.

#### Answer:

Let us mark the points M and N on line BC and point G on line ED.

Since line AB || line PF (or line CF) and line MN is their transversal, then

∠PCN = ∠ABC (Corresponding angles) ....(1)

Since line MN || line EG and line PF is their transversal, then

∠PCN = ∠CDG (Corresponding angles) ....(2)

Since EG and PF are straight lines intersecting at D, then

∠CDG = ∠FDE (Vertically opposite angles) ....(3)

∴ from (2) and (3), we get

∠PCN = ∠FDE ....(4)

∴ from (1) and (4), we get

∠ABC = ∠FDE

#### Page No 23:

#### Question 8:

In the given figure, line PS is a transversal of parallel line AB and line CD. If Ray QX, ray QY, ray RX, ray RY are angle bisectors, then prove that $\overline{){123}}$ QXRY is a rectangle.

#### Answer:

Given : AB and CD are two parallel lines which are cut by a transversal PS at the points Q and R respectively. The bisectors of the interior angles intersect at points X and Y.

To prove : Quadrilateral QXRY is a rectangle.

Proof : Since AB || CD and PS is a transversal, then

∠AQR = ∠DRQ (Alternate interior angles)

⇒$\frac{1}{2}$∠AQR = $\frac{1}{2}$∠DRQ ....(1)

Since QX bisects ∠AQR and RY bisects ∠DRQ, then

∠XQR = $\frac{1}{2}$∠AQR and ∠YRQ = $\frac{1}{2}$∠DRQ

∴ from (1), we get

∠XQR = ∠YRQ

But ∠XQR and ∠YRQ are alternate interior angles formed by the transversal QR with QX and RY respectively.

∴ QX || RY (Alternate angles test)

Similarly, we have RX || QY.

Hence, in quadrilateral QXRY, we have QX || RY and RX || QY.

It is known that, a quadrilateral is a parallelogram if kits opposite sides are parallel.

∴ QXRY is a parallelogram.

Since sum of the interior angles on the same side of transversal is 180^{∘}, then

∠BQR + ∠DRQ = 180^{∘}

⇒$\frac{1}{2}$∠BQR + $\frac{1}{2}$∠DRQ = 90^{∘} ....(2)

Since QY bisects ∠BQR and RY bisects ∠DRQ, then

∠YQR = $\frac{1}{2}$∠BQR and ∠YRQ = $\frac{1}{2}$∠DRQ

∴ from (2), we get

∠YQR + ∠YRQ = 90^{∘} ....(3)

In ∆QRY, we have

∠YQR + ∠YRQ + ∠QYR = 180^{∘} (Angle sum property of triangle)

⇒90^{∘} + ∠QYR = 180^{∘} [Using (3)]

⇒∠QYR = 180^{∘} − 90^{∘} = 90^{∘}

Since QXRY is a parallelogram, then

∠QXR = ∠QYR (Opposite angles of ||^{gm} are equal)

⇒∠QXR = 90^{∘} (∵ ∠QYR = 90^{∘})

Since adjacent angles in a parallelogram are supplementary, then

∠QXR + ∠XRY = 180^{∘}

⇒90^{∘} + ∠XRY = 180^{∘} (∵ ∠QXR = 90^{∘})

⇒∠XRY = 180^{∘} − 90^{∘} = 90^{∘}

Also, ∠XQY = ∠XRY = 90^{∘} (Opposite angles of ||^{gm} are equal)

Thus, QXRY is a parallelogram in which all the interior angles are right angles.

It is known that, a rectangle is a ||^{gm} in which each angle is a right angle.

Hence, $\square $QXRY is a rectangle.

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