Solutions-Class-9-Mathematics-2-Chapter-2-Parallel Lines-Maharashtra Board

Practice set 2.1 :

Question 1.1. In figure line RP || line MS and line DK is their transversal. ∠DHP = 85°. Find the measures of following angles.

(i) ∠RHD (ii) ∠PHG (iii) ∠HGS (iv) ∠MGK

Solution :

(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°

∴ ∠RHD = 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)

∠MG K= 85°

Question 1.2. In figure line p || line q and line l and line m are transversals. Measures of some angles are shown. Hence find the measures of ∠a, ∠b, ∠c, ∠d.

Solution :

∠a + 110° = 180° ... (Angles in a linear pair)

∴ ∠a = 180° - 110°

∴ ∠a = 70° ... (1)

Consider ∠x as shown in the figure.

line p || line q and line l is its transversal.

∴ ∠x ≅ ∠a ... (Corresponding angles)

∴ ∠x = 70° ... [From (1)]

∠b ≥ ∠x ... (Vertically opposite angles)

∴ ∠b = 70°

line p || line q ... (Given)

and line m is the transversal.

∴ ∠c = 115° ... (Corresponding angles)

∠d + 115° = 180° ... (Angles in a linear pair)

∴ ∠d = 180° - 115°

∴ ∠d = 65°

Answers is : ∠a = 70°, ∠b = 70°, ∠c = 115°, ∠d = 65°.

Question 1.3. In figure line l || line m and line n || line p. Find ∠a, ∠b, ∠c from the given measure of an angle.

Solution :

Consider d as shown in the figure.

line l || line m ... (Given)

and line p is the transversal.

∠d = 45° ... (Corresponding angles)

∠a + ∠d=180° ... (Angles in a linear pair)

∴ ∠a + 45° = 180°

∴ ∠a = 180° - 45° .. 4a= 135°

∴ ∠b ≅ ∠a ... (Vertically opposite angles)

v ∠b = 135°

line n || line p ... (Given)

and line m is the transversal.

v ∠c ≅ ∠a ... (Alternate angles)

∴ ∠c = 135°

Answers is : ∠a = 135°, ∠b = 135°, ∠c = 135°.

Question 1.4. In figure sides of ∠PQR and ∠XYZ are parallel to each other. Prove that, ∠PQR ≅ ∠XYZ

Solution :

Let ray XY intersect ray QR at point M such that, X-Y-M and Q-M-R.

Ray YZ || Ray QR .... (Given)

And line MX is their transversal.

∴ ∠XYZ ≅ ∠YMR ... (Corresponding angle)

ray XY || ray QP . ... (Given)

ray MX || ray QP ... (X-Y-M)

And line QR is their transversal.

∴ ∠YMR ≅ ∠PQM ... (Corresponding angle)

∴ ∠PQM ≅ ∠XYZ

∴ ∠PQR ≅ ∠XYZ .... (Q-M-R)

Question 1.5. In figure, line AB || line CD and line PQ is transversal. Measure of one of the angles is given. Hence find the measures of the following angles.

(i) ∠ART (ii) ∠CTQ (iii) ∠DTQ (iv) ∠PRB

Solution :

∠BRT = 105° ... (Given)

(i) ∠ART + ∠BRT = 180° ... (Angles in a linear pair)

∴ ∠ART + 105° = 180°

∴ ∠ART = 180° - 105°

∴ ∠ART = 75°

(ii) line AB || line CD ... (Given)

and line PQ is the transversal.

∴ ∠CTQ ≅ ∠ART ... (Corresponding angles)

∴ ∠CTQ = ∠ART

∴ ∠CTQ = 75°

(iii) ∠CTQ + ∠DTQ = 180° ... (Angles in a linear pair)

∴ 75° + ∠DTQ = 180°

∴ ∠DTQ = 180° - 75°

∴ ∠DTQ = 105°

(iv) ∠PRB + ∠BRT = 180° ... (Angles in a linear pair)

∴ ∠PRB + 105° = 180°

∴ ∠PRB = 180° - 105°

∠PRB = 75°.

Answer is : ∠ART = 75°, ∠CTQ = 75°, ∠DTQ = 105°, ∠PRB = 75°.

Practice set 2.2 :

Question 2.1. In figure, y = 108° and x = 71° Are the lines m and n parallel ? Justify ?

Solution :

We have, x = 71° and y = 108°

x + y = 71° + 108°

∴ x + y = 179°

So, x + y ≠ 180°

∴ The angles x and y are not supplementary.

∴ The angles do not satisfy the interior angles test for parallel lines.

∴ Line m and line n are not parallel lines.

Question 2.2. In figure, if ∠a ≅ ∠b then prove that line l || line m.

Solution :

Consider ∠c as shown in the figure.

∠a ≅ ∠b ... (Given) ... (1)

∠a ≅ ∠c ... (Vertically opposite angles) ... (2)

∴ ∠b ≅ ∠c ... [From (1) and (2)]

∴ line l || line m ... (Corresponding angles test for parallel lines)

Question 2.3. In figure, if ∠a ≅ ∠b and ∠x ≅ ∠y then prove that line l || line n.

Solution :

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 parallel to a third line in the plane, then those two lines are parallel to each other.

∴ line l || line n

Question 2.4. In figure, if ray BA || ray DE, ∠C = 50° and ∠D = 100°. Find the measure of ∠ABC.

(Hint : Draw a line passing through point C and parallel to line AB.)

Solution :

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

ray DE || ray BA ... (Given)

ray BA || line XY ... (Construction)

ray DE || ray BA || line XY ... (1)

∠EDC = 100° ... (Given)

ray DE || line XY ... [From (1)]

and line DC is the transversal.

∴ ∠EDC = ∠DCX ... (Alternate angles)

∴ ∠EDC = ∠DCX

∴ ∠DCX = 100° ... (Given : ∠EDC = 100°)

∠DCB + ∠BCX = ∠DCX ... (Angles addition postulate)

∴ 50° + ∠BCX = 100°

∴ ∠BCX = 100° - 50°

∴ ∠BCX = 50° ... (2)

Ray BA || line XY ... [From (1)]

and line BC is the transversal.

∴ ∠ABC + ∠BCX = 180° ... (Interior angles)

∴ ∠ABC + 50° = 180° ... [From (2)]

∴ ∠ABC = 180° - 50°

∴ ∠ABC = 130°

Answer is : ∠ABC = 130°.

Question 2.5. In figure, ray AE || ray BD, ray AF is the bisector of ∠EAB and ray BC is the bisector of ∠ABD. Prove that line AF || line BC.

Solution :

Proof : Ray AE || ray BD ... (Given)

and line AB is the transversal.

∴ ∠EAB ≅ ∠ABD ... (Alternate angles)

∴ ∠EAB = ∠ABD

Multiplying both the sides by \(\frac{1}{2}\) we get,

\(\frac{1}{2}\)∠EAB = \(\frac{1}{2}\)∠ABD ... (1)

\(\frac{1}{2}\)∠FAB = \(\frac{1}{2}\)∠EAB ... [Ray AF bisects ∠EAB] ... (2)

\(\frac{1}{2}\)∠ABC = \(\frac{1}{2}\)∠ABD ... [Ray BC bisects ∠ABD] ... (3)

∴ ∠FAB = ∠ABC ... [From (1), (2) and (3)]

∴ ∠FAB ≅ ∠ABC

∴ line AF || line BC ... (Alternate angles test for parallel lines)

Question 2.6. 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 of ∠BPQ and ∠PQC respectively. Prove that line AB || line CD.