Maharashtra Board Class 9 Math Part 2 Chapter 2 Parallel Lines Solution

Maharashtra Board Class 9 Math Part 2 Solution Chapter 2 – Parallel Lines

Balbharati Maharashtra Board Class 9 Math Part 2 Solution Chapter 2: Parallel Lines for Marathi/ English Medium Students of Class 9 get here Parallel Lines full Exercise Solution.

Std

Maharashtra Class 9

Subject

Math Part 2 Solution

Chapter

Parallel Lines


 

Practice Set – 2.1

 

(1) In figure 2.5, 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) RHD

We know,

∠RHD + ∠DHP = 180°

= ∠RHD = 180° – 85°

= 95°

 

(ii) PHG

We know,

∠DHP + ∠PHG = 180°

= ∠PHG = 180° – 85°

= 95°

Or ∠RHD = ∠PHG (Vertically opposite angels)

∴ ∠PHG = 95°

 

(iii) HGS

We know,

∠HGS = ∠DHP (corresponding angels)

= 85°

 

(iv) MGK

We know,

∠MGK = ∠HGS (Vertically opposite angles)

= 85°

 

(2) In figure 2.6, 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:

We know,

∠a + 110° = 180 (l is a straight line)

= ∠a = 70°

∴ ∠b = 70°

∠c = 115° (Corresponding angles)

Also, ∠d + 115° = 180°

= ∠d = 65°

 

(3) In figure 2.7, line l || line m and line n || line p. Find a, b, c from the given measure of an angle. 

Solution:

Given, ∠ABC = 45°

Now, ∠PBC = ∠BDE (Corresponding angles)

∠BDE = 45°

∴ ∠a + ∠BDE = 180°

= ∠a = 180° – 45° = 135°

∠a = ∠b (Vertically opposite angles)

∴ ∠b = 135°

∠b = ∠c (Corresponding angles)

∠c = 135°

∴ ∠a = ∠b = ∠c = 135°

 

(4) In figure 2.8, sides of PQR and XYZ are parallel to each other. Prove that, PQR XYZ

Solution:

We expand XY to A and ZY to B.

∵ BZ || QR and XA is transversal which cut QR at C.

∴ ∠XYZ = ∠XCR (Corresponding angles) —— (1)

For XA || PQ and QR transversal,

∠XCR = ∠PQR (Corresponding angles) —— (2)

From (1) & (2) =>

∴ ∠XYZ = ∠PQR

∴ ∠PQR ≅ ∠XYZ

 

(5) In figure 2.9, 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:

(i) We know,

∠ART + ∠TRB = 180°

= ∠ART = 180° – 105° = 75°

 

(ii) ∠ART = ∠CTQ (Corresponding angles)

∠CTQ = 75°

 

(iii) ∠TRB = ∠QTD (Corresponding angles)

= ∠DTQ = 105°

 

(iv) ∠ART = ∠PRB (Vertically opposite angles)

∠PRB = 75°

 

Practice Set – 2.2

 

(1) In figure 2.18, y = 108° and x = 71° Are the lines m and n parallel? Justify?

Solution:

Given, x = 71°

y = 108°

∴ x + y = 108° + 71°

= 179° <180°

∴ l ∦ m as the sum of pair of interior angles on the same side of transversal congruent.

 

(2) In figure 2.19, if a ≅ ∠b then prove that line l || line m. 

Solution:

Given, ∠a ≅ ∠b

∠AKP + ∠AKL = 180°

= ∠AKL + ∠a = 180° = ∠a = 180° – ∠AKL

∠KLD + ∠DLQ = 180° = ∠KLD +∠D = 180°

= ∠b = 180° – ∠KLD

∴ ∠a ≅ ∠b = 180° – ∠AKL ≅ 180 – ∠KLD

= ∠AKL ≅ ∠KLD

Line l || line m

 

(3) In figure 2.20, if a b and x y then prove that line l || line n. 

Solution:

∠a ≅ ∠b

= ∠a = ∠b (Corresponding)

∴ Line l || line m

∠x ≅ ∠y

= ∠x = ∠y (Alternate angles)

∴ Line m || line n

∴ Line l || line m || n = Line l || line n

 

(4) In figure 2.21, if ray BA || ray DE, ÐC = 50° and ÐD = 100°. Find the measure of ∠ABC.  

Solution:

We draws line PQ passing through C

AB || DE || PQ

∵ DE || CQ and DC transversal

∴ 100° + ∠DCQ = 180°

= ∠DCQ = 80°

∴ ∠BCQ = ∠BCD + ∠DCQ

= 80° + 50°

= 130°

∵ AB || PQ and BC transversal

∴ ∠ABC = ∠BCQ (All angle)

= ∠ABC = 130°

 

(5) In figure 2.22, 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:

∵ AE || BD and AB transversal

∠EAB = ∠ABD

= ½ ∠ EAB = ½ ∠ABD

= ∠FAB = ∠ABC (Alternative angles)

∴ FA || BC

 

(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.

Solution:

∴ PR and QS are parallel and EF transversal

∴ ∠RPQ = ∠SQP (Alternative angles) —— (i)

Also, PR and QS are angle bisectors

∴ ∠PRQ = 1/2 ∠BPQ and ∠SQP = ½ ∠CQP

(i) => 1/2 ∠BPQ = 1/2 ∠CQP => ∠BPQ => ∠CQP

Which are alternative angles for AB || CD and EF transversal

∴ AB || CD

 

Practice Set 2

 

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

(A) 0°

(B) 90°

(C) 180°

(D) 360°

Ans: (c) 180°

 

(ii) The number of angles formed by a transversal of two lines is …………

(A) 2

(B) 4

(C) 8

(D) 16

Ans: (C) 8.

 

(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 ………….

(A) 40°

(B) 140°

(C) 50°

(D) 180°

Ans: (A) 40°

 

(iv) In ABC, A = 76°, B = 48°, C =

(A) 66°

(B) 56°

(C) 124°

(D) 28°

Ans: (B) 56°

 

(v) Two parallel lines are intersected by a transversal. If measure of one of the alternate interior angles is 75° then the measure of the other angle is ………….

(A) 105°

(B) 15°

(C) 75°

(D) 45°

Ans: (C) 75°

 

(2) Ray PQ and ray PR are perpendicular to each other. Points B and A are in the interior and exterior of ÐQPR respectively. Ray PB and ray PA are perpendicular to each other. Draw a figure showing all these rays and write –

(i) A pair of complementary angles

(ii) A pair of supplementary angles

(iii) A pair of congruent angles

Solution:

(i) Complementary angles: ∠BPQ and ∠BPR or ∠BPQ and ∠QPA

(ii) Supplementary angles: ∠RPQ and ∠BPA

(iii) Congruent angles: ∠RPQ and ∠BPA

 

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

Solution:

Let, AB || CD be two lines and PQ be a perpendicular of AB and intersect CD at M.

To show: ∠PMD = 90°

∵ PL ⊥ AB

∴ ∠PLB = 90°

Again, ∵ AB || CD and PQ intersect AB, and CD, therefore PQ is a transversal of AB & CD.

∴ ∠PLB = ∠PMD = 90° (Corresponding angles)

∴ ∠PMD = 90°

∴ PQ ⊥ CD

 

(4) In figure 2.24, measures of some angles are shown. Using the measures find the measures of x and y and hence show that line l || line m.

Solution:

∵ PQ is a straight line

∴ ∠ALP + ∠ALQ = 180°

= 130° + x = 180° = x = 50°

Again, CD is a straight line

∴ ∠PMC + ∠CMQ = 180°

= 50° + y = 180° = y = 130°

∴ ∠ALP = ∠CMQ and ∠ALQ = ∠PMC which are alternate angles, therefore line l || m.

 

(5) Line AB || line CD || line EF and line QP is their transversal. If y : z = 3 : 7 then find the measure of x. 

Solution:

∵ Line AB || line CD || line EF and QF and QP is transversal.

∴ ∠x + ∠y = 180° (Pair of interior angles on same side) —– (i)

∠QMC = ∠DMN (vertically opposite angles)

= ∠DMN = y (ii)

∴ ∠DMN + ∠MNF = 180° (Pair of interior angles on same side)

= y + 3 = 180° —— (iv)

Given, y : 3 = 3 : 7

= y = 3/7 z. —– (iii)

Putting in (iv),

3/7 z + z = 180°

= 10z = 7×180°

= z = 126°

∴ y = 3/7 × 126° = 3×18° = 54°

(i) => x + y = 180°

=> x = 180° – 54° = 126°

 

(6) In figure 2.26, if line q || line r, line p is their transversal and if a = 80° find the values of f and g.  

Solution:

a = 80°

∠a = ∠c (Vertically opposite angles)

And ∠c = ∠g (Corresponding angle)

∴ ∠a = ∠g

= g = 80°

P is a straight line

∴ ∠f + ∠g = 180°

= ∠f = 180° – 80° = 100°

∴ f = 100°, g = 80°

 

(7) In figure 2.27, if line AB || line CF and line BC || line ED then prove that ABC = FDE.

Solution:

∵ AB || CF I.e, AB|| PF and BC transversal

∴ ∠ABC = ∠PCR (Corresponding angles)

Also, QR || ES and PF transversal

∠PCR = ∠PDS (Corresponding angle)

Again, ∠PDS = ∠EDF (vertically opposite angle)

Combining all, we get,

∠ABC = ∠PCR = ∠PDS = ∠EDF

= ∠ABC = ∠FDE

 

(8) In figure 2.28, 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 QXRY is a rectangle.

Solution:

∵ AB || CD and PS is transversal

∴ ∠AQR = ∠QRD (Alternate angles)

= ½ ∠AQR = ∠QRD

= ∠XQR = ∠YRQ —– (i) (∵ QX and RY are angle bisectors) which are the alternate angles formed by transversal QR with parallel lines.

∴ QX || RY

Similarly, QY || XR

∴ QYRX is a parallelogram (∵ QX || RY & QY || XR)

Again, ∠BQR + ∠QRD = 180°

= ½ ∠BQR + ½ ∠QRD = ½ × 180°

= ∠YQR + ∠YRQ = 90°

= ∠YQR + ∠XQR = 90° (by (i))

= ∠XQY = 90°

Similarly, ∠XRY = 90°

∴ QXRY is a rectangle.

 

 

Here is your solution of Maharashtra Board Class 9 Math Part 2 Chapter 2 Parallel Lines
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Updated: August 23, 2021 — 2:15 pm

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