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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