Rs Aggarwal 2020 2021 for Class 7 Maths Chapter 14 - Properties Of Parallel Lines

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Rs Aggarwal 2020 2021 Solutions for Class 7 Maths Chapter 14 Properties Of Parallel Lines are provided here with simple step-by-step explanations. These solutions for Properties Of Parallel Lines are extremely popular among Class 7 students for Maths Properties Of Parallel Lines Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 7 Maths Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

Page No 177:

Answer:

$Given : l ∥ m t is a transversal . ∠ 5 = 70 ° ∠ 5 = ∠ 3 = 70 ° (alternate interior angles) ∠ 5 + ∠ 8 = 180 ° (linear pair) or 70 ° + ∠ 8 = 180 ° ∠ 8 = 110 ° ∠ 1 = ∠ 3 = 70 ° (vertically opposite angles) ∠ 3 + ∠ 4 = 180 ° (linear pair) or ∠ 4 = 180 - ∠ 3 = 180 - 70 = 110 °$

Page No 178:

Answer:

$Given : l ∥ m t is a transversal . ∠ 1 : ∠ 2 = 5 : 7 Let the angles measure 5 x and 7 x . ∠ 1 + ∠ 2 = 180 ° (linear pair) ∴ 5 x + 7 x = 180 or 12 x = 180 or x = 15 ∴ ∠ 1 = 5 x = 5 (15) = 75 ° and ∠ 2 = 7 x = 7 (15) = 105 ° ∠ 2 + ∠ 3 = 180 ° (linear pair) ∠ 3 = 180 - 105 = 75 ° ∠ 3 + ∠ 6 = 180 (interior angles on the same side of t h e transversal are supplementary) ∠ 6 = 180 - ∠ 3 = 105 ° and ∠ 6 = ∠ 8 = 105 ° (vertically opposite angles) ∴ ∠ 1 = 75 ° ∠ 2 = 105 ° ∠ 3 = 75 ° ∠ 8 = 105 °$

Page No 178:

Answer:

$Given : l ∥ m t is a transversal . Let : ∠ 1 = (2 x - 8) ° ∠ 2 = (3 x - 7) ° We know that the consecutive interior angles are supplementary . ∴ ∠ 1 + ∠ 2 = 180 ° or (2 x - 8) + (3 x - 7) = 180 or 5 x - 15 = 180 or 5 x = 195 or x = 39 ∠ 1 = (2 x - 8) = (2 \times 39 - 8) = 70 ° ∠ 2 = (3 x - 7) = (3 \times 39 - 7) = 110 °$

Page No 178:

Answer:

From the given figure:

$∠ 1 = ∠ 3 = 50 ° (corresponding angles) and ∠ 1 + x ° = 180 ° (linear pair) or x ° = 180 ° - 50 ° = 130 ° or x = 130 ∠ 2 = ∠ 4 = 65 ° (corresponding angles) and ∠ 2 + y ° = 180 ° (linear pair) or y ° = 180 ° - 65 ° = 115 ° or y = 115$

Page No 178:

Answer:

$Given : ∠ B = 65 ° ∠ C = 45 ° DAE ∥ BC The given lines are parallel . ∴ x ° = ∠ B = 65 ° (alternate angles when AB is taken as the transversal) y ° = ∠ C = 45 ° (alternate angles when AC is taken as the transversal) ∴ x = 65 y = 45$

Page No 178:

Answer:

$Given : CE ∥ BA ∠ BAC = 80 °, ∠ ECD = 35 ° (i) ∠ BAC = ∠ ACE = 80 ° (alternate angles with AC as a transversal) (ii) ∠ ACB + ∠ ACD = 180 ° (linear pair) or ∠ ACB + ∠ ACE + ∠ ECD = 180 ° or ∠ ACB + 80 ° + 35 ° = 180 ° or ∠ ACB = 65 ° (iii) In ∆ ABC : ∠ BAC + ∠ ACB + ∠ ABC = 180 ° (angle sum property) 80 ° + 65 ° + ∠ ABC = 180 ° ∠ ABC = 35 °$

Page No 178:

Answer:

$Given : AO ∥ CD OB ∥ CE ∠ AOB = 50 ° ∠ AOD = ∠ CDB = 50 ° (when AO ∥ CD and OB is the transversal) ∠ ECD + ∠ CDB = 180 ° (consecutive interior angles are supplementary, DB ∥ CE and C D is t h e transversal) ∠ ECD = 180 ° - 50 ° = 130 °$

Page No 178:

Answer:

$Given : AB ∥ CD ∠ ABO = 50 ° ∠ CDO = 40 ° Construction : Through O, draw EOF ∥ AB . ∠ ABO = ∠ BOF = 50 ° (alternate angles, when AB ∥ EF and OB is a transversal) ∠ FOD = ∠ ODC = 40 ° (alternate angles, when CD ∥ EF and OD is a transversal) ∠ BOD = ∠ BOF + ∠ FOD ∠ BOD = 50 ° + 40 ° = 90 °$

Page No 178:

Answer:

$Given : AB ∥ CD GL and HM are angle bisectors of ∠ AGH and ∠ GHD, respectively . ∠ AGH = ∠ GHD (alternate angles) or \frac{1}{2} ∠ AGH = \frac{1}{2} ∠ GHD or ∠ LGH = ∠ GHM (given) Therefore, GL ∥ HM as we know that if the angles of any pair of alternate interior angles are equal, then the lines are parallel .$

Page No 179:

Answer:

$Given : AB ∥ CD ∠ ABE = 120 ° ∠ ECD = 100 ° ∠ BEC = x ° Construction : FEG ∥ AB Now, \sin c e A B ∥ F E G and A B ∥ C D, F E G ∥ C D ∴ E F G ∥ A B ∥ C D ∠ ABE = ∠ BEG = 120 ° (alternate angles) or x ° + y ° = 120 ° . . . . (i) ∠ DCE = ∠ CEF = 100 ° (alternate angles) or x ° + z ° = 100 ° . . . . . (ii) Also, x ° + y ° + z ° = 180 ° (FEG is a s traight line) . . . (iii) Adding (i) and (ii) : 2 x ° + y ° + z ° = 220 ° or, x ° + 180 ° = 220 ° (substi tuting (iii)) x ° = 40 ° ∴ x = 40$

Page No 179:

Answer:

$Given : AB ∥ CD AD ∥ BC ∠ 1 + ∠ 2 = 180 ° (AB ∥ CD and AD is the transversal) . . . (i) ∠ 2 + ∠ 3 = 180 ° (AD ∥ BC and AB is the transversal) . . . (ii) From (i) and (ii) : ∠ 1 + ∠ 2 = 180 ° = ∠ 2 + ∠ 3 ∠ 1 = ∠ 3 ∠ ADC = ∠ ABC$

Page No 179:

Answer:

$Given : l ∥ m p ∥ q ∠ 1 = 65 ° ∴ ∠ 1 = ∠ a = 65 ° (vertically opposite angles) ∠ a + ∠ d = 180 ° (consecutive interior angles on the same side of a transversal are supplementary) or ∠ d = 180 ° - 65 ° = 115 ° ∠ c + ∠ d = 180 ° (consecutive interior angles on the same side of a transversal are supplementary) or ∠ c = 180 ° - 115 ° = 65 ° ∠ c + ∠ b = 180 ° (consecutive interior angles on the same side of a transversal are supplementary) or ∠ b = 180 ° - 65 ° = 115 ° ∴ ∠ a = 65 ° ∠ b = 115 ° ∠ c = 65 ° ∠ d = 115 °$

Page No 179:

Answer:

$Given : AB ∥ DC AD ∥ BC ∠ BAC = 35 ° ∠ CAD = 40 ° ∴ ∠ BAC = y = 35 ° (alternate angles when AB ∥ DC) ∠ CAD = x = 40 ° (alternate angles when AD ∥ BC) ∴ x = 40 y = 35$

Page No 179:

Answer:

$Given : AB ∥ CD ∠ BAE = 125 ° ∠ CAB + ∠ BAE = 180 ° or 125 ° + x ° = 180 ° or x = 55 x + z = 180 ° (c onsecutive interior angles on the same side of transversal are s upplementary) z = 180 - x = 180 - 55 = 125 y + x = 180 ° (c onsecutive interior angles on the same side of transversal are s upplementary) y = 180 - x = 180 - 55 = 125$

Page No 179:

Answer:

$(i) ∠ 1 + ∠ 2 = 180 (linear pair) or 130^{°} + ∠ 2 = 180 ° or ∠ 2 = 50 ° \neq 40 ° = ∠ 3 ∴ l ∦ m (ii) ∠ 2 + ∠ 3 = 180 ° (linear pair) 35 ° + ∠ 3 = 180 ° ∠ 3 = 145 ° = 145 ° = ∠ 1 ∴ l ∥ m (iii) ∠ 2 + ∠ 3 = 180 (linear pair) ∠ 3 = 180 ° - 125 ° = 55 ° ∠ 3 = 55 ° \neq 60 ° = ∠ 1 ∴ l ∦ m$

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