Study MaterialsCBSE NotesLines and Angles Class 7 Extra Questions Maths Chapter 5

Lines and Angles Class 7 Extra Questions Maths Chapter 5

Lines and Angles Class 7 Extra Questions Maths Chapter 5

Extra Questions for Class 7 Maths Chapter 5 Lines and Angles

Lines and Angles Class 7 Extra Questions Very Short Answer Type

Question 1.
Find the angles which is \(\frac { 1 }{ 5 }\) of its complement.
Solution:
Let the required angle be x°
its complement = (90 – x)°
As per condition, we get
Lines and Angles Class 7 Extra Questions Maths Chapter 5 Q1
Lines and Angles Class 7 Extra Questions Maths Chapter 5 Q1.1

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    Question 2.
    Find the angles which is \(\frac { 2 }{ 3 }\) of its supplement.
    Solution:
    Let the required angle be x°.
    its supplement = (180 – x)°
    As per the condition, we get
    \(\frac { 2 }{ 3 }\) of (180 – x)° = x°
    Lines and Angles Class 7 Extra Questions Maths Chapter 5 Q2

    Question 3.
    Find the value of x in the given figure.
    Lines and Angles Class 7 Extra Questions Maths Chapter 5 Q3
    Solution:
    ∠POR + ∠QOR = 180° (Angles of linear pair)
    ⇒ (2x + 60°) + (3x – 40)° = 180°
    ⇒ 2x + 60 + 3x – 40 = 180°
    ⇒ 5x + 20 = 180°
    ⇒ 5x = 180 – 20 = 160
    ⇒ x = 32
    Thus, the value of x = 32.

    Question 4.
    In the given figure, find the value of y.
    Lines and Angles Class 7 Extra Questions Maths Chapter 5 Q4
    Solution:
    Let the angle opposite to 90° be z.
    z = 90° (Vertically opposite angle)
    3y + z + 30° = 180° (Sum of adjacent angles on a straight line)
    ⇒ 3y + 90° + 30° = 180°
    ⇒ 3y + 120° = 180°
    ⇒ 3y = 180° – 120° = 60°
    ⇒ y = 20°
    Thus the value of y = 20°.

    Question 5.
    Find the supplements of each of the following:
    (i) 30°
    (ii) 79°
    (iii) 179°
    (iv) x°
    (v) \(\frac { 2 }{ 5 }\) of right angle
    Solution:
    (i) Supplement of 30° = 180° – 30° = 150°
    (ii) Supplement of 79° = 180° – 79° = 101°
    (iii) Supplement of 179° = 180° – 179° = 1°
    (iv) Supplement of x° = (180 – x)°
    (v) Supplement of \(\frac { 2 }{ 5 }\) of right angle
    = 180° – \(\frac { 2 }{ 5 }\) × 90° = 180° – 36° = 144°

    Question 6.
    If the angles (4x + 4)° and (6x – 4)° are the supplementary angles, find the value of x.
    Solution:
    (4x + 4)° + (6x – 4)° = 180° (∵ Sum of the supplementary angle is 180°)
    ⇒ 4x + 4 + 6x – 4 = 180°
    ⇒ 10x = 180°
    ⇒ x = 18°
    Thus, x = 18°

    Question 7.
    Find the value of x.
    Lines and Angles Class 7 Extra Questions Maths Chapter 5 Q7
    Solution:
    (6x – 40)° + (5x + 9)° + (3x + 15) ° = 180° (∵ Sum of adjacent angles on straight line)
    ⇒ 6x – 40 + 5x + 9 + 3x + 15 = 180°
    ⇒ 14x – 16 = 180°
    ⇒ 14x = 180 + 16 = 196
    ⇒ x = 14
    Thus, x = 14

    Question 8.
    Find the value of y.
    Lines and Angles Class 7 Extra Questions Maths Chapter 5 Q8
    Solution:
    l || m, and t is a transversal.
    y + 135° = 180° (Sum of interior angles on the same side of transversal is 180°)
    ⇒ y = 180° – 135° = 45°
    Thus, y = 45°

    Lines and Angles Class 7 Extra Questions Short Answer Type

    Question 9.
    Find the value ofy in the following figures:
    Lines and Angles Class 7 Extra Questions Maths Chapter 5 Q9
    Solution:
    (i) y + 15° = 360° (Sum of complete angles round at a point)
    ⇒ y = 360° – 15° = 345°
    Thus, y = 345°
    (ii) (2y + 10)° + 50° + 40° + 130° = 360° (Sum of angles round at a point)
    ⇒ 2y + 10 + 220 = 360
    ⇒ 2y + 230 = 360
    ⇒ 2y = 360 – 230
    ⇒ 2y = 130
    ⇒ y = 65
    Thus, y = 65°
    (iii) y + 90° = 180° (Angles of linear pair)
    ⇒ y = 180° – 90° = 90°
    [40° + 140° = 180°, which shows that l is a straight line]

    Question 10.
    In the following figures, find the lettered angles.
    Lines and Angles Class 7 Extra Questions Maths Chapter 5 Q10
    Solution:
    (i) Let a be represented by ∠1 and ∠2
    ∠a = ∠1 + ∠2
    ∠1 = 35° (Alternate interior angles)
    ∠2 = 55° (Alternate interior angles)
    ∠1 + ∠2 = 35° + 55°
    ∠a = 90°
    Thus, ∠a = 90°

    Question 11.
    In the given figure, prove that AB || CD.
    Lines and Angles Class 7 Extra Questions Maths Chapter 5 Q11
    Solution:
    ∠CEF = 30° + 50° = 80°
    ∠DCE = 80° (Given)
    ∠CEF = ∠DCE
    But these are alternate interior angle.
    CD || EF ……(i)
    Now ∠EAB = 130° (Given)
    ∠AEF = 50° (Given)
    ∠EAB + ∠AEF = 130° + 50° = 180°
    But these are co-interior angles.
    AB || EF …(ii)
    From eq. (i) and (ii), we get
    AB || CD || EF
    Hence, AB || CD
    Co-interior angles/Allied angles: Sum of interior angles on the same side of transversal is 180°.

    Question 12.
    In the given figure l || m. Find the values of a, b and c.
    Lines and Angles Class 7 Extra Questions Maths Chapter 5 Q12
    Solution:
    (i) We have l || m
    ∠b = 40° (Alternate interior angles)
    ∠c = 120° (Alternate interior angles)
    ∠a + ∠b + ∠c = 180° (Sum of adjacent angles on straight angle)
    ⇒ ∠a + 40° + 120° = 180°
    ⇒ ∠a + 160° = 180°
    ⇒ ∠a = 180° – 160° = 20°
    Thus, ∠a = 20°, ∠b = 40° and ∠c = 120°.
    (ii) We have l || m
    ∠a = 45° (Alternate interior angles)
    ∠c = 55° (Alternate interior angles)
    ∠a + ∠b + ∠c = 180° (Sum of adjacent angles on straight line)
    ⇒ 45 + ∠b + 55 = 180°
    ⇒ ∠b + 100 = 180°
    ⇒ ∠b = 180° – 100°
    ⇒ ∠b = 80°

    Question 13.
    In the adjoining figure if x : y : z = 2 : 3 : 4, then find the value of z.
    Lines and Angles Class 7 Extra Questions Maths Chapter 5 Q13
    Solution:
    Let x = 2s°
    y = 3s°
    and z = 4s°
    ∠x + ∠y + ∠z = 180° (Sum of adjacent angles on straight line)
    2s° + 3s° + 4s° = 180°
    ⇒ 9s° = 180°
    ⇒ s° = 20°
    Thus x = 2 × 20° = 40°, y = 3 × 20° = 60° and z = 4 × 20° = 80°

    Question 14.
    In the following figure, find the value of ∠BOC, if points A, O and B are collinear. (NCERT Exemplar)
    Lines and Angles Class 7 Extra Questions Maths Chapter 5 Q14
    Solution:
    We have A, O and B are collinear.
    ∠AOD + ∠DOC + ∠COB = 180° (Sum of adjacent angles on straight line)
    (x – 10)° + (4x – 25)° + (x + 5)° = 180°
    ⇒ x – 10 + 4x – 25 + x + 5 = 180°
    ⇒ 6x – 10 – 25 + 5 = 180°
    ⇒ 6x – 30 = 180°
    ⇒ 6x = 180 + 30 = 210
    ⇒ x = 35
    So, ∠BOC = (x + 5)° = (35 + 5)° = 40°

    Question 15.
    In given figure, PQ, RS and UT are parallel lines.
    (i) If c = 57° and a = \(\frac { c }{ 3 }\), find the value of d.
    (ii) If c = 75° and a = \(\frac { 2 }{ 5 }\)c , find b.
    Lines and Angles Class 7 Extra Questions Maths Chapter 5 Q15
    Solution:
    (i) We have ∠c = 57° and ∠a = \(\frac { \angle c }{ 3 }\)
    ∠a = \(\frac { 57 }{ 3 }\) = 19°
    PQ || UT (given)
    ∠a + ∠b = ∠c (Alternate interior angles)
    19° + ∠b = 57°
    ∠b = 57° – 19° = 38°
    PQ || RS (given)
    ∠b + ∠d = 180° (Co-interior angles)
    38° + ∠d = 180°
    ∠d = 180° – 38° = 142°
    Thus, ∠d = 142°
    (ii) We have ∠c = 75° and ∠a = \(\frac { 2 }{ 5 }\) ∠c
    ∠a = \(\frac { 2 }{ 5 }\) × 75° = 30°
    PQ || UT (given)
    ∠a + ∠b = ∠c
    30° + ∠b = 75°
    ∠b = 75° – 30° = 45°
    Thus, ∠b = 45°

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