Class 9th Mathematics Lines and Angles Case Studies


Case Study 1: Road Crossing (Vertically Opposite Angles)

Two roads intersect each other at a point and form angles. At the intersection, ∠AOC = 70°.

Questions:

  1. Vertically opposite angle of ∠AOC is:
    (a) ∠AOB (b) ∠BOD (c) ∠COD (d) none
    Answer: (c)
  2. Value of ∠COD is:
    (a) 70° (b) 110° (c) 90° (d) 100°
    Answer: (a)
  3. ∠AOC + ∠BOC form:
    (a) linear pair (b) vertically opposite (c) supplementary (d) both (a) and (c)
    Answer: (d)
  4. Sum of ∠AOC + ∠COD =
    (a) 70° (b) 140° (c) 180° (d) 90°
    Answer: (c)
  5. Which property is applied?
    (a) Vertically opposite angles are equal (b) Angles in triangle (c) Parallel line rule (d) none
    Answer: (a)

Case Study 2: Railway Tracks (Parallel Lines)

A railway line is crossed by a transversal bridge. The corresponding angles formed are ∠1 = 65°.

Questions:

  1. If lines are parallel, ∠5 = ?
    (a) 115° (b) 65° (c) 45° (d) 75°
    Answer: (b)
  2. Alternate interior angles are:
    (a) equal (b) supplementary (c) complementary (d) none
    Answer: (a)
  3. ∠1 + ∠2 on same side of transversal =
    (a) 90° (b) 120° (c) 180° (d) 270°
    Answer: (c)
  4. Which theorem is used?
    (a) Parallel line theorem (b) Pythagoras (c) Triangle sum (d) none
    Answer: (a)
  5. If ∠1=65°, ∠3=?
    (a) 65° (b) 115° (c) 75° (d) none
    Answer: (b)

Case Study 3: Linear Pair

Two adjacent angles form a straight line. If ∠PQR = 120°, find ∠RQS.

Questions:

  1. ∠PQR + ∠RQS =
    (a) 90° (b) 120° (c) 180° (d) 270°
    Answer: (c)
  2. If ∠PQR=120°, ∠RQS=?
    (a) 50° (b) 60° (c) 70° (d) 80°
    Answer: (b)
  3. Linear pair are always:
    (a) complementary (b) supplementary (c) equal (d) none
    Answer: (b)
  4. Linear pair theorem states:
    (a) sum = 180° (b) sum = 90° (c) equal angles (d) none
    Answer: (a)
  5. ∠PQR and ∠RQS are:
    (a) vertically opposite (b) linear pair (c) alternate interior (d) none
    Answer: (b)

Case Study 4: Triangle Angles

In a triangle, ∠A = 50°, ∠B = 60°.

Questions:

  1. ∠C=?
    (a) 60° (b) 70° (c) 80° (d) 90°
    Answer: (b)
  2. Theorem used:
    (a) Triangle angle sum = 180° (b) Pythagoras (c) Exterior angle theorem (d) none
    Answer: (a)
  3. Exterior angle at C = ?
    (a) 110° (b) 120° (c) 130° (d) 140°
    Answer: (a)
  4. Exterior angle =
    (a) sum of interior opposite (b) difference (c) product (d) none
    Answer: (a)
  5. If ∠A=50°, ∠B=60°, ∠C=?
    Answer: 70°

Case Study 5: Clock Problem

At 3 o’clock, hands of a clock form an angle.

Questions:

  1. Angle between hour and minute hands =
    (a) 60° (b) 90° (c) 75° (d) 120°
    Answer: (b)
  2. Hands of a clock are examples of:
    (a) rays (b) lines (c) line segments (d) none
    Answer: (a)
  3. At 6 o’clock, angle formed =
    (a) 180° (b) 90° (c) 120° (d) 60°
    Answer: (a)
  4. At 12 o’clock, angle formed =
    (a) 0° (b) 180° (c) 90° (d) 60°
    Answer: (a)
  5. At 9 o’clock, angle formed =
    (a) 180° (b) 90° (c) 120° (d) 60°
    Answer: (b)

Case Study 6: Street Lamp

Two streets meet at right angles.

Questions:

  1. Angle formed =
    (a) 45° (b) 90° (c) 120° (d) 180°
    Answer: (b)
  2. Right angle =
    (a) 90° (b) 180° (c) 270° (d) 360°
    Answer: (a)
  3. Complement of 90° =
    (a) 0° (b) 45° (c) 90° (d) none
    Answer: (a)
  4. Supplement of 90° =
    (a) 45° (b) 90° (c) 180° (d) none
    Answer: (c)
  5. Right angle triangle has one angle =
    (a) 45° (b) 90° (c) 60° (d) 30°
    Answer: (b)

Case Study 7: Paper Fold

When a paper is folded diagonally, it creates intersecting lines at the center.

Questions:

  1. Opposite angles formed are:
    (a) equal (b) unequal (c) complementary (d) none
    Answer: (a)
  2. Such angles are called:
    (a) vertically opposite (b) linear pair (c) alternate (d) none
    Answer: (a)
  3. Angle sum around point =
    (a) 90° (b) 180° (c) 270° (d) 360°
    Answer: (d)
  4. If one angle=120°, opposite angle=?
    (a) 60° (b) 120° (c) 90° (d) none
    Answer: (b)
  5. Adjacent angles form:
    (a) linear pair (b) equal pair (c) alternate (d) none
    Answer: (a)

Case Study 8: Exterior Angle

In ΔABC, ∠A=40°, ∠B=60°. Find exterior angle at C.

Questions:

  1. ∠C=?
    (a) 80° (b) 70° (c) 90° (d) 100°
    Answer: (b)
  2. Exterior angle at C=?
    (a) 100° (b) 110° (c) 120° (d) 130°
    Answer: (a)
  3. Exterior angle property states:
    (a) ext = sum of interior opposite (b) ext=diff (c) ext=product (d) none
    Answer: (a)
  4. Here, ∠A+∠B=?
    (a) 100° (b) 110° (c) 120° (d) 130°
    Answer: (a)
  5. Exterior angle theorem is valid for:
    (a) all triangles (b) only isosceles (c) only right angle (d) none
    Answer: (a)

Case Study 9: Z-pattern Rule

Two parallel lines are cut by transversal, forming Z-shaped angles.

Questions:

  1. These angles are:
    (a) corresponding (b) alternate interior (c) vertically opposite (d) none
    Answer: (b)
  2. If one angle=75°, alternate interior=?
    (a) 105° (b) 75° (c) 90° (d) none
    Answer: (b)
  3. If transversal is ⟂, angle=?
    (a) 90° (b) 120° (c) 180° (d) none
    Answer: (a)
  4. Rule is called:
    (a) Z-rule (b) C-rule (c) F-rule (d) none
    Answer: (a)
  5. Alternate angles formed are:
    (a) equal (b) unequal (c) supplementary (d) none
    Answer: (a)

Case Study 10: Polygon Angles

In quadrilateral ABCD, ∠A=80°, ∠B=90°, ∠C=100°, find ∠D.

Questions:

  1. Sum of all angles in quadrilateral =
    (a) 180° (b) 270° (c) 360° (d) none
    Answer: (c)
  2. ∠D=?
    (a) 80° (b) 85° (c) 90° (d) 95°
    Answer: (d)
  3. Property used:
    (a) Sum of interior angles of quadrilateral = 360° (b) exterior angle theorem (c) parallel lines (d) none
    Answer: (a)
  4. In polygon with n sides, sum of angles =
    (a) (n−2)×180° (b) n×180° (c) (n+2)×180° (d) none
    Answer: (a)
  5. For triangle (n=3), sum=
    (a) 120° (b) 180° (c) 360° (d) none
    Answer: (b)

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