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About Practical Geometry — Class 6 Olympiad

Construct geometric figures using compass and ruler; solve construction-based reasoning problems. This topic is part of the Olympiad Class 6 mathematics syllabus (chapter: Module 12). On this page you can practice 38 questions across three difficulty levels — 10 easy, 9 medium, and 19 hard — each with a visual step-by-step solution, plus a timed 27-question mastery test with a live leaderboard. Worked examples and sample questions from the topic are below.

Practical Geometry — solved examples for Class 6 Olympiad

Example 1easy

A line segment AB is 8 cm long. If a point P lies on the perpendicular bisector of AB, which of the following statements must be true?
  1. A)PA + PB = 8 cm
  2. B)PA = PB
  3. C)PA = 8 cm
  4. D)PB = 4 cm

Step-by-step solution

  1. A perpendicular bisector is a line that divides a line segment into two equal halves at a 90° angle.
  2. A key property of a perpendicular bisector is that any point lying on it is equidistant from the two endpoints of the line segment.
  3. Therefore, if P lies on the perpendicular bisector of AB, the distance from P to A (PA) must be equal to the distance from P to B (PB).

Answer: PA = PB

Example 2medium

A point P is located outside a line segment AB. A perpendicular is constructed from P to AB, meeting AB at point M. If another point N is chosen on AB such that N ≠ M, what is always true about the distances?
  1. A)A) PN < PM
  2. B)B) PN = PM
  3. C)C) PN > PM
  4. D)D) PN = PM + MN

Step-by-step solution

  1. The perpendicular distance from a point to a line is the shortest distance between them.
  2. Since PM is the perpendicular from P to line AB, PM represents the shortest distance.
  3. For any other point N on AB (where N ≠ M), the segment PN forms the hypotenuse of the right-angled triangle ΔPMN (with the right angle at M).
  4. In a right-angled triangle, the hypotenuse is always longer than either of the other two sides. Thus, PN > PM.

Answer: C) PN > PM

Example 3hard

What is the minimum number of distinct arcs required to construct an angle of 150° using only a compass and a straightedge, starting from a given ray?
  1. A)5
  2. B)6
  3. C)7
  4. D)8

Step-by-step solution

  1. 1. Draw a ray OA and an arc with O as center, intersecting OA at P. (1st arc)
  2. 2. With P as center and the same radius, draw an arc intersecting the first arc at Q (forms 60°). (2nd arc)
  3. 3. With Q as center and the same radius, draw an arc intersecting the first arc at R (forms 120°). (3rd arc)
  4. 4. The angle between the 120° mark (R) and the 180° mark (let's call it S, which is on the initial line extension) is 60°. To get 150°, we need to bisect this 60° angle to add 30° to 120° or subtract 30° from 180°.
  5. 5. With R and S as centers, draw two arcs (using the same radius, greater than half RS) intersecting at a point T. (4th and 5th arcs)
  6. 6. Joining O to T gives 150°. Total distinct arcs: 1 (initial) + 1 (60°) + 1 (120°) + 2 (for bisection) = 5 arcs.

Answer: 5

Practice questions on Practical Geometry

  1. Q1.easy

    To construct an angle of 45° using only a compass and ruler, which of the following sequences of steps is the most efficient and standard method?
    1. A)Construct 60°, then bisect it twice.
    2. B)Construct 90°, then bisect it.
    3. C)Construct 120°, then subtract 75°.
    4. D)Construct 30°, then add 15°.
    Show answer

    Answer: Construct 90°, then bisect it.

    Hint: Think about which basic constructible angle is closest to 45° and how bisection can help you achieve it.

  2. Q2.easy

    Two parallel lines, L and M, are intersected by a transversal T. If one of the interior angles on the same side of the transversal is 70°, what is the measure of the other interior angle on the same side?
    1. A)70°
    2. B)110°
    3. C)20°
    4. D)180°
    Show answer

    Answer: 110°

    Hint: Recall the relationship between interior angles on the same side of a transversal when lines are parallel.

  3. Q3.easy

    Point P lies on the bisector of ∠ABC. If the perpendicular distance from P to the arm AB is 5 cm, what is the perpendicular distance from P to the arm BC?
    1. A)Less than 5 cm
    2. B)Greater than 5 cm
    3. C)Exactly 5 cm
    4. D)Cannot be determined
    Show answer

    Answer: Exactly 5 cm

    Hint: Consider the unique property that all points on an angle bisector possess regarding the arms of the angle.

  4. Q4.medium

    A student wants to construct a line parallel to a given line L passing through an external point P. They draw a transversal through P intersecting L at Q. Which geometric principle must they apply next to correctly draw the parallel line?
    1. A)A) Constructing an angle equal to a corresponding angle at P
    2. B)B) Constructing a perpendicular from P to L
    3. C)C) Bisecting the angle ∠PQL
    4. D)D) Extending the transversal through P
    Show answer

    Answer: A) Constructing an angle equal to a corresponding angle at P

    Hint: Think about the angle properties that define parallel lines when intersected by a transversal.

  5. Q5.medium

    A triangle XYZ needs to be constructed with XY = 6 cm, ∠Y = 75°, and YZ = 8 cm. A student starts by drawing the line segment XY. What is the most logical and correct next step using a compass and straightedge?
    1. A)A) Construct an arc of radius 8 cm from X
    2. B)B) Construct an angle of 75° at point Y
    3. C)C) Construct an arc of radius 8 cm from Y
    4. D)D) Construct an angle of 75° at point X
    Show answer

    Answer: B) Construct an angle of 75° at point Y

    Hint: Consider the Side-Angle-Side (SAS) construction criterion. The angle must be included between the two given sides.

  6. Q6.medium

    A student attempts to construct a triangle ABC where ∠A = 60°, ∠B = 70°, and side AC = 5 cm. Which statement correctly describes the feasibility or method for this construction?
    1. A)A) The third angle ∠C must first be calculated to use ASA.
    2. B)B) It can be constructed directly using the ASA criterion.
    3. C)C) It cannot be constructed as given.
    4. D)D) It requires knowing side AB.
    Show answer

    Answer: A) The third angle ∠C must first be calculated to use ASA.

    Hint: The ASA criterion requires two angles and the *included* side. If the given side is not included, you might need to find the third angle.

  7. Q7.hard

    A line segment XY is 8 cm long. A point P is constructed such that it lies on the perpendicular bisector of XY and is 3 cm away from the midpoint of XY. What is the length of PX?
    1. A)3 cm
    2. B)4 cm
    3. C)4.5 cm
    4. D)5 cm
    Show answer

    Answer: 5 cm

    Hint: Recall the properties of a perpendicular bisector and how it relates to distances from the endpoints. Consider the right-angled triangle formed.

  8. Q8.hard

    An angle ∠ABC is bisected by ray BD. If a point P lies on BD such that the perpendicular distance from P to ray BA is 6 cm, what is the perpendicular distance from P to ray BC?
    1. A)6 cm
    2. B)9 cm
    3. C)12 cm
    4. D)Cannot be determined
    Show answer

    Answer: 6 cm

    Hint: Remember the fundamental property of an angle bisector: any point on the angle bisector is equidistant from the two arms of the angle.

  9. Q9.hard

    You are given a line 'l' and a point 'A' not on 'l'. To find the shortest distance from point A to line 'l', which geometric construction is essential?
    1. A)Drawing a line parallel to 'l' through A.
    2. B)Constructing a perpendicular from A to 'l'.
    3. C)Drawing a line segment connecting A to a point on 'l' at 45°.
    4. D)Extending line 'l' indefinitely.
    Show answer

    Answer: Constructing a perpendicular from A to 'l'.

    Hint: Think about the definition of the shortest distance between a point and a line. What kind of line segment represents this shortest path?

These are 9 of the 38 questions available for Practical Geometry. Start practicing above to unlock visual solutions, AI coaching, and the topic leaderboard — completely free.