IMO Maths Olympiad Class 10 Chapter 11 Constructions MCQs with Answers

Answer: (b) Step II

Explanation:
1. Step I is correct: a circle of radius 6 cm and a concentric circle of radius 8 cm are drawn, and B is a point on the larger circle.
2. In Step II the circle on OB as diameter is drawn, so the mid-point must be a new point M of OB with BM = MO = 4 cm. Calling it A is wrong, because A is already a point of the 6 cm circle (OA = 6 cm), so A cannot be the mid-point of OB.
3. Step III is correct: P and Q lie on the circle on OB as diameter, so ∠OPB = 90° and BP, BQ are tangents.
4. So only Step II is incorrect.

Teacher's Note:
1. The circle on the segment joining the centre and the external point as diameter gives the points of contact (angle in a semicircle = 90°).
2. The mid-point of OB = 8 cm lies 4 cm from O, not 6 cm.

Answer: (d) Only Step IV

Explanation:
1. The angle between the tangents is 60°, so the angle between the two radii to the points of contact is 180° − 60° = 120°. Steps I, II and III are therefore correct.
2. The perpendiculars at A and B meet at C and give tangents CA and CB with ∠ACB = 360° − 90° − 90° − 120° = 60°.
3. Step IV says that the tangents are inclined at 120°, but they must be inclined at 60°. So Step IV is wrong.

Teacher's Note:
1. In quadrilateral OACB the angles at A and B are 90°, so ∠AOB + ∠ACB = 180°.
2. Do not confuse the angle between the radii (120°) with the angle between the tangents (60°).

Answer: (a) IV, VI, V, I, III, II

Explanation:
1. First draw the base: Step IV (AB = 4 cm). Then the angle at B: Step VI (∠ABP = 60°).
2. Next draw the line at 3 cm from AB to locate C: Step V, and join CA: Step I. This completes △ABC.
3. Now for the enlarged triangle, extend AB to D (Step III) and finally draw DE || BC (Step II).
4. The order is IV, VI, V, I, III, II.

Teacher's Note:
1. Build the given triangle completely before enlarging it.
2. The ray BP must exist before the line GH can intersect it.

Answer: (a) Step III

Explanation:
1. Step I, II (with the angle at C equal to 180° − (60° + 105°) = 15°), IV, V and VI follow the standard method of constructing a similar triangle with scale factor 3/2.
2. Step III asks for an obtuse angle ∠CBZ on the side opposite to A. To construct a triangle larger than △ABC with ratio 3/2 we need an acute angle ∠CBZ.
3. So Step III is incorrect.

Teacher's Note:
1. The auxiliary ray is always drawn at an acute angle, so that equal parts can be marked easily.
2. Mark the larger of 3 and 2, i.e. 3 equal parts on BZ.

Answer: (c) IV, III, VII, VI, I, V, II

Explanation:
1. First construct △ABC: take BC = 5 cm (IV), draw arcs of 2.3 cm and 2.9 cm to get A (III), and join AB and AC (VII).
2. Then make the acute angle ∠CBE (VI), cut 3 equal parts on BE (I), join B₃C and draw B₂C′ parallel to it (V).
3. Finally draw C′A′ parallel to CA (II).
4. The order is IV, III, VII, VI, I, V, II.

Teacher's Note:
1. Always construct the given triangle first and the similar triangle second.
2. For the ratio 2/3, mark 3 equal parts and join the 3rd point; the line is drawn from the 2nd point.