CBSE Class 9 Mathematics Constructions MCQs Set 12

Practice MCQs for Class 9 Mathematics Chapter 11 Constructions

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Multiple Choice Questions

Question 1: To draw the perpendicular bisector of a line segment AB, the arcs drawn from A and from B must have a radius
(a) more than half of AB
(b) equal to half of AB
(c) less than half of AB
(d) exactly equal to AB only
Show Answer & Explanation

Answer: (a) more than half of AB

Explanation:
1. The arcs from A and B must cross at points P and Q on both sides of AB.
2. This happens only when the common radius is more than AB/2.

Teacher's Note:
1. With radius exactly AB/2 the arcs just touch at the midpoint.

Question 2: In the construction of the perpendicular bisector PQ of AB, triangles APQ and BPQ are proved congruent using the criterion
(a) SAS
(b) SSS
(c) ASA
(d) RHS
Show Answer & Explanation

Answer: (b) SSS

Explanation:
1. AP = BP and AQ = BQ by construction (equal radii) and PQ is common.
2. So △APQ ≅ △BPQ by SSS.

Teacher's Note:
1. Then SAS in △APO and △BPO gives O as the midpoint and PQ ⟂ AB.

Question 3: To bisect ∠ABC, an arc centred at B cuts BA at D and BC at E. Arcs of equal radius from D and E meet at F. The radius of these arcs must be
(a) equal to BD
(b) less than half of DE
(c) more than half of DE
(d) equal to DE only
Show Answer & Explanation

Answer: (c) more than half of DE

Explanation:
1. The arcs from D and E intersect only if the radius exceeds DE/2.
2. Then BF is the bisector because △BDF ≅ △BEF (SSS).

Teacher's Note:
1. Same reasoning as in the perpendicular bisector.

Question 4: To construct an angle of 60°, an arc is drawn from B cutting BC at D. A second arc of the same radius from D cuts the first arc at E. Then ∠EBC = 60° because
(a) BD = BE only
(b) DE is a diameter
(c) ∠BDE = 90°
(d) △BDE is equilateral
Show Answer & Explanation

Answer: (d) △BDE is equilateral

Explanation:
1. BD = BE (radii of the first arc) and DE = BD (same radius).
2. All three sides are equal, so each angle of △BDE is 60°.

Teacher's Note:
1. The compass opening is never changed.

Question 5: The angle of 120° is constructed by marking two successive arcs of the same radius on the first arc. The arc is cut at E (from D) and F (from E). Then ∠FBC =
(a) 120°
(b) 60°
(c) 90°
(d) 150°
Show Answer & Explanation

Answer: (a) 120°

Explanation:
1. Each arc step adds 60° at the centre: ∠DBE = 60° and ∠EBF = 60°.
2. So ∠FBC = 120°.

Teacher's Note:
1. Each cut off equals the radius, which gives 60° at the centre.

Question 6: The angle of 105° is obtained by bisecting the angle between the rays making angles of
(a) 60° and 150°
(b) 90° and 120° with BC
(c) 90° and 150°
(d) 30° and 120°
Show Answer & Explanation

Answer: (b) 90° and 120° with BC

Explanation:
1. The bisector of the angle between 90° and 120° makes (90° + 120°) ÷ 2 = 105° with BC.
2. The 90° and 120° angles are already constructible.

Teacher's Note:
1. Bisector angle = average of the two angles.

Question 7: The angle of 135° is obtained by bisecting the angle between the rays making angles of
(a) 90° and 150°
(b) 60° and 120°
(c) 120° and 150° with BC
(d) 150° and 180° only
Show Answer & Explanation

Answer: (c) 120° and 150° with BC

Explanation:
1. Average of 120° and 150° is 135°.
2. Both 120° and 150° are constructible.

Teacher's Note:
1. Also 135° = 90° + 45°.

Question 8: Which of the following angles cannot be constructed directly with a ruler and compass by repeated bisecting and combining 60° and 90° angles?
(a) 75°
(b) 15°
(c) 165°
(d) 55°
Show Answer & Explanation

Answer: (d) 55°

Explanation:
1. 15° = 30° ÷ 2, 75° = (60° + 90°) ÷ 2, 165° = (150° + 180°) ÷ 2 are all possible.
2. 55° cannot be obtained by bisecting and adding these standard angles.

Teacher's Note:
1. Standard constructible angles are multiples of 7.5° in this method.

Question 9: To construct an equilateral triangle of side 5.4 cm without using a protractor, after drawing BC = 5.4 cm one draws arcs from B and C of radius
(a) 5.4 cm
(b) 2.7 cm
(c) 10.8 cm
(d) any radius
Show Answer & Explanation

Answer: (a) 5.4 cm

Explanation:
1. All sides of an equilateral triangle are equal.
2. The arcs of radius 5.4 cm from B and C meet at A with AB = AC = BC.

Teacher's Note:
1. Alternatively construct two 60° angles at B and C.

Question 10: To construct △ABC with base BC = 3 cm, ∠B = 60° and AB + AC = 8 cm, a point D is marked on the ray from B making 60° with BC such that BD =
(a) 3 cm
(b) 8 cm
(c) 5 cm
(d) 11 cm
Show Answer & Explanation

Answer: (b) 8 cm

Explanation:
1. D is chosen so that BD = AB + AC = 8 cm.
2. Then A is where the perpendicular bisector of CD meets BD.

Teacher's Note:
1. A lies on the bisector, so AC = AD.

Question 11: In constructing a triangle from its base, a base angle and the sum of the other two sides, A lies on the perpendicular bisector of CD. Therefore
(a) AB = AD
(b) AB = AC always
(c) AC = AD
(d) BD = BC
Show Answer & Explanation

Answer: (c) AC = AD

Explanation:
1. Every point on the perpendicular bisector of a segment is equidistant from its ends.
2. So AC = AD, giving BD = BA + AD = AB + AC.

Teacher's Note:
1. That is why BD is the required sum.

Question 12: A triangle with base 5 cm, one base angle 45° and sum of the other two sides 4 cm cannot be constructed because
(a) the base angle must be 60°
(b) the base must be less than 4 cm
(c) the sum must equal the base
(d) the sum of two sides must be greater than the third side
Show Answer & Explanation

Answer: (d) the sum of two sides must be greater than the third side

Explanation:
1. In any triangle AB + AC > BC.
2. Here AB + AC = 4 < 5 = BC, so no such triangle exists.

Teacher's Note:
1. Check the triangle inequality before constructing.

Question 13: To construct a right triangle ABC with AB = 3.8 cm, ∠B = 90° and BC + AC = 6 cm, one cuts BD = 6 cm on the perpendicular at B. Then C is the point where the perpendicular bisector of AD meets BD. The length BC is nearly
(a) 1.80 cm
(b) 2.20 cm
(c) 3.40 cm
(d) 6.00 cm
Show Answer & Explanation

Answer: (a) 1.80 cm

Explanation:
1. Let BC = x, so AC = 6 − x and (6 − x)² = x² + 3.8².
2. 36 − 12x = 14.44 gives x ≈ 1.80 cm.

Teacher's Note:
1. CA = CD because C is on the perpendicular bisector of AD.

Question 14: For a triangle with base BC, ∠B and the difference of the other two sides AC − AB (so AB < AC), the point D with BD equal to the difference is taken on
(a) the segment BA
(b) the ray BP produced backwards (opposite to A)
(c) the segment BC
(d) the perpendicular at C
Show Answer & Explanation

Answer: (b) the ray BP produced backwards (opposite to A)

Explanation:
1. Here BD = AD − AB = AC − AB with D beyond B, away from A.
2. If AB > AC instead, D lies on BA itself.

Teacher's Note:
1. Compare the two cases: D is on BA if AB − AC is given.

Question 15: To construct a triangle with perimeter 12 cm and base angles 45° and 60°, the angle bisectors at P and Q of the 12 cm segment PQ meet at A. If the base angles at P and Q are 45° and 60°, then ∠PAQ is
(a) 75°
(b) 105°
(c) 127.5°
(d) 135°
Show Answer & Explanation

Answer: (c) 127.5°

Explanation:
1. The bisectors make 22.5° at P and 30° at Q.
2. ∠PAQ = 180° − 22.5° − 30° = 127.5°.

Teacher's Note:
1. Triangle APQ is formed by the two half-angles.

Question 16: In the construction with the given perimeter, B lies on the perpendicular bisector of AP and C on that of AQ. This gives
(a) PB = BC
(b) AB = AC
(c) PQ = AB
(d) PB = AB and CQ = AC
Show Answer & Explanation

Answer: (d) PB = AB and CQ = AC

Explanation:
1. Points on a perpendicular bisector are equidistant from the ends.
2. Hence PQ = PB + BC + CQ = AB + BC + AC, the required perimeter.

Teacher's Note:
1. B is on the bisector of AP, C on that of AQ.

Question 17: A triangle has base angles 60°, 60° and perimeter 10 cm. Its side is
(a) 3.33 cm each
(b) 5 cm each
(c) 2.5 cm each
(d) 3 cm each
Show Answer & Explanation

Answer: (a) 3.33 cm each

Explanation:
1. Base angles 60° and 60° make the third angle 60°, so it is equilateral.
2. Each side = 10 ÷ 3 ≈ 3.33 cm.

Teacher's Note:
1. Perimeter ÷ 3.

Question 18: The side of an equilateral triangle whose altitude is 3.2 cm is nearly
(a) 2.77 cm
(b) 3.70 cm
(c) 4.80 cm
(d) 6.40 cm
Show Answer & Explanation

Answer: (b) 3.70 cm

Explanation:
1. Altitude = (√3/2) × side, so side = 2 × 3.2 ÷ √3 ≈ 3.70 cm.

Teacher's Note:
1. Altitude of the equilateral triangle splits the base equally.

Question 19: For a right triangle with hypotenuse 8 cm and one side 6 cm, the other side is nearly
(a) 2 cm
(b) 7 cm
(c) 5.29 cm
(d) 10 cm
Show Answer & Explanation

Answer: (c) 5.29 cm

Explanation:
1. √(8² − 6²) = √28 ≈ 5.29 cm.

Teacher's Note:
1. The construction uses a semicircle on the hypotenuse because the angle in a semicircle is 90°.

Question 20: In constructing a right triangle on hypotenuse AC = 8 cm, the vertex B is found on the semicircle with AC as diameter. This works because
(a) the radius equals AB
(b) the semicircle has radius 6 cm
(c) AB is the diameter
(d) the angle in a semicircle is a right angle
Show Answer & Explanation

Answer: (d) the angle in a semicircle is a right angle

Explanation:
1. B lies on the circle with diameter AC, so ∠ABC = 90° (angle in a semicircle).
2. An arc of 6 cm from A fixes AB = 6 cm.

Teacher's Note:
1. Thales' theorem.

Question 21: To construct △ABC with BC = 6.4 cm, altitude 3.2 cm and median AL = 4 cm, A lies on a line parallel to BC at a distance 3.2 cm and on an arc of radius 4 cm about L. The distance from L to the foot of the altitude is
(a) 2.4 cm
(b) 0.8 cm
(c) 3.2 cm
(d) 4.0 cm
Show Answer & Explanation

Answer: (a) 2.4 cm

Explanation:
1. In the right triangle with hypotenuse AL = 4 and height 3.2: foot distance = √(16 − 10.24) = 2.4 cm.

Teacher's Note:
1. The arc cuts the parallel line at two points on either side.

Question 22: In △ABC, ∠B = 30°, ∠C = 60° and the altitude from A is 4.8 cm. The length BC is nearly
(a) 8.31 cm
(b) 11.08 cm
(c) 9.6 cm
(d) 5.54 cm
Show Answer & Explanation

Answer: (b) 11.08 cm

Explanation:
1. ∠A = 90°. BC = h(cot 30° + cot 60°) = 4.8 × (1.732 + 0.577).
2. BC ≈ 4.8 × 2.309 ≈ 11.08 cm.

Teacher's Note:
1. Split the base at the foot of the altitude.

Question 23: To construct a triangle whose medians are 6 cm, 7 cm and 6 cm, one first draws a triangle APQ with sides
(a) 3 cm, 3.5 cm and 3 cm
(b) 12 cm, 14 cm and 12 cm
(c) 6 cm, 7 cm and 6 cm
(d) 6 cm, 6 cm and 12 cm
Show Answer & Explanation

Answer: (c) 6 cm, 7 cm and 6 cm

Explanation:
1. Triangle APQ has sides equal to the three medians.
2. Its medians AE and PF meet at G; the required triangle is completed by producing AE to B with EB = GE.

Teacher's Note:
1. The centroid divides each median in the ratio 2 : 1.

Question 24: The altitude of an equilateral triangle of side 5 cm (as in the construction of an equilateral triangle with side 5 cm) is nearly
(a) 2.5 cm
(b) 5 cm
(c) 3.75 cm
(d) 4.33 cm
Show Answer & Explanation

Answer: (d) 4.33 cm

Explanation:
1. Altitude = (√3/2) × 5 ≈ 4.33 cm.

Teacher's Note:
1. The altitude also bisects the base.

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