Class 9 Mathematics Chapter 9 Constructions: 30 MCQs with Solutions

The rationalized CBSE Class 9 syllabus now allocates significant weightage to multiple-choice questions, especially in periodic tests and summative assessments. Constructions, being a purely skill-based chapter, features heavily in MCQ formats because examiners can test whether you identify the correct construction method (SSS vs. SAS), recognize impossibility conditions, or recall angle/line properties instantly. MCQs also eliminate the need for lengthy construction diagrams in a timed exam—you prove understanding through logical selection. Studies show students who practice 30–50 targeted MCQs improve by 15–20 percentage points. Why? MCQs force precision: you cannot hide behind partial working. Every wrong option is a trap—understanding why Option C fails teaches you the theorem boundaries better than solving a full construction. The new pattern also emphasizes assertion-reason MCQs, where you justify both a statement and its reason. Constructions lend themselves perfectly: 'Assertion: A triangle cannot be constructed with sides 2 cm, 3 cm, 6 cm. Reason: The sum of any two sides must exceed the third.' This dual-check ensures deep conceptual learning, not rote memorization.

These questions test foundational recall: construction criteria, tool usage, and basic impossibility conditions. Each is worth 1 mark and appears in Section A of most CBSE papers. **Q1.** To construct a triangle given three sides, which criterion is used? (A) SAS (B) SSS (C) ASA (D) RHS **Answer:** (B) SSS **Reason:** Three sides determine a unique triangle via the Side-Side-Side criterion. **Q2.** If sides of a triangle are 4 cm, 5 cm, and 10 cm, can the triangle be constructed? (A) Yes (B) No (C) Only if angle is 90° (D) Insufficient data **Answer:** (B) No **Reason:** 4 + 5 = 9 < 10; the triangle inequality fails. **Q3.** To construct a triangle with two sides and the included angle, which criterion applies? (A) SSS (B) SAS (C) ASA (D) AAA **Answer:** (B) SAS **Reason:** Side-Angle-Side uniquely determines the triangle. **Q4.** Which tool is essential for drawing a perpendicular to a line from a point on it? (A) Compass alone (B) Straightedge alone (C) Compass and straightedge (D) Protractor **Answer:** (C) Compass and straightedge **Reason:** Both are needed to bisect the angle or arc systematically. **Q5.** To construct a line parallel to a given line, which property is most commonly used? (A) Corresponding angles (B) Alternate angles (C) Either (A) or (B) (D) Perpendicular bisector **Answer:** (C) Either (A) or (B) **Reason:** Both angle relationships guarantee parallel lines via Euclid's postulate. **Q6.** A triangle is to be constructed with sides 3 cm, 4 cm, and 5 cm. Is it a right triangle? (A) Yes, by Pythagorean theorem (B) No (C) Cannot determine (D) Only if angle C = 90° **Answer:** (A) Yes, by Pythagorean theorem **Reason:** 3² + 4² = 9 + 16 = 25 = 5². **Q7.** When constructing a triangle with two angles and the side between them, which criterion is used? (A) AAS (B) AAA (C) ASA (D) SAS **Answer:** (C) ASA **Reason:** Angle-Side-Angle (the included side) uniquely constructs the triangle. **Q8.** What is the maximum number of perpendiculars that can be drawn to a line from an external point? (A) One (B) Two (C) Infinite (D) None **Answer:** (A) One **Reason:** By Euclid's axiom, exactly one perpendicular exists from an external point. **Q9.** To construct a right triangle given hypotenuse and one leg, which criterion applies? (A) RHS (B) SAS (C) SSS (D) AAS **Answer:** (A) RHS **Reason:** Right angle-Hypotenuse-Side uniquely constructs a right triangle. **Q10.** A line l and a point P on it are given. How many lines parallel to l can be drawn through P? (A) One (B) Two (C) Zero (D) Infinite **Answer:** (C) Zero **Reason:** If P is on l, any line through P is either l itself or intersects l; Euclid's postulate forbids a parallel through a point on the line.

These require identifying construction methods in varied contexts and understanding when construction is impossible. Typical for Section B (2–3 marks converted to MCQ format). **Q11.** A triangle ABC is to be constructed with AB = 5 cm, ∠A = 45°, and ∠B = 60°. Which criterion will you use? (A) SSS (B) SAS (C) ASA (D) RHS **Answer:** (C) ASA **Reason:** Two angles and the included side AB are known; ∠C = 75° is determined automatically. **Q12.** To construct a line parallel to BC through point A (where A is not on BC), which method is most efficient? (A) Draw a transversal, construct alternate angles equal (B) Construct a perpendicular from A to BC, then perpendicular to that (C) Use a compass to copy BC (D) Measure and mark equal distances **Answer:** (A) Draw a transversal, construct alternate angles equal **Reason:** Alternate angle construction is the most direct Euclidean method. **Q13.** Triangle PQR has PQ = 4 cm, QR = 6 cm, and PR = 7 cm. If you construct a congruent triangle, how many such triangles are possible with the same orientation? (A) One (B) Two (C) Three (D) Infinite **Answer:** (A) One **Reason:** SSS determines a unique triangle up to congruence. **Q14.** You need to construct a triangle with sides 5 cm, 7 cm, and 13 cm. Is this possible? (A) Yes, by SSS (B) No, because 5 + 7 < 13 (C) Yes, if one angle is obtuse (D) No, insufficient data **Answer:** (B) No, because 5 + 7 < 13 **Reason:** Triangle inequality: sum of any two sides must exceed the third. **Q15.** Point M lies on segment AB. A perpendicular to AB is drawn at M. Which of the following is true? (A) The perpendicular bisects AB (B) The perpendicular makes 90° with AB (C) The perpendicular passes through the midpoint of AB (D) Multiple perpendiculars can exist at M **Answer:** (B) The perpendicular makes 90° with AB **Reason:** A perpendicular at a point means exactly one line at 90°; it doesn't imply bisection unless M is the midpoint. **Q16.** To construct an angle of 60°, the most direct method uses which property? (A) Angle bisector theorem (B) Equilateral triangle (C) Corresponding angles with parallel lines (D) Perpendicular bisector **Answer:** (B) Equilateral triangle **Reason:** An equilateral triangle has all angles = 60°; compass arc with radius = side length creates this. **Q17.** A triangle ABC is constructed with AB = 6 cm, AC = 6 cm, and ∠A = 50°. Triangle DEF is to be congruent to ABC with DE = 6 cm and DF = 6 cm. How many non-congruent triangles satisfy this? (A) One (B) Two (C) Infinite (D) Zero **Answer:** (B) Two **Reason:** With two equal sides (6 cm, 6 cm) and ∠D unknown, the third vertex F can lie on either side of DE—two possible positions. **Q18.** To bisect a given line segment XY without measuring, which construction is used? (A) Draw perpendicular at midpoint (B) Draw perpendicular bisector using equal arcs from X and Y (C) Fold the paper (D) Use a ruler to measure half-length **Answer:** (B) Draw perpendicular bisector using equal arcs from X and Y **Reason:** Equal radii (> XY/2) from X and Y create intersection points that determine the perpendicular bisector. **Q19.** In triangle ABC, ∠B = 90°. If you know AB = 5 cm, BC = 12 cm, and AC = 13 cm, which criterion validates the RHS construction? (A) The Pythagorean triple (5, 12, 13) (B) Two sides and a right angle (C) Hypotenuse AC and leg AB (D) All three sides form a right triangle **Answer:** (C) Hypotenuse AC and leg AB **Reason:** RHS requires right angle, hypotenuse, and one side; here AC (hypotenuse) and AB (leg) are sufficient. **Q20.** Two lines l₁ and l₂ are parallel. A transversal cuts both. To construct a line parallel to both using a compass, how many steps are minimized? (A) 2–3 steps (one angle construction) (B) 4–5 steps (two angle constructions) (C) 6–8 steps (D) Cannot be minimized **Answer:** (A) 2–3 steps (one angle construction) **Reason:** Using corresponding or alternate angle equality once is sufficient due to the transitive property of parallelism.

These demand higher-order thinking: why a construction fails, limiting conditions, and proof-based reasoning. Typical of Section C (3+ mark questions re-cast as assertion-reason MCQs). **Q21.** **Assertion (A):** A triangle with sides 3 cm, 4 cm, 7 cm cannot be constructed. **Reason (R):** The sum of any two sides of a triangle must be strictly greater than the third side. (A) Both A and R are true; R is the correct reason for A (B) Both A and R are true; R is not the correct reason for A (C) A is true; R is false (D) A is false; R is true **Answer:** (A) Both A and R are true; R is the correct reason for A **Reason:** 3 + 4 = 7 ≯ 7; the triangle inequality is violated, making construction impossible. **Q22.** **Assertion (A):** To construct a triangle given two sides and an angle not included between them, the construction is ambiguous. **Reason (R):** This is known as the SSA (Side-Side-Angle) case, which may yield 0, 1, or 2 triangles. (A) Both A and R are true; R is the correct reason for A (B) Both A and R are true; R is not the correct reason for A (C) A is true; R is false (D) A is false; R is true **Answer:** (A) Both A and R are true; R is the correct reason for A **Reason:** SSA is indeed ambiguous; the second side can rotate to two positions, creating zero, one, or two valid triangles. **Q23.** **Assertion (A):** A perpendicular to a line at a point on the line is unique. **Reason (R):** If two perpendiculars existed at the same point, they would be the same line by Euclid's first postulate. (A) Both A and R are true; R is the correct reason for A (B) Both A and R are true; R is not the correct reason for A (C) A is true; R is false (D) A is false; R is true **Answer:** (A) Both A and R are true; R is the correct reason for A **Reason:** Uniqueness of perpendicular follows from Euclid's axioms; two lines cannot both be perpendicular to the same line at the same point. **Q24.** **Assertion (A):** To construct a triangle, it is sufficient to know one side and two angles. **Reason (R):** If two angles are known, the third angle is determined by the angle-sum property. (A) Both A and R are true; R is the correct reason for A (B) Both A and R are true; R is not the correct reason for A (C) A is true; R is false (D) A is false; R is true **Answer:** (A) Both A and R are true; R is the correct reason for A **Reason:** One side and two angles (ASA or AAS) uniquely determine the triangle; the third angle is indeed redundant information. **Q25.** **Assertion (A):** Two triangles constructed with sides (5, 5, 6) and (5, 5, 8) are not congruent. **Reason (R):** For an isosceles triangle with equal sides a, changing the base changes the triangle's shape and size. (A) Both A and R are true; R is the correct reason for A (B) Both A and R are true; R is not the correct reason for A (C) A is true; R is false (D) A is false; R is true **Answer:** (A) Both A and R are true; R is the correct reason for A **Reason:** Different base lengths in isosceles triangles with fixed equal sides create non-congruent triangles; they differ in angles and altitudes. **Q26.** **Assertion (A):** When constructing parallel lines, corresponding angles must be equal, but alternate interior angles need not be. **Reason (R):** Alternate interior angles are equal only when both lines are perpendicular to the transversal. (A) Both A and R are true; R is the correct reason for A (B) Both A and R are true; R is not the correct reason for A (C) A is true; R is false (D) A is false; R is true **Answer:** (D) A is false; R is true **Reason:** Alternate interior angles are always equal when lines are parallel (Euclid's theorem), not just in the perpendicular case. R is true but doesn't support A. **Q27.** **Assertion (A):** An angle of 15° can be constructed using a compass and straightedge. **Reason (R):** 15° = 45° – 30°, and both 45° and 30° can be constructed. (A) Both A and R are true; R is the correct reason for A (B) Both A and R are true; R is not the correct reason for A (C) A is true; R is false (D) A is false; R is true **Answer:** (A) Both A and R are true; R is the correct reason for A **Reason:** Angle subtraction using bisectors allows construction of 15° from 60° bisected twice (60° → 30° → 15°), or from 45° – 30°. **Q28.** **Assertion (A):** A right triangle can always be constructed given the hypotenuse and one acute angle. **Reason (R):** The RHS criterion requires hypotenuse, one side, and the right angle; an acute angle determines the other sides uniquely. (A) Both A and R are true; R is the correct reason for A (B) Both A and R are true; R is not the correct reason for A (C) A is true; R is false (D) A is false; R is true **Answer:** (B) Both A and R are true; R is not the correct reason for A **Reason:** Both statements are true, but R doesn't directly justify A. The acute angle determines the triangle, but RHS typically refers to two sides + right angle, not angle-based construction. **Q29.** **Assertion (A):** If a triangle has sides a, b, c where a + b = c, no such triangle can be constructed. **Reason (R):** The points with sides a, b, and c would be collinear, forming a degenerate triangle. (A) Both A and R are true; R is the correct reason for A (B) Both A and R are true; R is not the correct reason for A (C) A is true; R is false (D) A is false; R is true **Answer:** (A) Both A and R are true; R is the correct reason for A **Reason:** When a + b = c (equality in triangle inequality), the three vertices lie on a straight line; this violates the definition of a non-degenerate triangle. **Q30.** **Assertion (A):** To construct a triangle given two sides and an included angle, the compass radius in the initial arc must be exactly equal to one of the given side lengths. **Reason (R):** In the SAS construction, the compass opening determines the arc radius, which must match the side length to ensure accuracy. (A) Both A and R are true; R is the correct reason for A (B) Both A and R are true; R is not the correct reason for A (C) A is true; R is false (D) A is false; R is true **Answer:** (A) Both A and R are true; R is the correct reason for A **Reason:** SAS requires marking two sides of exact length; the compass radius is set to the given length to cut arcs precisely on the rays from the angle vertex.

CBSE examiners deliberately plant plausible-sounding distractors. Recognizing these traps boosts your score by 10–15%. **Trap 1: Confusing SSA with SAS** Students often think 'two sides and an angle' always constructs a unique triangle. Wrong—if the angle is not between the two sides (SSA case), the triangle is ambiguous. Always check: is the angle *included* between the two given sides? **Trap 2: The Degenerate Triangle Trap** When a + b = c (not <), the three sides are collinear. Many students miss this boundary case and mistakenly say 'yes, construct the triangle.' Remember: strict inequality (a + b > c) is required. **Trap 3: Confusing Perpendicular at a Point with Perpendicular Bisector** A perpendicular at point M on line AB doesn't bisect AB unless M is the midpoint. Examiners slip in options like 'The perpendicular bisects AB'—only true if M is the midpoint. **Trap 4: Thinking Multiple Perpendiculars Can Exist** From a point on a line, only one perpendicular can be drawn (Euclid's axiom). From a point not on the line, also only one. Many students, confusing this with 'there are infinite lines through a point,' choose 'infinite perpendiculars'—wrong. **Trap 5: Misapplying Parallel Line Criteria** Corresponding angles are equal ⟺ lines are parallel. Alternate interior angles are equal ⟺ lines are parallel. But students sometimes mix these up or think 'corresponding angles' means angles in the same position on both sides (incorrect phrasing). Use the precise definitions: corresponding angles are on the same side of the transversal and in the same position relative to their respective lines. **Trap 6: The 'Always Possible' Assumption** Not all angle and side combinations yield a valid triangle. Options like 'Yes, by SSS' for sides (1, 2, 5) are traps. Always check the triangle inequality first. **Trap 7: Equating Congruence with Construction Uniqueness** If two triangles are congruent, they can be constructed uniquely up to position/orientation. But SSA can produce two non-congruent triangles. Don't assume 'two SSA triangles are always congruent'—they're not. **Trap 8: RHS vs. Right-Angled with Other Criteria** RHS specifically means Right angle + Hypotenuse + one Side. If an option says 'RHS because we know two legs and a right angle,' that's not RHS—it's SAS applied to a right triangle. Read the criterion name carefully.

In a typical 80-minute CBSE exam, Constructions usually has 4–6 questions (1–3 marks each). Here's how to maximize your score: **Pre-Exam Prep (1 week before)** Memorize the four construction criteria: SSS, SAS, ASA, RHS. Write them on a flashcard with a small diagram. Spend 5 minutes daily reviewing which criterion applies when. Know the triangle inequality cold—it eliminates 20% of trap options instantly. **During the Exam (Section A: MCQs, 20 minutes for 6 questions)** 1. **Scan the question type (0–5 seconds).** Is it a direct 'construct with...' (easy), a 'can we construct...' (medium), or assertion-reason (hard)? 2. **For direct construction questions:** Identify given data. Count: how many sides? How many angles? Which angle (included or not)? Match to criterion. Move on. (40–50 seconds per question) 3. **For 'Is it possible?' questions:** Run the triangle inequality or criterion check. 3–4 second mental math. (60–70 seconds per question) 4. **For assertion-reason:** Read assertion, check if true. Then read reason. Is the reason the *correct justification* for the assertion? This is the trickiest part; spend 80–90 seconds here. 5. **Eliminate obviously wrong options first.** If Option D says 'Infinite perpendiculars from a point on a line,' it's wrong—always eliminate it instantly. **Time Allocation (rough)** - Easy (Q1–Q2): 1 minute total (30 seconds each) - Medium (Q3–Q4): 2 minutes total (60 seconds each) - Hard assertion-reason (Q5–Q6): 2.5 minutes total (85–90 seconds each) - Buffer: 0.5 minutes for review **Common Time Wasters to Avoid** - Do not re-construct the full triangle on paper. Constructions MCQs don't require drawing; use mental imaging. - Do not measure with a ruler. Trust the given measurements and the criterion. - Do not overthink 'which method is most efficient?' unless explicitly asked. If SSS and SAS both work, either is correct; the question usually specifies one. **Confidence Scoring** If you're 90%+ certain, mark and move. If 60–70% certain, mark with a light pencil and revisit in the last 2 minutes. If <50% certain, use the elimination method: kill two obviously wrong options, then guess between the remaining two. Statistically, this yields 50% of points on uncertain questions. **Post-Exam Review** When you return home, revisit questions you guessed. Did you miss a keyword like 'included angle'? Did you forget the triangle inequality? Document these in a 'Trap Log' and review 2–3 days before the exam. At cbsetutor.ai, our AI tutors flag these exact patterns for you; a 3-day free trial exposes your blind spots and accelerates your weak areas.

While CBSE exams test theoretical construction, understanding real-world relevance deepens retention and satisfies conceptual curiosity. **Parallel Lines in Architecture & Engineering** When architects design a building's load-bearing walls, they must be parallel to ensure equal stress distribution. Using the corresponding angles method (or alternate interior angles), they verify parallelism without measurement. In road construction, lane markings are drawn parallel using the same principle: set a transversal (reference line), ensure equal angles, and you've guaranteed parallel lanes. **Triangle Constructions in Surveying** Surveyors measure land plots using triangulation. Given three side lengths (SSS) of a triangular plot, they reconstruct it accurately on a map. If a surveyor knows two sides and the angle between them (SAS), they can field-verify the plot before finalizing documents. Understanding construction criteria helps surveyors detect impossible measurements—if a plot's given sides violate the triangle inequality, the data is faulty before they waste time in the field. **Perpendiculars in Building Layout** When laying the foundation of a house, construction workers must ensure the corner is exactly 90°. Using a compass and straightedge (or a plumb line in practice), they construct perpendiculars to align walls. This ensures structural integrity; even a 2–3° deviation can cause cracks over time. **Right Triangle in Slope & Stability** Roof pitches, ramp slopes, and embankment angles all rely on right triangle geometry. Using the RHS criterion, engineers verify that a given slope angle and horizontal distance uniquely determine the rise—critical for drainage and safety. These applications remind you why the chapter exists: constructions are not abstract puzzles but tools for precision in the physical world. When you encounter a construction MCQ, imagine the surveyor or architect relying on your correct answer.