Class 9 Mathematics Chapter 9 Constructions Important Questions — With Complete Answers

Constructions is one of the few hands-on, application-heavy chapters in Class 9 Mathematics. Unlike theorem-proof chapters, it demands practical precision: you must draw accurate figures using compass, ruler, and protractor, understand the geometric logic behind each step, and justify why a construction works. The 2026–27 board examination pattern now emphasizes not just the final figure, but also step-by-step justification and the underlying principle (e.g., why SSS guarantees a unique triangle, or why a perpendicular bisector locates the centre of a circle). Examiners expect students to construct triangles under four key conditions (SSS, SAS, ASA, RHS), draw parallel lines using angles and compass, and construct perpendiculars and angle bisectors. This chapter carries 4–6 marks in most board papers and tests both conceptual clarity and technical accuracy. Mastery here directly boosts your geometry score and builds spatial reasoning for higher mathematics.

**Q1.** To construct a triangle when three sides are given, which criterion is used? (a) ASA (b) SAS (c) SSS (d) RHS **Answer:** (c) SSS — Side-Side-Side criterion requires three known sides to construct a unique triangle. **Q2.** When constructing a triangle using the SAS criterion, which two elements must be known first? (a) Two sides and included angle (b) Two angles and a side (c) Three sides (d) One side and one angle **Answer:** (a) Two sides and included angle — SAS (Side-Angle-Side) requires the angle between the two known sides. **Q3.** To draw a line parallel to a given line AB through an external point P, which property is commonly used? (a) Perpendicular distance (b) Alternate interior angles (c) Corresponding angles (d) Vertically opposite angles **Answer:** (c) Corresponding angles — Equal corresponding angles formed by a transversal guarantee parallel lines. **Q4.** What is the first step in constructing a perpendicular from a point to a line? (a) Draw the line (b) Mark the point (c) Use a compass to find equidistant points on the line (d) Draw the perpendicular directly **Answer:** (c) Use a compass to find equidistant points on the line — This locates the midpoint for the perpendicular bisector. **Q5.** In triangle construction by ASA (Angle-Side-Angle), how many elements are given? (a) Two angles and the included side (b) Two angles and a non-included side (c) Two sides and an angle (d) Three angles **Answer:** (a) Two angles and the included side — ASA requires the side between two known angles.

**Q1.** Draw a triangle ABC where AB = 5 cm, BC = 4.5 cm, and AC = 6 cm using the SSS criterion. Write the steps. **Solution:** Steps: (1) Draw base AB = 5 cm. (2) With centre A and radius 6 cm, draw an arc above AB. (3) With centre B and radius 4.5 cm, draw another arc to intersect the first arc at C. (4) Join AC and BC. Triangle ABC is constructed. (2 marks: 1 for correct steps, 1 for accurate diagram) **Q2.** Construct triangle PQR with PQ = 5.5 cm, angle Q = 60°, and QR = 4 cm using SAS criterion. **Solution:** Steps: (1) Draw QR = 4 cm. (2) At Q, construct angle RQP = 60° using a protractor. (3) On the ray QP, mark point P such that QP = 5.5 cm. (4) Join PR. Triangle PQR is constructed. (2 marks: 1 for construction method, 1 for accuracy) **Q3.** Draw a line parallel to line l through an external point X using the compass-and-ruler method. **Solution:** Steps: (1) Draw line l and mark point X above it. (2) Draw a transversal through X intersecting l at point Y. (3) At Y, measure angle XYl using a compass. (4) At X, construct an equal corresponding angle on the opposite side of the transversal. (5) Draw the line through X along this angle; it is parallel to l. (2 marks: 1 for method, 1 for correct parallel line) **Q4.** What is the condition for a triangle to be uniquely constructible by the RHS criterion? Give an example. **Solution:** RHS (Right-Hypotenuse-Side) requires: a right angle, the hypotenuse length, and one other side. Example: Right triangle with hypotenuse = 10 cm and one leg = 6 cm is uniquely constructible because these three elements determine a unique right triangle. (2 marks: 1 for condition explanation, 1 for example) **Q5.** Draw a perpendicular to line AB at point P on AB using compass and ruler. **Solution:** Steps: (1) With P as centre and any convenient radius, mark two arcs on AB equidistant from P (at Q and R). (2) With Q and R as centres and radius > QP, draw two arcs above AB intersecting at C. (3) Join CP; this is perpendicular to AB. (2 marks: 1 for compass technique, 1 for accuracy)

**Q1.** Construct triangle ABC where BC = 6 cm, angle ABC = 50°, and angle ACB = 70°. Justify why this uses the ASA criterion. **Solution:** Steps: (1) Draw BC = 6 cm. (2) At B, construct angle ABC = 50° using protractor. (3) At C, construct angle ACB = 70° using protractor. (4) Extend the rays from B and C to meet at point A. Triangle ABC is formed. **Justification:** Two angles (50° and 70°) and the included side BC (6 cm) are given; by ASA criterion, this uniquely determines the triangle. The third angle is 180° − 50° − 70° = 60°. (3 marks: 1 for construction, 1 for diagram accuracy, 1 for justification) **Q2.** Construct a right triangle XYZ with right angle at Y, hypotenuse XZ = 7.5 cm, and side XY = 5 cm. Name the criterion used. **Solution:** Steps: (1) Draw XZ = 7.5 cm (hypotenuse). (2) Find the midpoint M of XZ. (3) With M as centre and radius = XZ/2 = 3.75 cm, draw a semicircle. (4) With X as centre and radius 5 cm, draw an arc to intersect the semicircle at Y. (5) Join XY and YZ. **Criterion:** RHS (Right-Hypotenuse-Side) — the right angle is at Y, the hypotenuse XZ and one leg XY are known. (3 marks: 1 for construction steps, 1 for accuracy, 1 for criterion identification) **Q3.** Draw a line parallel to AB and passing through point P (above AB) using the alternate interior angle method. Explain why the constructed line is parallel to AB. **Solution:** Steps: (1) Draw a transversal l through P intersecting AB at Q. (2) At Q, mark angle AQl on the right side of the transversal. (3) At P, on the transversal, construct an equal angle on the left side (alternate interior angle). (4) Draw line m through P along this angle. **Explanation:** By the alternate interior angle theorem, if two alternate interior angles are equal, the lines are parallel. Thus, line m is parallel to AB. (3 marks: 1 for method, 1 for diagram, 1 for geometric justification) **Q4.** Construct triangle PQR with sides PQ = 4.5 cm, QR = 5.5 cm, and PR = 6 cm. Verify that your construction satisfies the triangle inequality: the sum of any two sides exceeds the third. **Solution:** Steps: (1) Draw base PQ = 4.5 cm. (2) With centre P and radius 6 cm, draw an arc. (3) With centre Q and radius 5.5 cm, draw an arc intersecting the first arc at R. (4) Join PR and QR. **Verification:** PQ + QR = 4.5 + 5.5 = 10 > 6 = PR ✓; QR + PR = 5.5 + 6 = 11.5 > 4.5 = PQ ✓; PR + PQ = 6 + 4.5 = 10.5 > 5.5 = QR ✓. All three conditions satisfied; triangle is valid. (3 marks: 1 for SSS construction, 1 for verification calculations, 1 for logical clarity)

**Q1.** Construct triangle ABC with AB = 7 cm, angle BAC = 45°, and angle ABC = 60°. Find angle ACB. Then construct a perpendicular from C to line AB, and mark the foot as D. Measure CD. **Solution:** **Step 1 — Construct Triangle ABC (ASA Criterion):** (i) Draw base AB = 7 cm using ruler. (ii) At A, using protractor, construct angle BAC = 45°. (iii) At B, using protractor, construct angle ABC = 60°. (iv) Extend the rays from A and B; they meet at C. **Step 2 — Calculate angle ACB:** Angle ACB = 180° − 45° − 60° = 75° **Step 3 — Construct Perpendicular CD from C to AB:** (i) With C as centre and any convenient radius r, draw arcs on line AB at two points, say E and F (equidistant from the foot). (ii) With E and F as centres and radius > r, draw two arcs on opposite sides of AB to meet at point D. (iii) Join CD; this is perpendicular to AB. (Verify: angle CDA = 90°) **Step 4 — Measure CD:** Using ruler, measure CD. (Typical answer: CD ≈ 4.2–4.5 cm, depending on diagram accuracy) **Marks Distribution:** 1 mark each for (1) base and angle construction, (2) correct angle measure, (3) perpendicular method, (4) perpendicular accuracy, (5) measurement and presentation. --- **Q2.** Construct a triangle with sides 5.5 cm, 6.5 cm, and 7 cm (SSS). Then construct the angle bisector of the angle opposite the longest side. Verify that the angle bisector divides the opposite side in the ratio of the adjacent sides. **Solution:** **Step 1 — Construct Triangle ABC (SSS Criterion) with BC = 5.5 cm (opposite to A), AC = 6.5 cm, AB = 7 cm (longest):** (i) Draw BC = 5.5 cm. (ii) With centre B and radius 7 cm, draw an arc above BC. (iii) With centre C and radius 6.5 cm, draw an arc to intersect the first arc at A. (iv) Join AB and AC. Triangle ABC is complete. **Step 2 — Construct Angle Bisector of angle BAC:** (i) With A as centre and any radius r, draw an arc cutting AB at P and AC at Q. (ii) With P and Q as centres and radius > r/2, draw two arcs intersecting at point R. (iii) Join AR; this is the angle bisector of angle BAC, meeting BC at point D. **Step 3 — Verify the Angle Bisector Theorem:** The angle bisector theorem states: BD/DC = AB/AC By construction, AB = 7 cm and AC = 6.5 cm, so BD/DC should equal 7/6.5 = 10/9.26 ≈ 1.08 Measure BD and DC using ruler; verify the ratio. (If BD ≈ 2.8 cm and DC ≈ 2.7 cm, ratio ≈ 1.04, confirming the theorem within measurement tolerance.) **Marks Distribution:** 1 mark for SSS construction, 1 for angle bisector method, 1 for accuracy, 1 for theorem statement, 1 for verification and conclusion. --- **Q3.** Construct a right triangle with the right angle at C, where the hypotenuse AB = 8 cm and one leg CA = 5 cm (RHS criterion). Then, construct a line parallel to CA through point B, and another line parallel to CB through point A. Verify that these two lines intersect to form a rectangle ACBD. **Solution:** **Step 1 — Construct Right Triangle ABC (RHS Criterion):** (i) Draw hypotenuse AB = 8 cm. (ii) Find midpoint M of AB using compass (equidistant arc method). (iii) With M as centre and radius MA = MB = 4 cm, draw a semicircle. (iv) With A as centre and radius 5 cm, draw an arc to intersect the semicircle at C. (v) Join CA and CB. Angle ACB = 90° (angle in a semicircle). Triangle ABC is complete. **Step 2 — Construct Line through B Parallel to CA:** (i) Draw a transversal through B and any point on CA extended. (ii) At the intersection with CA (or extended), measure an angle using compass. (iii) At B, on the transversal, construct an equal alternate interior angle. (iv) Draw line through B along this angle; this line is parallel to CA. Call it line l₁. **Step 3 — Construct Line through A Parallel to CB:** (i) Similarly, construct a line through A parallel to CB using the corresponding angles method. Call it line l₂. **Step 4 — Mark Intersection Point D:** Lines l₁ and l₂ meet at point D. By construction, ACBD is a parallelogram with: - AC || BD (by construction) - CB || AD (by construction) - All angles are 90° (since angle ACB = 90° and opposite sides are parallel) Hence, ACBD is a **rectangle**. **Marks Distribution:** 1 mark for RHS triangle construction, 1.5 marks for constructing both parallel lines, 1 mark for marking D, 1.5 marks for verification and proving the rectangle property.

**Case Study:** A civil engineer needs to plan a small park. She has a triangular plot with sides 80 m, 90 m, and 100 m. She wants to construct a perpendicular pathway from one corner to the opposite side, dividing the land for different uses. Additionally, she plans to extend two straight boundary walls (parallel to each other) to enclose a new garden section. **Questions:** (a) **(2 marks)** Using the SSS criterion, describe the steps to construct a scale model of the triangular plot (using 1 cm = 10 m). What criterion ensures a unique triangle from three known sides? (b) **(2 marks)** Explain how to construct a perpendicular from one vertex to the opposite side using compass and ruler. Why is this perpendicular important in the context of dividing the land? (c) **(2 marks)** If the engineer wants to construct a line parallel to an existing boundary (using angles and a transversal), which angle property guarantees the two lines are parallel? Provide a geometric justification. (d) **(1 mark)** If a rectangular section PQRS is to be formed inside the triangle with two sides parallel to boundaries, name the four key construction steps needed. **Answer Key:** **(a)** SSS Steps: (i) Draw base 8 cm (scaled 80 m). (ii) With compass, mark arcs of radius 9 cm and 10 cm from endpoints. (iii) Intersect to locate third vertex. **Criterion:** Three sides uniquely determine a triangle because the compass arcs from two endpoints meet at exactly one point (given all triangle inequality conditions are met). ✓ **(b)** Perpendicular Steps: (i) Mark two equidistant points on the base from the foot. (ii) Draw arcs from both points to meet above the base. (iii) Join the vertex to this intersection. **Importance:** The perpendicular divides the triangle into two right-angled regions, making it easier to calculate areas and assign land uses based on geometric proportions. ✓ **(c)** **Angle Property:** Alternate interior angles OR corresponding angles. When a transversal cuts two lines, equal alternate interior angles (or equal corresponding angles) guarantee the lines are parallel. **Justification:** By Euclid's parallel postulate, if these angle pairs are equal, the lines will never meet, hence are parallel. ✓ **(d)** **Four Steps:** (1) Construct side PQ. (2) Draw a perpendicular at P to form right angle. (3) Construct a parallel line to PQ through a suitable point. (4) Complete the rectangle by joining the endpoints. ✓

CBSETUTOR.ai is a premium AI-powered tutor designed specifically for CBSE Class 9 students. Here's how it reinforces Constructions mastery: **Adaptive Question Drills:** Every day, the AI generates personalized construction practice sessions. If you struggle with RHS triangles, the system serves you 5–7 RHS questions with increasing difficulty before moving to other criteria. Unlike static textbooks, the difficulty adapts to your pace. **Instant Visual Feedback:** You sketch your construction using a digital protractor and compass tool, and CBSETUTOR.ai's AI compares your diagram against NCERT-aligned benchmarks. It flags errors in angle measurement (e.g., "Your angle ABC is 52°; it should be 50°") and step sequencing in real time, so you correct mistakes immediately rather than learning a wrong method. **Step-by-Step Justification:** After each construction, the AI prompts: "Why does SSS guarantee a unique triangle?" and evaluates your textual or verbal answer against NCERT definitions. This embeds deeper understanding beyond rote drawing. **Daily Mini-Tests:** Every 3–4 days, a randomized 15-minute quiz spans all four construction criteria and parallel-line methods. Your performance triggers targeted revision—e.g., if you score <70% on ASA, the next session prioritizes ASA constructions. **Board Exam Simulation:** CBSETUTOR.ai's exam module mimics real CBSE board patterns. You solve 1-mark MCQs, 2-mark short answers, and 5-mark constructions under timed conditions, with detailed feedback mirroring how a board examiner would evaluate your work. **Parent Insights:** Your parent receives weekly summaries: "Your child has mastered SSS and SAS but needs practice on parallel lines. Recommended: 15 minutes daily for 2 weeks." With CBSETUTOR.ai, Constructions shifts from a chapter you fear to a skill you own—because you practice the exact patterns that appear on the board, every single day.