Class 9 Mathematics Chapter 8 Playing with Constructions: Important Questions & Complete Solutions

Constructions occupy 8–10% of Class 9 CBSE Mathematics papers and test three core competencies: understanding geometric properties (perpendicular bisectors, angle bisectors, circle tangents), executing multi-step compass-and-straightedge procedures accurately, and explaining *why* a construction works (proof of validity). Unlike pure theory questions, construction problems demand hands-on spatial reasoning—skills the CBSE now prioritises under its competency-based assessment framework. The 2026-27 rationalized syllabus emphasises construction of line segments and circles (with given radii and through given points), perpendicular bisectors of line segments, and constructions of triangles under various conditions (SSS, SAS, ASA, RHS). Examiners frequently blend constructions with coordinate geometry, circle properties, and angle bisector theorems, making deep conceptual clarity essential. Mastering these 17+ questions ensures you score confidently on construction sections, reduce careless errors, and build geometric intuition that transfers to higher geometry in Classes 10–12.

MCQs in constructions test quick recall of definitions, properties, and single-step procedures. Each question targets a specific concept: recognizing perpendicular bisectors, identifying correct compass angles, or naming the construction method. **Q1.** When constructing the perpendicular bisector of a line segment AB, what is the minimum radius of the compass? (A) Any radius less than AB (B) Any radius greater than half of AB (C) Exactly half of AB (D) Any radius less than or equal to half of AB **Answer: (B)** Any radius greater than half of AB. The compass radius must exceed AB/2 so that arcs drawn from both A and B intersect above and below the segment. **Q2.** To construct a triangle with sides 5 cm, 6 cm, and 7 cm, which construction method is most direct? (A) ASA (B) SSS (C) SAS (D) RHS **Answer: (B)** SSS. When all three side lengths are given, you construct using the side-side-side condition by drawing one side and using compass arcs to locate the third vertex. **Q3.** What is the first step when constructing an angle bisector of ∠PQR? (A) Draw arc QR (B) Draw arc with centre Q, cutting rays QP and QR (C) Draw perpendicular bisector of PR (D) Extend ray QR **Answer: (B)** Draw arc with centre Q, cutting rays QP and QR. This creates equal arc lengths, which allows equal radii later to find the bisector direction. **Q4.** To construct a circle passing through three non-collinear points A, B, C, which point must be found first? (A) The centroid (B) The circumcentre (C) The orthocentre (D) The incentre **Answer: (B)** The circumcentre. The circumcentre is equidistant from all three points, so it becomes the centre of the circle. **Q5.** If you construct a perpendicular to a line ℓ at point P on ℓ, and then construct a parallel to ℓ through an external point Q, the angle between these two lines is: (A) 45° (B) 60° (C) 90° (D) 0° **Answer: (C)** 90°. The perpendicular to ℓ is perpendicular by definition; a parallel to ℓ maintains that perpendicular relationship.

Two-mark questions require a clear construction diagram (or description) and brief justification. These often appear in mid-term papers and board exams as standalone or paired questions. **Q1. Construct the perpendicular bisector of line segment AB = 7 cm. Mark any point P on the bisector and verify that PA = PB.** *Solution:* 1. Draw line segment AB = 7 cm. 2. Open compass to more than half of AB (say 4 cm). 3. With centre A, draw arcs above and below AB. 4. With the same radius and centre B, draw arcs intersecting the previous arcs at points M and N. 5. Draw line MN—this is the perpendicular bisector of AB. 6. Mark point P anywhere on MN. Measure PA and PB using a ruler; both are equal, confirming that every point on the perpendicular bisector is equidistant from the endpoints. **Q2. Construct a triangle with sides AB = 5 cm, BC = 6 cm, and CA = 5.5 cm. Measure and write the largest angle.** *Solution:* 1. Draw base BC = 6 cm. 2. With centre B and radius 5 cm, draw an arc. 3. With centre C and radius 5.5 cm, draw another arc intersecting the first arc at A. 4. Join AB and AC to complete triangle ABC. 5. Measure the three angles using a protractor. The angle opposite the longest side (BC = 6 cm) is ∠BAC and will be the largest (approximately 72–74°, depending on accuracy). **Q3. Construct an angle of 60° without using a protractor, using only a compass and straightedge.** *Solution:* 1. Draw a ray OA. 2. With centre O and any radius r, draw an arc intersecting OA at point B. 3. With centre B and the same radius r, draw another arc intersecting the first arc at point C. 4. Join O and C to form ray OC. Angle AOC = 60°. (This works because triangle OBC is equilateral.) **Q4. Construct the angle bisector of ∠ABC where ∠ABC = 80°. Verify by measuring that the bisector divides the angle into two equal parts.** *Solution:* 1. With centre B and any radius, draw an arc cutting BA at P and BC at Q. 2. With centres P and Q and equal radius (greater than half PQ), draw arcs intersecting at R. 3. Draw ray BR—this is the angle bisector. 4. Measure ∠ABR and ∠RBC using a protractor; both should be approximately 40°, confirming the bisection. **Q5. Construct a circle passing through three given non-collinear points A, B, and C (with AB = 4 cm, BC = 5 cm, and AC = 6 cm).** *Solution:* 1. Draw triangle ABC with the given sides using SSS construction. 2. Construct the perpendicular bisector of side AB. 3. Construct the perpendicular bisector of side BC. 4. Mark the intersection of these two bisectors as O (the circumcentre). 5. With centre O and radius OA, draw the circle. It passes through A, B, and C.

Three-mark questions blend construction with explanation, often requiring a diagram plus reasoning or proof. These test deeper understanding of why constructions work. **Q1. Construct a triangle ABC such that AB = 6 cm, ∠A = 50°, and ∠B = 60°. Also find ∠C without measuring.** *Solution:* 1. Draw base AB = 6 cm. 2. At A, construct an angle of 50° using a compass and straightedge (or protractor for efficiency). Draw ray AC. 3. At B, construct an angle of 60°. Draw ray BC. 4. These rays intersect at C, forming triangle ABC. 5. By the angle sum property: ∠C = 180° − 50° − 60° = 70°. (1 mark: construction, 1 mark: diagram, 1 mark: angle calculation and reasoning) **Q2. Construct a right-angled triangle with hypotenuse AC = 8 cm and one leg AB = 5 cm. Name the right angle and calculate the third side using Pythagoras' theorem.** *Solution:* 1. Draw hypotenuse AC = 8 cm. 2. Find the midpoint M of AC by constructing its perpendicular bisector. 3. With centre M and radius MA = MC = 4 cm, draw a semicircle. 4. With centre A and radius 5 cm, draw an arc intersecting the semicircle at B. 5. Join AB and BC. Angle ABC = 90° (angle in a semicircle). 6. By Pythagoras' theorem: BC² = AC² − AB² = 64 − 25 = 39, so BC = √39 ≈ 6.24 cm. (1 mark: construction steps, 1 mark: identification of right angle with justification, 1 mark: calculation) **Q3. Construct triangle PQR with PQ = 7 cm, ∠P = 70°, and ∠Q = 50°. Construct the angle bisector from vertex P and show that it meets QR at point D. Measure PD.** *Solution:* 1. Draw base PQ = 7 cm. 2. At P, construct ∠P = 70° and draw ray PR. 3. At Q, construct ∠Q = 50° and draw ray QR intersecting PR at R. 4. Triangle PQR is complete; ∠R = 180° − 70° − 50° = 60°. 5. Construct the angle bisector of ∠P (35° on each side). It meets QR at point D. 6. Measure PD using a ruler (approximately 6–6.5 cm, exact value depends on construction accuracy). (1 mark: triangle construction, 1 mark: angle bisector construction and marking D, 1 mark: measurement of PD) **Q4. Construct a parallelogram ABCD with AB = 5 cm, BC = 6 cm, and ∠ABC = 70°. Verify that opposite sides are equal by measuring.** *Solution:* 1. Draw base AB = 5 cm. 2. At B, construct ∠ABC = 70° and mark point C such that BC = 6 cm. 3. From C, draw a line parallel to AB (using compass and straightedge or alternate angle method). On this line, mark D such that CD = AB = 5 cm. 4. Join AD. Verify: Measure CD (should be 5 cm, equal to AB) and AD (should be 6 cm, equal to BC). ✓ 5. Also verify: Opposite angles are equal: ∠A = ∠C and ∠B = ∠D. (1 mark: construction of parallelogram, 1 mark: construction of parallel line, 1 mark: verification of properties)

Five-mark questions are typically board exam standard, requiring detailed construction, clear reasoning, and often a multi-step proof or compound construction. **Q1. Construct triangle ABC where AB = 6 cm, BC = 7 cm, and the median from A to side BC is 5 cm. [Hint: If M is the midpoint of BC, then AM = 5 cm.]** *Full Solution:* Medians are interior line segments from a vertex to the midpoint of the opposite side. This is a complex construction because we're given two sides and a median. *Step-by-step:* 1. Draw BC = 7 cm. Mark the midpoint M of BC, so BM = MC = 3.5 cm. 2. With centre A and radius 5 cm, we need to find the locus of all positions of A such that AM = 5 cm. This is a circle with centre M and radius 5 cm. 3. With centre B and radius AB = 6 cm, we also need A to be 6 cm from B. This is a circle with centre B and radius 6 cm. 4. Draw the circle with centre M and radius 5 cm. 5. Draw the circle with centre B and radius 6 cm. 6. These circles intersect at two points; choose one as A (conventionally, the upper position). 7. Join AB (which will measure 6 cm) and AC. 8. Measure AC and the angles to complete the description. *Verification:* Measure AM using a ruler—it should be 5 cm (the median condition). Measure AB—it should be 6 cm. Both conditions are satisfied. (2 marks: correct construction method, 2 marks: clear diagram with all labels, 1 mark: verification or measurement) **Q2. Construct a triangle ABC with AB = 5.5 cm, BC = 6.5 cm, and CA = 7 cm. Then construct its incircle (the circle inscribed inside the triangle, touching all three sides).** *Full Solution:* The incircle's centre is the incentre, found at the intersection of the three angle bisectors. *Steps:* 1. **Construct triangle ABC:** Draw AB = 5.5 cm. With centre A, draw an arc of radius 7 cm. With centre B, draw an arc of radius 6.5 cm. These arcs meet at C. Join AC and BC. 2. **Construct angle bisectors:** - Bisect ∠A: With centre A and any radius, draw an arc cutting AB and AC at points P and Q. With centres P and Q (equal radius), draw intersecting arcs at R. Join AR. - Bisect ∠B: Similarly, with centre B, cut BA and BC at points S and T. With centres S and T, draw intersecting arcs at U. Join BU. - The bisectors AR and BU meet at the incentre I. 3. **Draw the incircle:** From I, drop a perpendicular to any side (say AB), meeting it at point X. Measure IX (this is the inradius r, approximately 1.5–2 cm). 4. **Draw the incircle:** With centre I and radius IX, draw the circle. It will be tangent to all three sides of triangle ABC. *Verification:* Measure the perpendicular distances from I to each side; all should equal IX, confirming the incircle touches all sides. (2 marks: triangle construction, 2 marks: angle bisector constructions and incentre location, 1 mark: incircle construction and labelling) **Q3. Construct a square ABCD with side length 5 cm. Then construct a circle circumscribing the square (passing through all four vertices). Find the radius of this circle.** *Full Solution:* A square has all sides equal and all angles 90°. Its circumradius (radius of the circumscribed circle) is half the diagonal. *Steps:* 1. **Construct the square:** - Draw AB = 5 cm. - At A, construct a 90° angle and mark D such that AD = 5 cm. - From D, draw a line parallel to AB. Mark C on this line such that DC = 5 cm. - Verify: BC = 5 cm and ∠BCD = ∠CDA = ∠DAB = ∠ABC = 90°. 2. **Find the circumcentre:** The centre of the circumscribed circle is at the intersection of the diagonals AC and BD. Construct (or draw) both diagonals and mark their intersection as O. 3. **Measure the radius:** Measure OA (or OB, OC, OD—all equal for a square). Using the Pythagorean theorem: diagonal = 5√2 cm, so radius = (5√2)/2 ≈ 3.54 cm. Alternatively, measure OA directly using a ruler. 4. **Draw the circumcircle:** With centre O and radius OA, draw the circle. It passes through A, B, C, and D. *Verification:* Measure the distances OA, OB, OC, OD. All should be equal and approximately 3.54 cm. (2 marks: square construction with all properties verified, 2 marks: circumcentre location and circle construction, 1 mark: radius calculation or measurement)

**Case Study:** A landscape architect is designing a garden with a circular flower bed. Three existing trees at points A, B, and C are at distances: AB = 8 m, BC = 9 m, and CA = 10 m. The architect wants to plant the circular flower bed such that it touches all three trees (i.e., the circle is tangent to the trees). **(a)** Construct a scale drawing (1 m = 1 cm on paper) of triangle ABC representing the positions of the three trees. **(b)** Find the centre of the circular flower bed (the incentre of triangle ABC) using compass-and-straightedge construction. **(c)** Measure the radius of the flower bed (the inradius) in your scale drawing. Then calculate the actual radius in metres. **(d)** If the landscape architect wants to enclose the flower bed with a fence, calculate the circumference of the fence in metres. (Take π ≈ 22/7.) **Solution:** **(a) Construct triangle ABC at scale 1 m = 1 cm:** - Draw AB = 8 cm. - With centre A, radius 10 cm, draw an arc. - With centre B, radius 9 cm, draw an arc intersecting the first arc at C. - Join AC and BC. **(b) Find the incentre:** - Bisect ∠A by drawing an arc from A (radius r₁) cutting AB and AC, then bisecting the arc as described in Q2 above. Draw the bisector ray. - Bisect ∠B similarly and draw the bisector ray. - These bisectors meet at point I, the incentre (centre of the flower bed). **(c) Measure and calculate the inradius:** - Drop a perpendicular from I to side AB, meeting AB at point X. - Measure IX in your scale drawing (say, 2.4 cm). - Actual radius = 2.4 cm × (1 m / 1 cm) = 2.4 m. (Alternatively, use the formula: r = Area / semi-perimeter. Semi-perimeter s = (8 + 9 + 10)/2 = 13.5 m. Using Heron's formula: Area = √[13.5(13.5−8)(13.5−9)(13.5−10)] = √[13.5 × 5.5 × 4.5 × 3.5] ≈ √1163.0625 ≈ 34.1 m². So r = 34.1 / 13.5 ≈ 2.53 m—close to the measured value.) **(d) Calculate the circumference:** - Circumference = 2πr = 2 × (22/7) × 2.4 ≈ (44/7) × 2.4 ≈ 6.286 × 2.4 ≈ 15.09 m. - The actual circumference of the fence is approximately 15.09 metres. **Why This Matters (HOTS Connection):** This problem integrates construction (finding the incentre), measurement, scaling, and real-world application. It tests whether students understand that constructions are not abstract—they solve practical design problems. The incentre is equidistant from all sides, making it the ideal centre for a circle tangent to a triangle's sides.

At cbsetutor.ai, our AI tutoring system is engineered to master Class 9 geometry constructions through adaptive, real-time drilling. Here's how it works: **1. Pattern Recognition & Question Randomisation:** Our AI generates unlimited variations of the 17+ core question types you've studied here. One day you construct a triangle with AB = 5 cm, BC = 6 cm, CA = 5.5 cm; the next, you construct with different values. Each randomisation preserves the underlying method, ensuring you internalise the step-by-step procedure rather than memorising a single answer. **2. Real-Time Visual Feedback:** When you submit a construction (via a drawing tool or step-by-step description), the AI checks: - Compass radii are correct (e.g., > AB/2 for perpendicular bisectors). - Arc intersections are marked accurately. - Angles are bisected correctly. - Final measurements align with theoretical expectations (e.g., PA = PB on the perpendicular bisector). Incorrect steps trigger immediate, targeted hints—not the answer, but the *next step you should check*. **3. Difficulty-Adaptive Path:** You start with 2-mark MCQs and 2-mark short-answer constructions. As you master these, the AI unlocks: - 3-mark compound constructions (e.g., triangle + angle bisector). - 5-mark multi-step problems (e.g., triangle + incircle). - HOTS case studies requiring measurement, scaling, and real-world reasoning. The system never overwhelms you—it paces difficulty based on your accuracy and response time. **4. Concept-Linked Drilling:** Constructions don't exist in isolation. The AI links: - Perpendicular bisector geometry → circle constructions (points equidistant from endpoints). - Angle bisectors → incentres and inradii. - Median and altitude constructions → triangle properties and area formulas. When you make an error on a 5-mark incircle problem, the AI can trace it back to a missing angle bisector concept and drill that explicitly before re-assigning the incircle problem. **5. Board-Exam Simulation:** Our AI simulates actual CBSE exam conditions: - Mixed difficulty: 1-mark MCQ, 2-mark short-answer, 3-mark mid-length, 5-mark essay, and HOTS in a single 90-minute mock test. - Time management: The system tracks how long you spend on each construction and alerts you if you're falling behind. - Performance analytics: After each mock, you receive a detailed report showing which constructions you rushed, which you over-thought, and which concept gaps caused errors. **6. Parent Dashboard & Progress Tracking:** Parents can log in to see real-time progress: How many construction types has my child mastered? What's their accuracy on perpendicular bisectors vs. angle bisectors? When will they be board-ready? This transparency ensures both student and parent stay accountable. **Start a 3-day free trial at cbsetutor.ai** to experience adaptive construction drilling—no credit card required. See firsthand how our AI transforms hesitant compass-users into confident geometry masters.

Before your next test or board exam, tick off these essentials: ☐ **Construction of Line Segments:** - Drawing a line segment of a given length. - Constructing a line segment congruent to a given segment (using compass). - Dividing a line segment into a given ratio (using parallel line method or direct division). ☐ **Perpendicular Bisector:** - Definition: the locus of all points equidistant from the endpoints of a segment. - Construction method: equal arcs from both endpoints, intersecting above and below. - Key property: every point on it is equidistant from the endpoints. ☐ **Angle Constructions:** - Constructing a 60° angle (equilateral triangle method). - Constructing a 90° angle (perpendicular to a line or in a semicircle). - Bisecting any given angle (equal arc method). - Constructing angles of 30°, 45°, 120°, etc. (by bisecting or combining). ☐ **Triangle Constructions:** - SSS (three sides given): using compass arcs from two vertices. - SAS (two sides and included angle given): draw two sides at the given angle, then complete with the third side. - ASA (two angles and included side given): draw the base, construct angles at both ends, complete at intersection. - RHS (right angle, hypotenuse, and one side given): use semicircle method for the right angle. ☐ **Circle Constructions:** - Circle through three non-collinear points: find circumcentre (intersection of perpendicular bisectors of two sides). - Incircle of a triangle: find incentre (intersection of angle bisectors), then draw circle with inradius. - Circle with given centre and radius. - Tangent to a circle at a given point. ☐ **Proof & Justification:** - Explain *why* the perpendicular bisector is perpendicular (can include angle proofs, congruent triangles). - Justify why an inscribed angle is half the central angle (Thales' theorem context). - Verify constructions by measurement (PA = PB, angle bisector splits angle equally). ☐ **Common Pitfalls to Avoid:** - Using too small a compass radius for perpendicular bisectors (must be > half the segment). - Forgetting to extend arcs far enough so they actually intersect. - Mixing up angle bisector (∠ bisector) with perpendicular bisector (side bisector). - Measuring angles with a protractor in an exam when the construction method is expected. - Not labelling all points clearly, confusing examiners about which intersection is which. ☐ **Time Management (for 5-mark construction questions):** - Allocate 1 minute to understand the question and sketch roughly. - Allocate 3–4 minutes to the precise construction using compass and straightedge. - Allocate 0.5–1 minute to label points, verify measurements, and write any brief justification. Review these points daily for 1 week before your exam, and your Chapter 8 score will soar.