Constructions: CBSE Class 9 Maths NCERT Guide

Foundations: The 'Why' Behind Basic Constructions

In geometry, a 'construction' has a very specific meaning. It means creating a figure using only a straightedge and a compass. The straightedge draws straight lines, and the compass draws circles or arcs of a fixed radius. Every construction you perform is based on a solid geometric proof.

Let's explore two fundamental constructions:

1. Perpendicular Bisector of a Line Segment: When you construct a perpendicular bisector of a line segment AB, you place the compass at A and B to draw intersecting arcs above and below the line. Why does the line joining these intersection points (say P and Q) perfectly bisect AB at a 90° angle? The logic lies in triangle congruence. Any point on the arcs you draw from A is equidistant from A. Any point on the arcs from B is equidistant from B. The intersection points P and Q are therefore equidistant from both A and B. In ΔPAQ and ΔPBQ, PA = PB (arcs of same radius), QA = QB (arcs of same radius), and PQ is common. By SSS congruence, ΔPAQ ≅ ΔPBQ. This proves that ∠APQ = ∠BPQ, and by CPCTC, the line PQ bisects the angle at P. This ultimately proves that PQ is the perpendicular bisector.

1. Bisector of an Angle: To bisect an angle ∠ABC, you draw an arc from B that cuts the arms BA and BC. Then, from these new points, you draw two more intersecting arcs. The line from B through this intersection point bisects the angle. The proof again uses SSS congruence, proving that the two smaller triangles formed are congruent, which means their corresponding angles at the vertex B are equal.

Step-by-Step: How to Construct Key Angles

1. Step 1: Constructing a 60° Angle — This is the most fundamental angle construction. Draw a ray OA. With O as the center and any radius, draw an arc cutting OA at point P. Now, with P as the center and the same radius, cut the previous arc at a point Q. Join OQ and extend it to form a ray OB. The angle ∠AOB is exactly 60°. Justification: If you join PQ, the triangle ΔOPQ is an equilateral triangle because OP = OQ = PQ (all are equal to the radius used). All angles in an equilateral triangle are 60°.
2. Step 2: Constructing a 120° Angle — Start by constructing a 60° angle as described above (let the point on the arc be Q). Now, without changing the compass radius, place the compass point at Q and draw another arc that intersects the original arc at a new point, R. Join OR. The angle formed with the original line is 120°. This is simply doing the 60° construction twice (60° + 60°).
3. Step 3: Constructing a 90° Angle (Right Angle) — A 90° angle is the bisector of the angle between 60° and 120°. First, construct a 120° angle, marking the 60° point (Q) and the 120° point (R) on your initial arc. Now, with Q and R as centers and a radius more than half of the arc length QR, draw two intersecting arcs. Let the intersection point be S. Join OS. The angle formed is 90°. Justification: You have bisected the 60° angle between the 60° and 120° marks, so the angle is 60° + (120°-60°)/2 = 90°.
4. Step 4: Constructing a 45° Angle — A 45° angle is simply half of a 90° angle. First, follow all the steps to construct a 90° angle. Let the ray for the 90° angle intersect your initial arc at a point T. Your 90° angle is between the initial ray (point P on the arc) and the new ray (point T on the arc). Now, bisect the angle by placing your compass at P and T, drawing intersecting arcs, and drawing a line from the vertex through this intersection. This new angle will be 45°.

Worked Example: Constructing a Triangle (Given Base, Base Angle, and Sum of Sides)

• Problem: Construct a triangle ABC in which the base BC = 7 cm, the base angle ∠B = 75°, and the sum of the other two sides AB + AC = 13 cm.

  Steps of Construction:

  1. Draw the base BC of length 7 cm.
  2. At point B, construct an angle ∠XBC = 75°. (You can do this by first constructing 90° and 60°, and then bisecting the angle between them).
  3. From the ray BX, cut a line segment BD equal to the sum of the other two sides, i.e., BD = AB + AC = 13 cm.
  4. Join points D and C to form the line segment DC.
  5. Now, construct the perpendicular bisector of the line segment DC. Let this bisector intersect the line segment BD at a point A.
  6. Join AC.
  7. ΔABC is the required triangle.

  Justification (Why this works):
  Point A lies on the perpendicular bisector of DC. A key property of a perpendicular bisector is that any point on it is equidistant from the endpoints of the segment.
  Therefore, AD = AC.
  We constructed BD such that BD = 13 cm.
  We can also see from the figure that BD = BA + AD.
  By substituting AD = AC into this equation, we get BD = BA + AC.
  Since we constructed BD = 13 cm, it follows that BA + AC = 13 cm, which satisfies the given condition.

Exam Tips: Scoring Full Marks in Constructions

1. Do Not Erase Construction Arcs: The arcs are your 'rough work' that shows the examiner you followed the correct procedure. Faint, thin arcs are essential for full marks. A final figure without construction lines will get zero marks.

2. Use a Sharp Pencil and Tight Compass: A blunt pencil or a loose compass hinge will cause inaccuracies. This can lead to lines not meeting correctly or angles being slightly off, resulting in marks being deducted. Always use an HB pencil sharpened to a fine point.

3. Label Everything Clearly: Label all points (A, B, C), rays (BX), and measurements (7 cm, 13 cm) as you go. This makes your diagram easy to understand and helps you write the steps of construction correctly.

4. Write Steps of Construction: Unless specified otherwise, you must write down the steps you took to create the final figure. List them sequentially and be precise (e.g., "With B as center and radius 13 cm, cut an arc on ray BX at D").

5. Understand the Justification: Even if not asked, understanding why a construction works helps you remember the steps. For complex triangle constructions, the justification is usually based on SSS or RHS congruence criteria.

Practice Questions with Solutions

• Q: Construct an angle of 75° using a ruler and compass and write the steps of construction.
  A: Step 1: Draw a ray OA. Construct a 90° angle (∠XOA) and a 60° angle (∠YOA) on this ray.
  Step 2: The rays OX (for 90°) and OY (for 60°) now form a 30° angle between them.
  Step 3: Bisect this 30° angle. Place the compass on the points where the 60° and 90° lines cut the initial arc. Draw two intersecting arcs.
  Step 4: Draw a ray OB from O through this intersection point. The angle ∠BOA is 75° (since 60° + 15° = 75°).
  Final answer: The angle ∠BOA is the required 75° angle.

• Q: Construct the perpendicular bisector of a line segment of length 8.4 cm.
  A: Step 1: Draw a line segment AB of length 8.4 cm.
  Step 2: With A as the center and a radius more than half of AB (e.g., 5 cm), draw arcs on both sides of AB.
  Step 3: With B as the center and the same radius, draw arcs that intersect the previous arcs at points P and Q.
  Step 4: Join P and Q. The line PQ intersects AB at a point M. The line PQ is the required perpendicular bisector of AB.
  Final answer: The line PQ is the perpendicular bisector of AB, and M is the midpoint of AB.

• Q: Construct a triangle PQR where QR = 6 cm, ∠Q = 60°, and PR - PQ = 2 cm.
  A: Step 1: Draw the base QR of length 6 cm.
  Step 2: At point Q, construct an angle of 60° and extend the ray downwards to form a line XQY (since PR > PQ, the arc is cut below the base).
  Step 3: From the ray QY, cut a line segment QS = PR - PQ = 2 cm.
  Step 4: Join S and R.
  Step 5: Construct the perpendicular bisector of the line segment SR. Let it intersect the ray QX at point P.
  Step 6: Join P and R. ΔPQR is the required triangle.
  Final answer: ΔPQR is the constructed triangle with the given conditions.

• Q: Construct an equilateral triangle with a side length of 5.5 cm.
  A: Step 1: Draw a line segment AB of length 5.5 cm.
  Step 2: With A as the center and radius 5.5 cm, draw an arc.
  Step 3: With B as the center and the same radius 5.5 cm, draw another arc that intersects the first arc at point C.
  Step 4: Join AC and BC.
  Final answer: ΔABC is the required equilateral triangle with all sides equal to 5.5 cm.

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