Construction: CBSE Class 9 Maths NCERT Guide

The Logic Behind Geometric Constructions

Why do we restrict ourselves to just a compass and a straightedge? This tradition goes back to ancient Greek mathematics, particularly the work of Euclid. The idea was to build all of geometry from the simplest possible axioms and tools. A straightedge lets you draw a straight line between any two points. A compass lets you draw a circle with a given center and radius. These two actions correspond to fundamental geometric postulates. By using only these, every figure you create is a direct consequence of logical deduction, not measurement approximation. For example, when you construct a 60° angle, you are actually creating an equilateral triangle, where all angles are proven to be 60°. Every arc and line you draw has a purpose, forming a visual proof. This process forces you to think about the properties of shapes, making you a stronger mathematician.

How to Construct a 90° Angle

1. Step 1: Draw a Base Ray — Draw a ray, let's call it AB. Let A be the initial point where you want to construct the 90° angle.
2. Step 2: Draw a Semicircle — With A as the center and any convenient radius, draw an arc that cuts the ray AB at a point, say C. Continue the arc to form a semicircle.
3. Step 3: Make the First Arc (60°) — With C as the center and the same radius as before, draw an arc that intersects the semicircle at a point, let's call it D. This creates a 60° angle (∠DAB = 60°).
4. Step 4: Make the Second Arc (120°) — With D as the center and the same radius, draw another arc that intersects the semicircle at a point E. This creates a 120° angle (∠EAB = 120°).
5. Step 5: Bisect the 60°-120° Arc — Now, you have the points D (representing 60°) and E (representing 120°). With D and E as centers and a radius greater than half of DE, draw two arcs that intersect each other at a point, say F.
6. Step 6: Join to Get 90° — Draw a ray from A passing through F. Let's call it AG. The angle ∠GAB is the required 90° angle. This works because you have bisected the angle between 60° and 120°, and (60° + 120°)/2 = 90°.

Worked Example: Constructing a Triangle (Sum of Sides)

• Problem: Construct a triangle ABC in which BC = 7 cm, ∠B = 75°, and 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 constructing 90° and 60°, and then bisecting the angle between them). 3. From the ray BX, cut a line segment BD equal to AB + AC = 13 cm. 4. Join D to C. 5. Now, construct the perpendicular bisector of the line segment DC. Let this bisector intersect BD at a point A. 6. Join A to C. 7. Triangle ABC is the required triangle. Justification: Point A lies on the perpendicular bisector of DC. Therefore, by the perpendicular bisector theorem, any point on it is equidistant from the endpoints D and C. So, AD = AC. From our construction, we have BD = 13 cm. We can write BD as BA + AD. So, BA + AD = 13 cm. Substituting AD = AC, we get: BA + AC = 13 cm. This proves that our construction is correct.

Exam Tip: Justification is Key!

Many students lose marks in exams because they only perform the construction steps but forget to write the justification. The justification is a short proof that explains why your construction method results in the correct figure. It connects your drawing to the geometric theorems you've learned. For example, when you bisect an angle, the justification involves proving the congruence of the two small triangles formed by your arcs (using SSS congruence). Always include two parts for every construction question: Steps of Construction and Justification.

Practice Questions with Solutions

• Q: Construct an angle of 45° at the initial point of a given ray and write the justification. A: Step 1: Construct a 90° angle (let's call it ∠XAB) using a compass and ruler as described in the process section. Step 2: With A as the center, draw an arc that intersects the rays AX and AB at points P and Q respectively. Step 3: With P and Q as centers and a radius greater than half of PQ, draw two arcs that intersect at a point R. Step 4: Join A to R. The angle ∠RAB is the required 45° angle. Justification: By joining PR and QR, we get two triangles, ΔAPR and ΔAQR. In these triangles, AP = AQ (radii of the same arc), PR = QR (arcs of equal radii), and AR is common. By SSS congruence, ΔAPR ≅ ΔAQR. By CPCTC, ∠PAR = ∠QAR. Since ∠PAQ = 90°, we have ∠RAB = 1/2 * 90° = 45°. Final answer: The constructed angle ∠RAB is 45°.
• 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 the line segment AB. Step 3: With B as the center and the same radius (5 cm), draw arcs on both sides of AB, intersecting the previous arcs at points P and Q. Step 4: Join the points P and Q. The line PQ intersects AB at a point M. Final answer: The line PQ is the required perpendicular bisector of AB. M is the midpoint of AB, and ∠PMA = 90°.
• Q: Construct a triangle PQR given QR = 6 cm, ∠Q = 60°, and PR - PQ = 2 cm. A: Step 1: Draw the base QR = 6 cm. Step 2: At point Q, construct an angle ∠XQR = 60°. Step 3: Since PR - PQ is positive, the side opposite to ∠Q (PR) is longer. Extend the ray QX downwards to QX'. Step 4: From the ray QX', cut a line segment QS = PR - PQ = 2 cm. Step 5: Join S to R. Step 6: Draw the perpendicular bisector of the line segment SR. Let it intersect the ray QX at point P. Step 7: Join P to R. Final answer: ΔPQR is the required triangle. Justification: P lies on the perpendicular bisector of SR, so PS = PR. We know QS = 2 cm. Also, QS = PS - PQ. Substituting PS=PR, we get QS = PR - PQ, which matches the given condition.
• Q: Construct a triangle ABC in which the perimeter is 12 cm and the base angles are ∠B = 60° and ∠C = 45°. A: Step 1: Draw a line segment XY of length 12 cm (equal to the perimeter AB + BC + CA). Step 2: At point X, construct an angle ∠LXY = 60° (same as ∠B). At point Y, construct an angle ∠MYX = 45° (same as ∠C). Step 3: Bisect the angles ∠LXY and ∠MYX. Let these bisectors intersect at a point A. So, ∠AXY = 30° and ∠AYX = 22.5°. Step 4: Draw the perpendicular bisector of AX. Let it intersect XY at point B. Step 5: Draw the perpendicular bisector of AY. Let it intersect XY at point C. Step 6: Join AB and AC. Final answer: ΔABC is the required triangle with perimeter 12 cm and the given base angles.

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