Exercise 13.1 — Basic Constructions

Basic constructions of angles.

Lesson Notes PDF

1 / —

Loading PDF…

Class 9 Mathematics · Chapter 13

Geometrical Constructions — Exercise 13.1

Complete step-by-step construction procedures and proofs for constructing angles of 90°, 45°, 30°, 22½°, 15°, 75°, 105°, 135°, an equilateral triangle and an isosceles triangle — using only a ruler and compass. Aligned with CBSE, Telangana, and Andhra Pradesh Class 9 syllabuses.

CBSE Class 9 Telangana Board Andhra Pradesh Board Chapter 13 · Ex 13.1

What is Geometrical Construction?

Geometrical construction means drawing precise geometric figures — angles, triangles, perpendiculars, bisectors — using only two tools: a ruler (unmarked straightedge) and a compass. No protractor is used for construction. A protractor may be used only to verify the result after the construction is complete.

The key idea is that a compass can replicate a fixed radius repeatedly and precisely. By exploiting the fact that equal arcs of the same circle subtend equal angles, you can build any angle that is a multiple or fraction of 60° using just these two tools.

📐 The Master Angle: Everything in this exercise grows from one seed — the 60° angle. When you draw three arcs of equal radius from O, P, and Q in sequence, you automatically create two equilateral triangles, giving you 60° for free. Bisecting it gives 30°, bisecting again gives 15°, and stacking 60° + 30° gives 90°, and so on. Every construction here is a variation of this single idea.

The foundation of all constructions: equal arcs create an equilateral △OPQ, giving ∠POQ = 60°.

📊 How Every Angle in This Exercise is Built from 60°

60°=One arc step → equilateral triangle angle

30°=½ × 60° → bisect the 60° angle

90°=60° + 30° → second equilateral triangle + bisect ∠QOR

45°=½ × 90° → bisect the 90° angle

15°=½ × 30° → bisect the 30° angle

22½°=½ × 45° → bisect the 45° angle

75°=60° + 15° → third arc step + bisect ∠QOM

105°=90° + 15° → use 90° region + bisect ∠ROM

135°=90° + 45° → four arc steps + two bisections

Question 1(a) — Constructing 90°

This is the most important construction in the chapter. The 90° angle is built by creating two consecutive equilateral triangles on the same ray and then bisecting the second one.

Q1(a) · 90°

Construct a 90° angle at the initial point of a ray and justify.

O–P–Q–R are equally spaced arcs; M bisects ∠QOR; ∠AOM = 90°.

1Draw ray OA with initial point O. With O as centre, draw an arc of fixed radius that cuts OA at P.
2With same radius and P as centre, draw an arc cutting the previous arc at Q. (This creates equilateral △OPQ, so ∠POQ = 60°.)
3With same radius and Q as centre, draw an arc cutting the previous arc at R. (This creates equilateral △OQR, so ∠QOR = 60°.)
4With Q and R as centres and same radius, draw two arcs that intersect at M. (OM bisects ∠QOR, giving ∠QOM = 30°.)
5Join O and M. Draw ray OM.
6∠AOM = 90° is the required angle.

In △OPQ: OP = PQ = OQ (same radius) → △OPQ is equilateral
∴ ∠POQ = 60°

Each angle of equilateral △ = 60°

In △OQR: OQ = QR = OR (same radius) → △OQR is equilateral
∴ ∠QOR = 60°

Each angle of equilateral △ = 60°

OM bisects ∠QOR → ∠QOM = ∠MOR = 60°/2 = 30°

Construction of angle bisector

∠POM = ∠POQ + ∠QOM = 60° + 30° = 90°

Angle addition

∴ ∠AOM = ∠POM = 90° ✓

Proved

Question 1(b) — Constructing 45°

Q1(b) · 45°

Construct a 45° angle at the initial point of a ray and justify. (45° = ½ × 90°)

OR bisects ∠POQ (→ 30°); ON bisects ∠ROQ (→ 15°); ∠PON = 30° + 15° = 45°.

1Draw ray OA; with O as centre draw an arc cutting OA at P.
2Same radius, P as centre → arc cuts at Q. (∠POQ = 60°.)
3Draw ray OR bisecting ∠POQ. → ∠POR = ∠ROQ = 30°.
4Draw ray ON bisecting ∠ROQ. → ∠RON = ∠NOQ = 15°.
5∠PON = ∠POR + ∠RON = 30° + 15° = 45° is the required angle.

△OPQ equilateral → ∠POQ = 60°

Equal arcs

OR bisects ∠POQ → ∠POR = ∠ROQ = 30°

Angle bisector construction

ON bisects ∠ROQ → ∠RON = ∠NOQ = 15°

Angle bisector construction

∠PON = 30° + 15° = 45° ✓

Angle addition

Question 2 — Six More Angles Using Ruler and Compass

For each of the six angles below, the pattern is always the same: start with arcs of equal radius from O and mark intersection points, then bisect as needed. Verify each result with a protractor after construction.

Q2(a) · 30°

Construct 30° | 30° = ½ × 60°

1Draw ray OA; arc from O cuts OA at P.
2Same radius, P as centre → arc cuts at Q. Now ∠POQ = 60°.
3Draw ray OR bisecting ∠POQ. → ∠POR = 30°.
4∠POR = 30° is the required angle.

△OPQ equilateral → ∠POQ = 60°

Equal arcs

OR bisects ∠POQ → ∠POR = 30° ✓

Angle bisector

Q2(b) · 22½°

Construct 22½° | 22.5° = ½ × 45°

1Draw ray OA; arc from O cuts OA at P.
2Same radius, P as centre → arc cuts at Q. (∠POQ = 60°.)
3Draw ray OR bisecting ∠POQ → ∠POR = ∠ROQ = 30°.
4Draw ray ON bisecting ∠ROQ → ∠RON = ∠NOQ = 15°. (So ∠PON = 45°.)
5Draw ray OL bisecting ∠PON → ∠POL = 22½°.
6∠POL = 22½° is the required angle.

✅ Key logic: 22½° = ½ × 45° = ½ × (30° + 15°). The three successive bisections each halve the available region until we reach 22.5°.

Q2(c) · 15°

Construct 15° | 15° = ½ × 30°

1Draw ray OA; arc from O cuts OA at P.
2Same radius, P as centre → arc cuts at Q. (∠POQ = 60°.)
3Draw ray OR bisecting ∠POQ → ∠POR = 30°.
4Draw ray ON bisecting ∠POR → ∠PON = 15°.
5∠PON = 15° is the required angle.

∠POQ = 60° → OR bisects → ∠POR = 30°

Equilateral △ + bisection

ON bisects ∠POR → ∠PON = 15° ✓

Second bisection

Q2(d) · 75°

Construct 75° | 75° = 60° + 15°

1Draw ray OA; arc from O cuts OA at P.
2Same radius, P as centre → arc cuts at Q. (∠POQ = 60°.)
3Same radius, Q as centre → arc cuts at R. (∠QOR = 60°, so ∠POR = 120°.)
4Draw ray OM bisecting ∠QOR → ∠QOM = ∠MOR = 30°. (So ∠POM = 90°.)
5Draw ray OL bisecting ∠QOM → ∠QOL = 15°. So ∠POL = 60° + 15° = 75°.
6∠POL = 75° is the required angle.

💡 Breakdown: 75° = 60° (from △OPQ) + 15° (quarter of the second 60° segment between Q and R).

Q2(e) · 105°

Construct 105° | 105° = 60° + 45° = 90° + 15°

1Draw ray OA; arc from O cuts OA at P.
2Same radius, P as centre → arc cuts at Q. (∠POQ = 60°.)
3Same radius, Q as centre → arc cuts at R. (∠POR = 120°.)
4Draw ray OM bisecting ∠QOR → ∠QOM = 30°, so ∠POM = 90°.
5Draw ray OL bisecting ∠ROM → ∠ROL = 15°. So ∠POL = 90° + 15° = 105°.
6∠POL = 105° is the required angle.

💡 Breakdown: 105° = 90° + 15°. After constructing the 90° ray (OM), the remaining gap between OM and OR is 30°. Bisecting it gives 15°, which is added to 90°.

Q2(f) · 135°

Construct 135° | 135° = 90° + 45°

1Draw ray OA; arc from O cuts OA at P.
2Same radius, P as centre → arc cuts at Q. (∠POQ = 60°.)
3Same radius, Q as centre → arc cuts at R. (∠POR = 120°.)
4Same radius, R as centre → arc cuts at K. (∠POK = 180° — K is on the opposite ray.)
5Draw ray OM bisecting ∠KOR → ∠KOM = 30°, so ∠POM = 150°.
6Draw ray OL bisecting ∠ROM → ∠ROL = 15°. So ∠POL = 120° + 15° = 135°.
7∠POL = 135° is the required angle.

💡 Breakdown: 135° = 90° + 45°. Four arc steps reach 180° (a straight line). Working back from K (180°) and R (120°), bisecting the 60° gap between them gives 30°, and bisecting that 30° region at the right place gives 135°.

Question 3 — Constructing an Equilateral Triangle (Side = 4.5 cm)

Question 3

Construct an equilateral triangle with side 4.5 cm and justify the construction.

Two arcs of radius 4.5 cm from A and B intersect at C — all sides equal 4.5 cm.

1Draw line segment AB = 4.5 cm.
2With A and B as centres and radius = 4.5 cm, draw two arcs above AB that intersect at C.
3Join A to C and B to C.
4△ABC is the required equilateral triangle with all sides = 4.5 cm.

By construction: AC = AB = 4.5 cm → ∠ABC = ∠ACB ...(1)

Angles opp. equal sides are equal

By construction: BC = AB = 4.5 cm → ∠BAC = ∠BCA ...(2)

Angles opp. equal sides are equal

From (1) & (2): ∠A = ∠B = ∠C → AC = BC = AB = 4.5 cm

All angles equal → all sides equal

∴ △ABC is an equilateral triangle ✓

Proved

Question 4 — Constructing an Isosceles Triangle (Base 6 cm, Base Angle 45°)

Question 4

Construct an isosceles triangle given its base (6 cm) and base angle (45°) and justify.

∠BAX = ∠ABY = 45°; rays AX and BY meet at C; AC = BC (isosceles).

1Draw line segment AB = 6 cm.
2At A, construct ray AX such that ∠BAX = 45° (using the 45° construction from Q1(b)).
3At B, construct ray BY such that ∠ABY = 45° on the same side.
4Mark the intersection of rays AX and BY as C.
5△ABC is the required isosceles triangle.

∠A = ∠B = 45° (by construction)

Both base angles equal

In △ABC: ∠A = ∠B → BC = AC

Sides opposite equal angles are equal

∴ △ABC is an isosceles triangle (AC = BC) ✓

Proved

All Constructions at a Glance

Angle / Figure	Built From	Arc Steps	Bisections	Formula
90°	60° + 30°	3	1 (bisect 2nd 60°)	60 + ½×60
45°	½ × 90°	2	2	30 + 15
30°	½ × 60°	2	1	½ × 60
22½°	½ × 45°	2	3	½ × (30+15)
15°	½ × 30°	2	2	¼ × 60
75°	60° + 15°	3	2	60 + ¼×60
105°	90° + 15°	3	2	90 + ¼×60
135°	90° + 45°	4	2	120 + ¼×60
Equilateral △	Side = 4.5 cm	2 arcs	—	3 equal sides
Isosceles △	Base 6 cm, ∠ = 45°	2 rays at 45°	—	∠A = ∠B → AC = BC

Common Mistakes to Avoid

Changing the compass radius mid-construction: Every arc in an angle construction must be drawn with the same fixed radius. Changing it breaks the equilateral triangle logic and produces a wrong angle.
Not stating "equilateral triangle" in the proof: The justification for ∠POQ = 60° specifically requires writing that "OP = PQ = OQ (same radius) → △OPQ is equilateral → ∠POQ = 60°." Skipping this step loses the proof mark.
Bisecting the wrong angle for 75°, 105°, 135°: These three are the trickiest because there are multiple arcs and it is easy to bisect from the wrong pair of points. Always re-identify which 60° region you are working in before bisecting.
Not using the compass to bisect — using the protractor instead: In construction questions, you must use ruler and compass only. Using a protractor to draw the bisector invalidates the construction method.
Forgetting to write the required angle at the end: Always conclude with "∴ ∠___ = ___° is the required angle." Board exams penalise incomplete statements.

⛔ Board Exam Alert (CBSE / Telangana / AP): Construction questions carry a justification mark separate from the construction mark. Drawing the correct figure earns 2 marks; writing the proof (using equilateral triangle + angle bisector logic) earns the remaining marks. Never skip the "Proof" section even if the diagram looks right.

What This Exercise Prepares You For

The ruler-and-compass constructions in Exercise 13.1 underpin the rest of Chapter 13, where you will construct triangles given three pieces of information (SAS, ASA, SSS, RHS conditions). The angle bisector technique you practised here reappears directly in the perpendicular bisector construction and in the construction of triangles with given altitudes and medians.

In Class 10, these constructions evolve into division of a line segment in a given ratio and the tangent-from-an-external-point construction — both of which require confident compass work built on the habits from this exercise. You can explore related geometry proofs in Chapter 7 (Triangles) and revisit angle properties in Chapter 8 (Quadrilaterals).

📌 Exam Tip: In Telangana and Andhra Pradesh board papers, construction questions are typically 4–5 marks each. The split is usually 3 marks for the correct diagram with all arcs and labels visible, and 1–2 marks for the written justification. Never erase your construction arcs — they are evidence of your method and are checked by the examiner.