Constructions

Why This Chapter Matters

Construction is geometry turned into action. Instead of proving that a point exists, you actually create it using allowed tools.

The key idea is locus even when the word is not emphasised. A compass arc represents all points at a fixed distance from its centre. The intersection of two arcs therefore finds a point satisfying two distance conditions at once.

The advanced triangle constructions in Class 9 look difficult because the data are unusual—such as the sum or difference of two sides. The trick is to convert that unusual information into an auxiliary segment, construct a simpler figure, and then recover the required triangle.

Bridge from Earlier Classes

Before starting, be comfortable with:

• Ruler and compass use
• Basic triangle construction
• Angle bisector
• Perpendicular bisector
• Triangle inequality

These are the tools this chapter reuses and deepens.

1. Why Compass Arcs Work

A circle consists of all points at the same distance from its centre. A compass arc therefore encodes a distance condition.

If two arcs intersect, their intersection point is at the required distance from both centres. This is the logic behind SSS triangle construction, perpendicular bisectors and many advanced steps.

2. Perpendicular Bisector

To construct the perpendicular bisector of AB, draw equal-radius arcs from A and B with radius greater than half of AB. Join the two arc intersections.

Every point on this line is equidistant from A and B. The construction works because the intersection points form congruent triangles with A and B.

3. Angle Bisector and Standard Angles

To bisect an angle, draw an arc cutting both arms, then equal arcs from those cut points. Joining the original vertex to the new arc intersection creates two congruent triangles and hence equal angles.

Angles such as 60°, 90°, 45° and 30° can be built from equilateral triangles, perpendiculars and repeated bisection.

4. Triangle with Base, Base Angle and Sum of Other Two Sides

Suppose base BC, angle B and AB+AC are given. On the ray from B making the required angle, mark D so that BD equals the given sum.

Now choose A on BD so that AD=AC, usually by using the perpendicular bisector of CD. Then BA+AC = BA+AD = BD, exactly matching the given sum.

The important insight is that an awkward sum becomes one measurable auxiliary segment.

5. Triangle with Base, Base Angle and Difference of Other Two Sides

The difference construction uses a similar transformation, but the auxiliary segment is chosen to encode one side minus the other.

You must first identify which of the two sides is larger because the geometric arrangement depends on the sign of the difference. A careful rough sketch prevents constructing the wrong case.

6. Triangle from Perimeter and Two Base Angles

When perimeter and two base angles are given, begin with an auxiliary segment equal to the full perimeter. Suitable angle constructions and bisectors then locate the actual base endpoints and recover the triangle.

Again the deeper pattern is transformation: convert an unusual set of data into standard lengths and angles that ruler-and-compass methods can handle.

7. Writing a Construction in the Exam

A complete answer normally includes a rough sketch, steps of construction and, where requested, justification.

Do not erase every arc. Arcs show how a point was obtained. Use a sharp pencil, label points consistently and avoid measuring from the rough sketch.

Visualise the Core Idea

Advanced constructions become manageable when unusual data is converted into an auxiliary segment.

Worked Examples — Follow the Reasoning

How to Think in the Exam

Common Mistakes

Quick Revision

Focused Practice

Final Chapter Summary