Constructions: Full Chapter Notes With Step-by-Step Diagrams

Constructions is the one chapter in Class 10 Maths where the answer is a drawing, and where a perfect-looking drawing can still lose marks if the steps and the justification are missing. It is also a chapter that looks harder than it is, because there are only three constructions to learn, and each rests on a theorem you already know.

This guide takes the three in turn, with the steps, a sketch where it helps, and the reason each one works. It also states clearly where this chapter stands: Constructions is part of the Karnataka SSLC syllabus, while CBSE has removed it from Class 10, so CBSE students can use it as an optional geometry skill. The reasoning in every construction leans on similar triangles, the basic proportionality theorem and the tangent to a circle, so a quick look at those chapters first will make everything below easier.

What You Need and How to Work

You need a sharp pencil, a ruler, a compass, a protractor and a set square. Three habits keep the marks safe.

Draw lightly, and darken only the final figure.
Do not rub out the construction arcs. They show how the figure was obtained.
Label every point, and write the steps of construction in order, followed by the justification where the question asks for it.

Construction One: Dividing a Line Segment in a Given Ratio

Problem. Divide a line segment AB = 6 cm in the ratio 3 : 2.

Steps

Draw AB = 6 cm.
Draw a ray AX making an acute angle with AB.
On AX, mark 5 points A₁, A₂, A₃, A₄, A₅ at equal distances, using the compass with the same opening each time. Five comes from 3 + 2.
Join A₅ to B.
Through A₃, draw a line parallel to A₅B by copying the angle ∠AA₅B at A₃. Let it meet AB at C.

Why it works

In triangle AA₅B, the line A₃C is parallel to A₅B. By the basic proportionality theorem, AC/CB = AA₃/A₃A₅. Since the marked points are equally spaced, AA₃ = 3 units and A₃A₅ = 2 units, so AC : CB = 3 : 2. The theorem is explained in detail in this guide to the common proof-writing mistakes in the basic proportionality theorem.

Rule of thumb: for the ratio m : n, mark m + n equal points, and draw the parallel through the m-th point.

Construction Two: A Triangle Similar to a Given Triangle

Problem. Construct a triangle similar to a given triangle ABC with sides equal to 3/4 of the corresponding sides of triangle ABC.

Steps for a scale factor less than 1 (3/4)

Draw the given triangle ABC, with BC as the base.
Draw a ray BX making an acute angle with BC on the side opposite to the vertex A.
Mark 4 equal points B₁, B₂, B₃, B₄ on BX. Four is the larger of 3 and 4.
Join B₄ to C.
Through B₃, draw a line parallel to B₄C, meeting BC at C′.
Through C′, draw a line parallel to CA, meeting BA at A′.

Triangle A′BC′ is the required triangle.

Steps for a scale factor greater than 1 (5/3)

Draw the triangle and the ray BX as before.
Mark 5 equal points on BX. Five is the larger of 5 and 3.
Join B₃ to C. Here 3 is the smaller number.
Through B₅, draw a line parallel to B₃C, meeting the extended line BC at C′.
Through C′, draw a line parallel to CA, meeting the extended line BA at A′.

Triangle A′BC′ is similar to ABC with sides 5/3 times as long.

Scale factor	Points to mark on BX	Which point to join to C	Which point to draw the parallel through
m/n with m < n (smaller)	n	The n-th point	The m-th point
m/n with m > n (larger)	m	The n-th point	The m-th point

Why it works

By the construction, BC′/BC equals BB₃/BB₄ = 3/4, and since C′A′ is parallel to CA, triangles A′BC′ and ABC have equal angles. By the AA criterion they are similar, so A′B/AB = BC′/BC = A′C′/AC = 3/4. You can read more about the ideas in similar triangles in real life.

Construction Three: Tangents From a Point Outside a Circle

Problem. Draw a circle of radius 3 cm with centre O. Take a point P at 7 cm from O, and draw the two tangents from P to the circle.

Steps

Draw the circle with centre O and radius 3 cm. Mark P with OP = 7 cm.
Join OP, and bisect it. Let M be its midpoint.
With M as centre and MO as radius, draw a circle. It passes through both O and P.
Let this circle cut the given circle at Q and R.
Join PQ and PR. These are the tangents.

Why it works

OP is a diameter of the circle drawn with centre M, so the angle in a semicircle, ∠OQP, is 90°. Thus PQ is perpendicular to the radius OQ at Q, and a line perpendicular to the radius at its end on the circle is a tangent. The same argument works for PR.

A check by calculation

The length of each tangent is PQ = √(OP² − OQ²) = √(7² − 3²) = √40, which is about 6.3 cm. Measure your tangent with a ruler. If it is close to 6.3 cm, the construction is right. For the properties behind this, see circles and tangent properties, with common proof mistakes.

Summary of the Three Constructions

Construction	Key idea	Theorem it rests on
Dividing a segment in a ratio	Equally spaced points on a ray, and a parallel line	Basic proportionality theorem
A similar triangle	The same trick on one side, then a second parallel	Basic proportionality theorem and AA similarity
Tangents from an outside point	A circle on OP as diameter	Angle in a semicircle is 90°, and tangent ⟂ radius

Slips That Cost Marks

Marking the equal points at different compass openings, so they are not equally spaced.
Drawing the ray BX on the same side as vertex A for a similar triangle, which makes the construction awkward.
Joining the wrong point to C, for example B₄ when the scale factor needs B₃.
Erasing the arcs, which removes the evidence of the steps.
Leaving out the steps of construction or the justification when the question asks for them.

Constructions in Your Syllabus

In the Karnataka SSLC Maths textbook, Constructions is a full chapter, and published listings show it carrying marks in the paper, so confirm the weightage with the latest KSEAB blueprint. CBSE has removed the chapter from Class 10 for 2026-27, according to published deleted-syllabus lists. For practice, use the chapter MCQs on constructions and the important questions on constructions. These ideas return in later geometry, but they are not a separate chapter in Class 11, so the skill is mostly about neatness and reasoning now.

Frequently Asked Questions

Q1. How do you divide a line segment in the ratio m : n?

Draw a ray from one end at an acute angle, mark m + n equal points on it, join the last point to the other end of the segment, and draw a line parallel to this through the m-th point. It meets the segment at the required point.

Q2. Which theorem justifies the division of a line segment?

The basic proportionality theorem, also called Thales’ theorem: a line parallel to one side of a triangle divides the other two sides in the same ratio.

Q3. How do you construct a triangle similar to a given triangle?

Draw a ray from one vertex, mark equal points on it according to the scale factor, join the appropriate point to the other end of the base, and draw parallels to get the new base and then the new vertex.

Q4. How are tangents drawn from an external point?

Join the point to the centre, bisect that line, draw a circle with the segment as diameter, and join the external point to the two points where the circle meets the given circle.

Q5. Is Constructions in the CBSE Class 10 syllabus?

Published deleted-syllabus lists show it removed from CBSE Class 10 for 2026-27, while published Karnataka SSLC listings still include it. Confirm with your own board’s latest syllabus.

A Drawing Is an Argument

Every construction is a small proof that has been drawn instead of written. Know the theorem behind each one, and the steps stop being a list to memorise and become a story you can retell.