CBSE Class 10 Mathematics Constructions Notes

About This Chapter

The chapter on Constructions in Class 10 Mathematics builds on the foundational geometric construction skills developed in earlier classes. It focuses on two primary construction tasks: dividing a line segment in a given ratio, and constructing tangents to a circle from an external point. These constructions rely on the properties of similar triangles and the tangent-radius relationship studied in the chapter on Circles.

Constructions have immense real-life relevance. Architects use geometric division to design proportionate structures. Engineers apply tangent constructions when designing gear systems, pulleys, and roads. Graphic designers use similar principles for precise scaling. Understanding these constructions also deepens conceptual clarity in geometry and trigonometry.

In the CBSE Class 10 Board Examination, the Constructions chapter carries a weightage of approximately 5 marks. Questions are typically of the 2-mark or 3-mark category, requiring students to perform a construction accurately and also write the justification (proof) for the construction. Accuracy and neatness of the diagram are awarded credit.

By the end of this chapter, students will have a strong grasp of the underlying theory, step-by-step procedures, and the geometric reasoning behind each construction. The PDF study material is attached below for download and offline reference.

What You Will Learn

•         How to divide a line segment in a given ratio using similar triangles.

•         How to construct a line segment whose length is a fraction of a given segment.

•         How to draw tangents to a circle from an external point.

•         The geometric justification (proof) behind each construction.

•         How to handle special cases such as right-angle tangent constructions and internal/external division.

The PDF study material for this chapter is attached below for easy download and offline revision.

1. Introduction and Definition

Constructions in geometry refer to drawing precise geometric figures using only a ruler (straightedge) and a compass, without the use of measuring scales or protractors. This classical approach to geometry, dating back to ancient Greek mathematicians, ensures that constructions are based purely on geometric properties rather than numerical measurement.

In Class 10, the focus shifts from basic constructions such as angle bisectors and perpendicular bisectors to more advanced problems involving ratios and circles. The two main constructions covered are:

•         Division of a line segment in a given ratio (both internally and as a related fraction).

•         Construction of tangents to a circle from an external point.

Why Only a Ruler and Compass?

The restriction to ruler and compass is not a limitation but a discipline. It ensures that the constructions are theoretically exact and rely entirely on Euclid's postulates. Using a protractor or scale introduces measurement error, whereas compass-ruler constructions are provably precise. This approach also reinforces understanding of theorems such as the Basic Proportionality Theorem (BPT), properties of tangents, and the midpoint theorem.

Scope in CBSE Class 10

The CBSE syllabus for Class 10 specifies the following construction tasks: dividing a line segment in a given ratio m:n, constructing a triangle similar to a given triangle with a specified scale factor, and drawing one or two tangents from an external point to a circle. Students are also expected to write the justification, which is the geometric proof demonstrating why the construction is valid.

2. Key Concepts and Components

Basic Tools Required

All constructions in this chapter require only two instruments: a sharpened pencil, a ruler (straightedge), and a compass. The ruler is used to draw straight line segments, while the compass is used to mark arcs and transfer lengths. A sharp pencil is essential for precision.

Fundamental Concepts

Basic Proportionality Theorem (BPT)

The Basic Proportionality Theorem, also called Thales Theorem, states that if a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio. This theorem is the backbone of the line segment division construction. If DE is parallel to BC in triangle ABC, then AD/DB = AE/EC.

Similar Triangles

Two triangles are similar if their corresponding angles are equal and their corresponding sides are proportional. The concept of similarity is used to construct a triangle similar to a given triangle with a given scale factor. If the scale factor is m/n, the new triangle will have sides that are m/n times the corresponding sides of the original triangle.

Tangent to a Circle

A tangent to a circle is a line that touches the circle at exactly one point, called the point of tangency. The key property used in the tangent construction is that the radius drawn to the point of tangency is always perpendicular to the tangent at that point. This allows us to construct a right angle at the point of contact, which in turn defines the tangent direction.

Angle in a Semicircle

The angle inscribed in a semicircle is always a right angle (90 degrees). This property is used in the construction of tangents from an external point. When the midpoint of the segment joining the external point to the centre of the circle is taken as the centre, and a semicircle is drawn through the external point and the centre, it intersects the circle at the exact points of tangency.

Key Terminology

Understanding the following terms is essential for this chapter:

•         Line Segment: A part of a line bounded by two distinct endpoints.

•         Ratio: A comparison of two quantities by division, written as m:n.

•         Scale Factor: The ratio of the sides of the new figure to the corresponding sides of the original figure.

•         Point of Tangency: The single point at which a tangent touches the circle.

•         External Point: A point lying outside the boundary of the circle.

•         Arc: A portion of the circumference of a circle.

3. Core Concepts with Derivations

3.1 Division of a Line Segment in a Given Ratio

Given a line segment AB, we wish to divide it in the ratio m:n, where m and n are positive integers. The construction is based on the Basic Proportionality Theorem.

Step-by-Step Procedure

1.       Draw the given line segment AB of the required length.

2.       From point A, draw a ray AX making any acute angle with AB (angle BAX is acute).

3.       Starting from A, mark (m + n) equal points A1, A2, ..., A(m+n) on the ray AX using the compass set to any convenient radius.

4.       Join the last marked point A(m+n) to point B.

5.       Through the point Am, draw a line parallel to A(m+n)B (using compass-and-ruler parallel line construction). Let this line meet AB at point C.

6.       C is the required point that divides AB in the ratio m:n, i.e., AC:CB = m:n.

Division Formula (Internal Division):

AC : CB = m : n

Justification

In triangle ABA(m+n), the line AmC is parallel to A(m+n)B (by construction). By the Basic Proportionality Theorem, AA(m)/ A(m)A(m+n) = AC/CB. Since we have marked equal lengths, AA(m) = m units and A(m)A(m+n) = n units. Therefore, AC/CB = m/n, which means AC:CB = m:n. This proves the construction is correct.

3.2 Construction of a Similar Triangle

To construct a triangle similar to a given triangle ABC with scale factor k = m/n, we use the following procedure. The scale factor determines whether the new triangle is larger (m > n) or smaller (m < n) than the original.

When Scale Factor is Less Than 1 (m < n)

1.       Draw the given triangle ABC.

2.       From vertex A, draw a ray AX making an acute angle with AB on the side opposite to C.

3.       Mark n equal arcs on AX to get points A1, A2, ..., An.

4.       Join An to B. Through Am, draw a line parallel to AnB to meet AB at B'.

5.       Through B', draw a line parallel to BC to meet AC at C'.

6.       Triangle AB'C' is the required triangle similar to ABC with ratio m:n.

Scale Factor Relationship:

AB'/AB = AC'/AC = B'C'/BC = m/n

3.3 Tangents from an External Point

To draw tangents from an external point P to a circle with centre O and radius r, we use the angle-in-a-semicircle property.

Step-by-Step Procedure

1.       Draw the given circle with centre O and radius r.

2.       Mark the external point P outside the circle.

3.       Join OP and find its midpoint M (bisect OP using compass).

4.       With M as centre and MO (= MP) as radius, draw a semicircle.

5.       This semicircle intersects the original circle at two points, say T1 and T2.

6.       Join PT1 and PT2. These are the two required tangents from P to the circle.

Length of Tangent from External Point:

PT = sqrt(OP^2 - r^2)

Justification

The angle OT1P is inscribed in a semicircle (since OP is the diameter of the construction circle), so angle OT1P = 90 degrees. This means OT1 is perpendicular to PT1. But OT1 is the radius of the original circle, and a line perpendicular to the radius at the point of tangency is indeed a tangent. Hence, PT1 is a tangent to the circle. By symmetry, PT2 is also a tangent.

4. Solved Examples

Example 1: Divide a line segment in ratio 3:5

Problem: Divide a line segment AB = 8 cm in the ratio 3:5.

Solution:

Step 1: Draw AB = 8 cm.

Step 2: From A, draw ray AX at an acute angle. Mark 3 + 5 = 8 equal points A1 through A8.

Step 3: Join A8 to B.

Step 4: Through A3, draw a line parallel to A8B to intersect AB at C.

Result: AC = 3 cm and CB = 5 cm, so AC:CB = 3:5. Verified!

Example 2: Construct a triangle with scale factor 2/3

Problem: Construct a triangle similar to triangle ABC (AB = 6 cm, BC = 7 cm, AC = 5 cm) with scale factor 2/3.

Solution:

Step 1: Construct triangle ABC with the given dimensions.

Step 2: From A, draw ray AX at an acute angle below AB. Mark 3 equal arcs: A1, A2, A3.

Step 3: Join A3 to B. Through A2, draw a line parallel to A3B to meet AB at B'.

Step 4: Through B', draw a line parallel to BC to meet AC at C'.

Result: AB' = (2/3) x 6 = 4 cm, AC' = (2/3) x 5 = 3.33 cm, B'C' = (2/3) x 7 = 4.67 cm.

Example 3: Draw tangents from external point

Problem: Draw tangents from an external point P to a circle of radius 3 cm, where OP = 5 cm.

Solution:

Step 1: Draw circle with centre O and radius 3 cm.

Step 2: Mark external point P such that OP = 5 cm.

Step 3: Find midpoint M of OP.

Step 4: With M as centre and radius MO = 2.5 cm, draw a semicircle intersecting the circle at T1 and T2.

Step 5: Join PT1 and PT2.

Verification: PT = sqrt(5^2 - 3^2) = sqrt(25 - 9) = sqrt(16) = 4 cm.

Example 4: Divide segment internally in ratio 2:3

Problem: Draw a line segment of length 5.6 cm and divide it in the ratio 2:3 internally.

Solution:

Step 1: Draw AB = 5.6 cm.

Step 2: Draw acute ray AX. Mark 5 equal arcs to get A1 through A5.

Step 3: Join A5 to B. Through A2, draw parallel to A5B meeting AB at C.

Result: AC = 2.24 cm (= 5.6 x 2/5) and CB = 3.36 cm (= 5.6 x 3/5). Ratio AC:CB = 2:3. Verified!

Example 5: Tangent from point on circle

Problem: Draw a tangent to a circle of radius 4 cm from a point on the circle.

Solution:

Step 1: Draw circle with centre O and radius 4 cm. Mark point P on the circle.

Step 2: Draw radius OP.

Step 3: At P, construct a perpendicular to OP using compass-and-ruler method.

Result: The perpendicular line at P is the tangent to the circle at P. Note: Only one tangent exists at a point on the circle.

5. Applications and Special Cases

Applications in Real Life

The constructions studied in this chapter have numerous practical applications. In architecture, dividing a line segment in a given ratio is used to design proportional structures and facades. In mechanical engineering, tangent constructions are used when designing belt-and-pulley systems, gear trains, and cam mechanisms where a belt wraps tangentially around circular components. Road and railway engineers use tangent calculations to determine curvature of tracks.

In computer graphics and animation, similar triangle constructions are used in scaling and transformation algorithms. Cartographers use proportional division to create accurate map scales. The principles also underlie the pantograph, a mechanical device that copies drawings at a specified scale.

Special Case 1: Point at Infinity

When the external point P moves farther from the circle, the two tangents become more nearly parallel, approaching the pair of tangents from a point at infinity, which would be two parallel tangents on opposite sides of the circle.

Special Case 2: Point on the Circle

When the external point coincides with a point on the circle, only one tangent can be drawn at that point. The construction simplifies to drawing a perpendicular to the radius at the point of tangency.

Special Case 3: Point Inside the Circle

No tangent can be drawn from a point inside the circle. This is because any line from an interior point will intersect the circle at two points (becoming a secant), not touch it at exactly one point. The construction method for external points does not apply here.

Scale Factor Greater Than 1

When constructing a similar triangle with scale factor m/n where m > n, the new triangle is larger than the original. In this case, the construction extends beyond the original triangle: we extend the sides of the original triangle and locate the new vertices on the extended lines rather than within the original triangle.

6. Formula Summary

The following table summarises all key formulas and relationships used in the Constructions chapter.

Division of Line Segment AB in ratio m:n:

AC/CB = m/n    (Point C divides AB internally)

Scale Factor for Similar Triangle:

New side / Original side = m/n

Length of Tangent from External Point P (circle centre O, radius r):

PT = sqrt(OP^2 - r^2)

Angle in Semicircle (used in tangent construction):

Angle OTP = 90 degrees  (T = point of tangency)

Two Tangents from External Point (equal in length):

PT1 = PT2

Basic Proportionality Theorem (BPT) for similar triangles:

AD/DB = AE/EC  (when DE is parallel to BC in triangle ABC)

7. Key Theorems and Properties

Theorem 1: Basic Proportionality Theorem (BPT / Thales Theorem)

Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

This theorem is the foundation of the line segment division construction. In triangle ABC, if DE is parallel to BC, then AD/DB = AE/EC.

Theorem 2: Converse of BPT

Statement: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. This converse is used in the justification of the construction: we construct a parallel line, then invoke the converse to confirm the ratio is correct.

Theorem 3: Tangent-Radius Perpendicularity

Statement: The tangent at any point of a circle is perpendicular to the radius drawn to the point of tangency.

This is the core property used in the tangent construction. Since angle OTP = 90 degrees, we can use the semicircle method (Thales theorem for a semicircle) to construct this right angle.

Theorem 4: Equal Tangent Lengths

Statement: The lengths of the two tangents drawn from an external point to a circle are equal.

Proof: In triangles OT1P and OT2P, OT1 = OT2 (radii), OP is common, and angle OT1P = angle OT2P = 90 degrees. By RHS congruence, the triangles are congruent, so PT1 = PT2.

Property: Angle Between Tangents

The angle between the two tangents from an external point P and the angle subtended at the centre by the chord T1T2 are supplementary (add up to 180 degrees). That is, angle T1PT2 + angle T1OT2 = 180 degrees.

8. Common Mistakes and Exam Tips

Common Mistakes to Avoid

•         Using a protractor to draw the acute angle in line segment division -- the angle should be drawn freehand at an approximate acute angle; exact measurement is not needed.

•         Marking unequal arcs on the ray AX instead of equal arcs. Always keep the compass opening fixed when marking multiple points.

•         Forgetting to write the justification. In CBSE boards, 1 mark is typically allocated specifically for the justification/proof.

•         Drawing the parallel line through the wrong point. In m:n division, the parallel must go through the m-th point, not the n-th.

•         Confusing internal and external division. All constructions in this chapter involve internal division unless stated otherwise.

•         For tangent construction, not finding the midpoint accurately, resulting in the construction circle being too large or too small to intersect the given circle at the correct points.

Exam Tips for Maximum Marks

•         Always use a sharp pencil for clean, precise arcs and lines. Thick lines lead to inaccuracy and may result in deduction of marks.

•         Show all construction lines -- do not erase the arcs, rays, and construction marks. These are evidence of method and carry marks.

•         Label all key points clearly: A, B, C, the division point, O, P, T1, T2, and so on.

•         Write the justification step-by-step. Mention the theorem used (BPT, converse of BPT, angle in semicircle, tangent-radius perpendicularity).

•         For the tangent length verification, always apply the formula PT = sqrt(OP^2 - r^2) to cross-check your measured answer.

•         Practise constructions freehand until you can complete them within 8 to 10 minutes, as board exams are time-constrained.

9. Practice Questions

1 Mark Questions (MCQ / Very Short Answer)

1.       To divide a line segment AB in the ratio 3:4, how many equal arcs must be marked on the ray from A?  Answer: 7 arcs (3 + 4 = 7).

2.       What is the length of a tangent drawn from a point 13 cm from the centre of a circle of radius 5 cm?  Answer: PT = sqrt(169 - 25) = sqrt(144) = 12 cm.

3.       From an external point, how many tangents can be drawn to a circle?  Answer: Two.

4.       State the theorem used to justify the division of a line segment using a parallel line.  Answer: Basic Proportionality Theorem (Thales Theorem).

5.       What angle does a tangent make with the radius at the point of tangency?  Answer: 90 degrees.

6.       If the scale factor for a similar triangle construction is 5/3, is the new triangle larger or smaller than the original?  Answer: Larger (scale factor > 1).

3 Mark Questions (Short Answer)

7.       Divide a line segment PQ = 7.2 cm in the ratio 3:5. Show the full construction and write the justification.

8.       Draw a circle of radius 3.5 cm. From an external point 7 cm from the centre, draw two tangents. Measure and verify the length of each tangent.

9.       Construct a triangle ABC with AB = 5 cm, BC = 6 cm, and angle B = 60 degrees. Then construct a similar triangle with scale factor 4/5, writing the justification.

10.   A point P is at a distance of 10 cm from the centre of a circle of radius 6 cm. Calculate the length of the tangent from P to the circle and verify by construction.

11.   To divide segment AB internally in ratio m:n, a student drew only m arcs instead of (m + n) arcs. Explain the error and describe its consequence on the resulting division point.

5 Mark Questions (Long Answer)

7.       Draw a line segment AB = 9 cm. Divide it in the ratio 2:7. Write all construction steps, draw the figure, and write the complete geometric justification using BPT. Also verify by measuring the two parts.

8.       Construct a triangle with sides 4 cm, 5 cm, and 6 cm. Then construct a similar triangle whose sides are 5/4 times the corresponding sides of the first triangle. Write all steps and the justification for why the new triangle is similar to the original.

9.       Draw a circle of radius 4 cm. From an external point P, 8 cm from the centre O, construct two tangents PT1 and PT2. Measure PT1 and PT2, and verify using the formula PT = sqrt(OP^2 - r^2). Also prove that PT1 = PT2 using the RHS congruence criterion.

10.   A student is asked to divide a segment in ratio 3:2 but accidentally constructs the division in ratio 2:3. Describe the difference in the two constructions step by step, identify where the error occurs, and explain how to correct it.

11.   Construct a right triangle with legs 3 cm and 4 cm. On this triangle as base, construct a similar triangle with scale factor 3/2. Write all steps, the justification, and identify the theorem that ensures the similarity of the two triangles.