Constructions Class 10 Maths Notes Chapter 11

Chapter 11 of the NCERT Class 10 Maths textbook delves into the world of Construction and covers various topics such as dividing line segments in a given ratio, without measuring with scale, drawing similar ratio triangles without measuring and drawing tangents to a circle from an external point. The chapter also provides the reason why that particular construction works. These notes are designed to give students a comprehensive summary of the entire chapter and include all the essential topics, and concepts needed to succeed in their exams.

Chapter 11 of NCERT Class 10 Maths deals with geometrical constructions using a compass and ruler. It focuses on creating specific geometrical shapes with given properties. The chapter provides step-by-step instructions with diagrams to guide you through each construction process. It also explains the reasoning behind the constructions, helping you understand the underlying geometrical principles.

What is Construction in Maths?

Construction is the drawing of various objects, with high efficiency. Basic geometry and angles can be made with the help of constructions, and by using different properties of lines, triangles, and circles, advanced geometry can also be drawn, which, in turn, helps develop highly efficient drawings of various objects. 

Prerequisites

Before starting the construction part, we need to understand some basic properties of triangles, and circles, along with some terms, which will be used while construction, and proving the construction.

Constructions in Geometry

Geometry constructions are the process of drawing shapes, angles, or lines using mathematical instruments like a compass, straightedge, and pencil.

Types of Geometric Constructions

The six basic constructions are copy a line segment, copy an angle, create a perpendicular bisector, create an angle bisector, create parallel lines and create a perpendicular line through a given point.

• Angle Bisector: A line that cuts an angle in half exactly, always going through the corner point (vertex). These cuts are important for solving geometry problems.
• Perpendicular Bisector: A way to create a new four-sided shape (quadrilateral) from another one by drawing lines exactly perpendicular to each side of the original shape.
• Tangnt a circle: A line that comes from outside a circle and meets the circle’s flat surface (plane) at only one exact spot. This meeting point is called the touch point.
• Parallel Lines: Two lines on the same flat surface (plane) that will always stay the same distance apart and never cross.
• Copying an Angle: Making a new angle that has the same size (measure) as another angle, but in a different location.
• Cutting an angle in half: Dividing an angle into two parts that are exactly the same size.

Basic Proportionality Theorem (BPT)

BPT states that, if a line is drawn parallel to the side of the triangle, then, the line cuts the other two sides of the triangle into equal proportion. For example, in △ABC, a line is drawn parallel to BC and cuts side AB at D, and BC at D. The ratio of AD, and DB, is equal to the ratio of AE, and EC. 

AD/DB = AE/EC

Angle in a Semicircle is 90°

The angle in a semicircle is a right angle. Consider a circle, with centre O, AB is the diameter of the circle. A point C, lies on the circle, on joining, A, B, and C, a right-angle triangle is formed, where, ∠ACB = 90°. 

Scale Factor 

It represents the ratio of the sides of the triangle to be constructed, with respect to the sides of the given triangle. For example, a given triangle ABC, for which a new triangle AB`C` is constructed. Here, the ratio of AB` and AB is the scale factor, which tells us by what ratio the sides of the new triangle scale with respect to a given triangle.

Constructions – Class 10 Maths

In this chapter, we have three different types of constructions that are:

• Dividing a Line in a given Ratio 
• Construct a Similar Triangle with the given Scale Factor 
• Construct Tangents of a Circle from an External Point 

Let’s learn how to learn these constructions in detail as follows:

Dividing a Line in a Given Ratio 

It is the most asked question, in the exam. We can divide a line in any ratio without measuring them with the scale, with the help of the construction. In general, we are given a line segment, and we have to divide the line into m:n ratio. The concept could better be understood with an example. Given a line segment, divide the line in the ratio of 4:5. 

The following are the steps of construction: 

Step 1: Draw a line segment AB. Draw a ray AX, while making an acute angle with line segment AB. 

Step 2: Divide the ray AX into 5 parts, such that, each part is of equal length. Keep the compass at position A, open a small length of the compass, and mark an arc on ray AX, and name that point A1. With the same length open, keep the tip of the compass on A1, and mark an arc on the ray AX again, name that point A2, similarly Mark 5 points on the ray AX i.e. A1, A2, A3, A4, A5. 

Dividing a Line in a given Ratio

Step 3: Now, join points A5 and point B. As, we want to divide the line in 4:5 ratio, at the fourth point A4, draw a line parallel to A₅B. To draw a parallel line, make your compass open to an angle equal to BA₅A, draw the same angle at point A4, and draw the parallel line to A₅B. The line cuts AB at C. The ratio of line AC: CB = 4:5. 

How does this Work?

Now, we will understand the mathematical reason why such a construction works. Line AA₅ is divided into a 4:5 ratio by AA₄ and AA₅. In triangle ABA₅, 

AA₄/A₄A₅ = AC/CB  (By BPT theorem)

⇒ AC/CB = 4/5

Construct a Similar Triangle with the Given Scale Factor 

As explained earlier scale factor is the ratio of the corresponding sides of the triangle to be constructed with respect to the given triangle. The general statements for such problems are that, we are given a triangle, and we need to construct a triangle of scale factor m:n. For example, construct a triangle similar to the given triangle ABC, with a scale factor of 4:5. 

The following are the steps of the construction: 

Step 1: Draw a triangle ABC. Draw a ray BX, while making an acute angle with line segment BC.

Step 2: Divide the ray BX into 5(greater one 4 or 5) parts, such that, each part is of equal length. Keep the compass at position B, open a small length of the compass, mark an arc on ray BX, and name that points B1. With the same length open, keep the tip of the compass on B1, and mark an arc on the ray BX again, name that point B2, and similarly mark 5 points on the ray BXi.e. B1, B2, B3, B4, B5. 

Step 3: Now, join point B5 and point C. As, we want to divide the triangle in 4:5 ratio, at the fourth point B4, draw a line parallel to B5C. To draw a parallel line, make your compass open to an angle equal to BB5C, draw the same angle at point B4, and draw the parallel line to B5C. The line cuts BC at C`. 

Construction of Similar Triangle with the given Scale Factor

Step 4: As, we want to divide the triangle in a 4:5 ratio, at the fourth point C, draw a line parallel to AC. To draw a parallel line, make your compass open to an angle equal to BCA, draw the same angle at point C`, and draw the parallel line to AC. The line cuts AB at B`. △B`BC` is the required triangle. 

How does it work?

Now, we will understand the mathematical reason why such a construction works. In triangle ABA5, 

BC`/C`C` = BB₄/BB₅  (By BPT theorem)

In triangle ABC, 

BB`/B`A = BC`/C`C  (By BPT theorem),

BB`/B`A = BC`/C`C = 4/5

Construct Tangents of a Circle from an External Point

As studied in the previous chapters, from an external point of a circle, exactly two tangents can be drawn. The general statements for such problems are, Given a circle of radius r, with a point outside the circle, with distance x units from the centre, construct tangents from that point. For example, Given a circle of radius 4cm, with a point outside the circle, with a distance of 10 units from the centre, construct tangents from that external point. 

The following are the steps for the construction: 

Step 1: Draw a circle of radius 4cm, with centre O. An external point P, is marked at a distance of 10 cm, from the centre O. The line cuts the circle at point Q. 

Step 2: Now, draw a perpendicular bisector of line segment OP. To draw a perpendicular bisector, keep your compass at point P, open the compass with a length less than OP, and mark an arc, on both the above and below the horizon, repeat the same with point O.  Now, join the line where the two arcs, have cut each other. The perpendicular bisector will cut the line segment OP, at point M. 

Step 3: Now, take the centre as M, and the radius as OM, and draw a circle with it. The circle with centre M, cuts the circle with centre O, at two points, name C, and D. Now, join C, and P, and also join P, and D. The tangents CP, and PD are successfully constructed. 

Construction of Tangents of a Circle from an External Point

How does it work?

To understand why, it works, draw a dotted line, joining points C, and O. Now, ∠PCO = 90°, because the angle formed is a part of the semicircle with centre M, and radius MO. Also, we have studied in previous chapters, the angle formed between the tangent and the radius of the circle is a right angle. Hence, it is confirmed at PC, and PD, is the tangent to the circle, with centre O.

Reason why Construction of Tangent Works

NCERT Solutions Class 10 Maths Chapter 11 Constructions

The NCERT Solutions for Class 10 Maths Chapter 11 – Constructions offer comprehensive assistance to students navigating the intricacies of geometric constructions. Despite the removal of this chapter from the CBSE Syllabus 2023-24, these solutions remain invaluable resources for students seeking clarity and proficiency in geometric constructions. Crafted under the guidance of subject experts and following NCERT guidelines, these solutions provide step-by-step explanations, aiding students in mastering construction techniques essential for their academic journey.

Covering diverse topics such as the division of line segments, construction of tangents to circles, and angle bisectors, these solutions foster a deeper understanding of geometric principles. By offering detailed insights and logical justifications for each construction, they empower students to not only solve problems effectively but also comprehend the underlying concepts. With these resources at hand, students can enhance their problem-solving skills and approach geometric constructions with confidence, thus enriching their learning experience in mathematics.

As per the lates NCERT rationalisation exercise, the chapter on “Constructions” in class 10th Maths has been removed and you won’t find it in regular NCERT textbooks for class 10 maths. You can still Download Chapter 11 Constructions PDF for class 10 mathematics.

FAQs on NCERT Notes for Class 10 Maths Chapter 11 Constructions 

What is Construction in Mathematics?

Constructions is the drawing of shapes, angles, lines, without actually measuring them. 

What are Some of the Common Constructions that we Learn in Class 10 Maths Chapter 11?

Some of the common constructions that we learnt in class 10 maths chapter 11 are: 

1. Dividing a line, in a ratio n:m. 

2. Drawing tangents to a circle, from an external point. 

How can I improve my skills in geometry constructions?

Geometry constructions can be mastered only if one knows every step. This can be achieved by drawing constructions, and writing each construction step in the copy. This increases the speed of drawing a construction, and reduces chances of getting a unexpected output from the construction. 

What is the Basic Proportionality Theorem?

BPT states that, if a line parallel to a side of a triangle, cuts the other two sides of the triangle, then, the ratio of their lengths is same. 

What is Scale Factor?

Scale factor is the ratio of the sides of the triangle to be constructed with respect to the ratio of the sides of the given triangle. 

What is an Isosceles Triangle?

An isosceles triangle, is a triangle whose two sides are equal. 

What is a Right-Angled Triangle?

A right-angle triangle is a triangle, which has one of it’s angle as 90°, and follows the pythagoras property.