Constructions Chapter 11 Class 10 Maths NCERT Textbook With Solutions PDF

NCERT Solutions for Class 10 Maths Chapter 11‘ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 10 Maths Chapter 8 Exercise Solution’ using the download button.

NCERT Class 10 Maths Textbook Chapter 11 With Answer Book PDF Free Download

Constructions class 10

Chapter 12: Constructions

11.1 Introduction

In Class IX, you have done certain constructions using a straight edge (ruler) and a compass, e.g., bisecting an angle, drawing the perpendicular bisector of a line segment, some constructions of triangles, etc., and also gave their justifications.

In this chapter, we shall study some more constructions by using the knowledge of the earlier constructions. You would also be expected to give the mathematical reasoning behind why such constructions work.

11.2 Division of a Line Segment

Suppose a line segment is given and you have to divide it in a given ratio, say 3 : 2. You may do it by measuring the length and then marking a point on it that divides it in the given ratio.

But suppose you do not have any way of measuring it precisely, how would you find the point? We give below two ways of finding such a point.

AuthorNCERT
Language English
No. of Pages7
PDF Size258 KB
CategoryMathematics
Source/ Creditsncert.nic.in

NCERT Solutions Class 10 Maths Chapter 11 Constructions

1. Draw a line segment of length 7.6 cm and divide it in the ratio of 5: 8. Measure the two parts.

Construction Procedure:

A line segment with a measure of 7.6 cm in length is divided in the ratio of 5:8 as follows.

1. Draw line segment AB with the length measure of 7.6 cm

2. Draw a ray AX that makes an acute angle with line segment AB.

3. Locate the points i.e.,13 (= 5+8) points, such as A1, A2, A3, A4 …….. A13, on the ray AX such that it becomes AA1 = A1A2 = A2A3 and so on.

4. Join the line segment and the ray, BA13.

5. Through the point A5, draw a line parallel to BA13 which makes an angle equal to ∠AA13B

6. point A5 which intersects line AB at point C.

7. C is the point divides line segment AB of 7.6 cm in the required ratio of 5:8.

8. Now, measure the lengths of the line AC and CB. It comes out to measures 2.9 cm and 4.7 cm respectively.

Ncert solutions class 10 Chapter 11-1

Justification:

The construction of the given problem can be justified by proving that

AC/CB = 5/ 8

By construction, we have A5C || A13B. From the Basic proportionality theorem for the triangle AA13B, we get

AC/CB =AA5/A5A13….. (1)

From the figure constructed, it is observed that AA5 and A5A13 contain 5 and 8 equal divisions of line segments respectively.

Therefore, it becomes

AA5/A5A13=5/8… (2)

Compare the equations (1) and (2), we obtain

AC/CB = 5/ 8

Hence, Justified.

2. Construct a triangle of sides 4 cm, 5 cm, and 6 cm, and then a triangle similar to it whose sides are 2/3 of

the corresponding sides of the first triangle.

Construction Procedure:

1. Draw a line segment AB which measures 4 cm, i.e., AB = 4 cm.

2. Take point A as center, and draw an arc of the radius of 5 cm.

3. Similarly, take point B as its center, and draw an arc of radius 6 cm.

4. The arcs drawn will intersect each other at point C.

5. Now, we obtained AC = 5 cm and BC = 6 cm, and therefore ΔABC is the required triangle.

6. Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C.

7. Locate 3 points such as A1, A2, A3 (as 3 is greater between 2 and 3) on line AX such that it becomes AA1= A1A2 = A2A3.

8. Join the point BA3 and draw a line through A2which is parallel to the line BA3 that intersects AB at point B’.

9. Through the point B’, draw a line parallel to the line BC that intersects the line AC at C’.

10. Therefore, ΔAB’C’ is the required triangle.

Ncert solutions class 10 Chapter 11-2

Justification:

The construction of the given problem can be justified by proving that

AB’ = (2/3)AB

B’C’ = (2/3)BC

AC’= (2/3)AC

From the construction, we get B’C’ || BC

∴ ∠AB’C’ = ∠ABC (Corresponding angles)

In ΔAB’C’ and ΔABC,

∠ABC = ∠AB’C (Proved above)

∠BAC = ∠B’AC’ (Common)

∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion)

Therefore, AB’/AB = B’C’/BC= AC’/AC …. (1)

In ΔAAB’ and ΔAAB,

∠A2AB’ =∠A3AB (Common)

From the corresponding angles, we get,

∠AA2B’ =∠AA3B

Therefore, from the AA similarity criterion, we obtain

ΔAA2B’ and AA3B

So, AB’/AB = AA2/AA3

Therefore, AB’/AB = 2/3 ……. (2)

From the equations (1) and (2), we get

AB’/AB=B’C’/BC = AC’/ AC = 2/3

This can be written as

AB’ = (2/3)AB

B’C’ = (2/3)BC

AC’= (2/3)AC

Hence, justified.

3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle

Construction Procedure:

1. Draw a line segment AB =5 cm.

2. Take A and B as centre, and draw the arcs of radius 6 cm and 7 cm respectively.

3. These arcs will intersect each other at point C and therefore ΔABC is the required triangle with the length of sides as 5 cm, 6 cm, and 7 cm respectively.

4. Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C.

5. Locate the 7 points such as A1, A2, A3, A4, A5, A6, A7 (as 7 is greater between 5 and 7), on line AX such that it becomes AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7

6. Join the points BA5 and draw a line from A7 to BA5 which is parallel to the line BA5 where it intersects the extended line segment AB at point B’.

7. Now, draw a line from B’ the extended line segment AC at C’ which is parallel to the line BC and it intersects to make a triangle.

8. Therefore, ΔAB’C’ is the required triangle.

Ncert solutions class 10 Chapter 11-3

Justification:

The construction of the given problem can be justified by proving that

AB’ = (7/5)AB

B’C’ = (7/5)BC

AC’= (7/5)AC

From the construction, we get B’C’ || BC

∴ ∠AB’C’ = ∠ABC (Corresponding angles)

In ΔAB’C’ and ΔABC,

∠ABC = ∠AB’C (Proved above)

∠BAC = ∠B’AC’ (Common)

∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion)

Therefore, AB’/AB = B’C’/BC= AC’/AC …. (1)

In ΔAA7B’ and ΔAA5B,

∠A7AB’=∠A5AB (Common)

From the corresponding angles, we get,

∠A A7B’=∠A A5B

Therefore, from the AA similarity criterion, we obtain

ΔA A2B’ and A A3B

So, AB’/AB = AA5/AA7

Therefore, AB /AB’ = 5/7 ……. (2)

From the equations (1) and (2), we get

AB’/AB = B’C’/BC = AC’/ AC = 7/5

This can be written as

AB’ = (7/5)AB

B’C’ = (7/5)BC

AC’= (7/5)AC

Hence, justified.

NCERT Class 10 Maths Textbook Chapter 11 With Answer Book PDF Free Download

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