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NCERT Solutions for Exercise 11.1 Class 10 Maths Chapter 11 - Constructions

NCERT Solutions for Exercise 11.1 Class 10 Maths Chapter 11 - Constructions

Edited By Ramraj Saini | Updated on Nov 27, 2023 04:09 PM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.1

NCERT Solutions for Exercise 11.1 Class 10 Maths Chapter 11 Constructions are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 Maths ex 11.1 enlists a few practice problems on the construction of triangles with given measurements base and heights. In the later section some more questions, terms related to the property of triangles, and how its construction can be made in two-dimensional geometry will be discussed.

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  1. NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.1
  2. Download Free Pdf of NCERT Solutions for Class 10 Maths chapter 11 exercise 11.1
  3. Assess NCERT Solutions for Class 10 Maths chapter 11 exercise 11.1
  4. More About NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1
  5. Benefits of NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1
  6. NCERT Solutions Subject Wise
NCERT Solutions for Exercise 11.1 Class 10 Maths Chapter 11 - Constructions
NCERT Solutions for Exercise 11.1 Class 10 Maths Chapter 11 - Constructions

10th class Maths exercise 11.1 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Constructions Class 10 Chapter 11 Exercise: 11.1

In each of the following, give the justification of the construction also:
Q1 Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.

Answer:

Steps of construction:-

(i) Draw a line segment AB of measurement 7.6 cm (length).

(ii) Now draw an acute angle AC with line segment AB.

(iii) Now cut 13 equal points on the line AC where the zeroth point is A.

(iv) Join the 13th point with point B. So the new line is BA 13 .

(v) Now, from point A 5 draw a line parallel to BA 13 on line AB. Name the point as D.

The point D is the required point which divides the line segment in the ratio of 5: 8.

The length of the two parts obtained is 2.9 cm and 4.7 cm for AD and DB respectively.


Constructions, 24670

Justification:- In the figure, we can see two similar triangles: \Delta ADA_5 and \Delta ABA_{13}

Thus \frac{AA_5}{A_5A_{13}}\ =\ \frac{AD}{DB}\ =\ \frac{5}{8} .

Q2 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.

Answer:

Steps of construction are:-

(i) Firstly draw a line segment AB of length 4 cm.

(ii) Now cut an arc of radius 5 cm from point A and an arc of 6 cm from point B.

(iii) Name the point of intersection of arcs to be point C.

(iv) Now join point AC and BC. Thus \Delta ABC is the required triangle.

(v) Draw a line AD which makes an acute angle with AB and is opposite of vertex C.

(vi) Cut three equal parts of line AD namely AA 1 , AA 2 , AA 3 .

(vii) Now join A 3 to B. Draw a line A 2 B' parallel to A 3 B.

(viii) And then draw a line B'C' parallel to BC.

Hence \Delta AB'C' is the required triangle.

1636090329369

Q3 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.

Answer:

Steps of construction are:-

(i) Firstly draw a line segment AB of length 5 cm.

(ii) Now cut an arc of radius 6 cm from point A and an arc of 7 cm from point B.

(iii) Name the point of intersection of arcs to be point C.

(iv) Now join point AC and BC. Thus \Delta ABC is the required triangle.

(v) Draw a line AD which makes an acute angle with AB and is opposite of vertex C.

(vi) Cut seven equal parts of line AD namely AA 1 , AA 2 , AA 3 , AA 4 , AA 5 , AA 6 , AA 7 ,.

(vii) Now join A 5 to B. Draw a line A 7 B' parallel to A 5 B.

(viii) And then draw a line B'C' parallel to BC.

Hence \Delta AB'C' is the required triangle.

1636090397260

Q4 Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 3/ 2 times the corresponding sides of the isosceles triangle.

Answer:

Steps of construction:-

(i) Draw a line segment AB of length 8 cm.

(ii) Cut arcs taking point A and point B as the center. Draw the line to intersect on line segment AB. Mark the intersecting point as point D.

(iii) Cut arc of length 4 cm on the same line which will be the altitude of the triangle.

(iv) Name the point as C. Then \Delta ABC is the isosceles triangle.

(v) Draw a line AX which makes an acute angle with AB and is opposite of vertex C.

(vi) Cut seven equal parts of line AX namely AA 1 , AA 2 , AA 3 .

(vii) Now join A 2 to B. Draw a line A 3 B' parallel to A 2 B.

(viii) And then draw a line B'C' parallel to BC.

Hence \Delta AB'C' is the required triangle.

1636090408707

Q5 Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and \angle ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.

Answer:

Steps of construction:-

(i) Draw a line segment BC with a measurement of 6 cm.

(ii) Now construct angle 60 o from point B and draw AB = 5 cm.

(iii) Join point C with point A. Thus \Delta ABC is the required triangle.

(iv) Draw a line BX which makes an acute angle with BC and is opposite of vertex A.

(v) Cut four equal parts of line BX namely BB 1 , BB 2 , BB 3 , BB 4.

(vi) Now join B 4 to C. Draw a line B 3 C' parallel to B 4 C.

(vii) And then draw a line B'C' parallel to BC.

Hence \Delta AB'C' is the required triangle.

1636090424003

Q6 Draw a triangle ABC with side BC = 7 cm, \angle B = 45°, \angle A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of D ABC.

Answer:

Steps of construction:-

(i) Draw a line segment BC.

(ii) Now draw an angle \angle B = 45 o and \angle C = 30 o and draw rays in these directions.

(iii) Name the intersection of these lines as A.

(iv) Thus \bigtriangleup ABC is the required triangle.

(v) Draw a line BX which makes an acute angle with BC and is opposite of vertex A.

(vi) Cut four equal parts of line BX namely BB 1 , BB 2 , BB 3 , BB 4.

(vii) Now join B 3 to C. Draw a line B 4 C' parallel to B 3 C.

(viii) And then draw a line B'C' parallel to BC.

Hence \Delta AB'C' is the required triangle.

1636090438268

Q7 Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

Answer:

Steps of construction:-

(i) Draw a line segment AB having a length of 4 cm.

(ii) Now, construct a right angle at point A and make a line of 3 cm.

(iii) Name this point C. Thus \Delta ABC is the required triangle.

(iv) Draw a line AX which makes an acute angle with AB and is opposite of vertex C.

(v) Cut four equal parts of line AX namely AA 1 , AA 2 , AA 3 , AA 4 , AA 5 .

(vi) Now join A 3 to B. Draw a line A 5 B' parallel to A 3 B.

(vii) And then draw a line B'C' parallel to BC.

Hence \Delta AB'C' is the required triangle.

1636090455562



More About NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1

This section contains the extra information the source and scenario of content present in exercise 11.1 Class 10 Maths. Starting from the first. This exercise 11.1 Class 10 is there to represent the problems based on building triangles over conditions provided and given measurements i.e. AB and AC in triangle ABC. As the number increases in ascending for the questions, concepts get changed like here will be discussing other questions based on the property of the triangle mentioned in the previous chapter. So in simple words, we say Class 10 Maths chapter 11 exercise 11.1 is basically a Platform in the roadmap to learn more about 2D Geometry for beginners.

Also Read| Constructions Class 10 Notes

Benefits of NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1

  • NCERT solutions for Class 10 Maths chapter 11 exercise 11.1 are important as to get better understanding over the subject.

  • Another benefit of exercise 11.1 Class 10 Maths is makes students understand terminologies like construction procedures and justification of the given diagram.

  • Student can solve most of the question provided to him/her in his HW, exams or test.

Also see-

NCERT Solutions Subject Wise

Frequently Asked Questions (FAQs)

1. What is the sum of all angles of the triangle?

The sum of all angles of triangle is 180 degrees 

2. A point A is 8 cm from the center of a circle with a radius of 10 cm. What is the maximum number of tangents that may be traced from point A to the circle?

The radius of the given circle is  10cm and point A is 8 cm from center of circle hence

The point lies inside of the circle hence no tangents can be traced from that point

3. A perpendicular line segment traced from the vertex of a triangle to the opposite side is called a

A perpendicular line segment traced from the vertex of a triangle to the opposite side is called an Altitude

4. How many Exercises are there in chapter 11 Class 10 Maths?

There are a total of 2 exercises in chapters 11 ie exercise 11.1 and 11.2

5. Is Exercise 11.1 Tough?

 Toughness of any chapter exercise and topic is based on the amount of hard work required to master that thing. As far as this chapter is concerned students need to follow definite path in order to solve each and every question of maths

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