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.

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.

## Assess NCERT Solutions for Class 10 Maths chapter 11 exercise 11.1

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.

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.

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

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

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.

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.

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.

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.

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.

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.

## 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

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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### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

The Sadhu Ashram in Aligarh is located in Chhalesar . The ashram is open every day of the week, except for Thursdays . On Mondays, Wednesdays, and Saturdays, it's open from 8:00 a.m. to 7:30 p.m., while on Tuesdays and Fridays, it's open from 7:30 a.m. to 7:30 p.m. and 7:30 a.m. to 6:00 a.m., respectively . Sundays have varying hours from 7:00 a.m. to 8:30 p.m. . You can find it at Chhalesar, Aligarh - 202127 .

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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