NCERT Solutions for Class 10 Maths Chapter 11 Constructions
NCERT solutions for class 10 maths chapter 11 Constructions- In this chapter, you have to deal with the problems related to geometric constructions. NCERT solutions for class 10 maths chapter 11 Constructions is covering the step by step explanation to each and every problem of practice exercises. Construction is one of the scoring chapters under the geometry unit. It is a part of geometry and it holds 15 marks in the final exams. There are 2 exercises with 14 questions in this chapter. NCERT solutions for class 10 maths chapter 11 Constructions will be helpful for doing homework as well for the board exam preparation. In class 9, we have done certain calculations and constructions using a compass and a straight edge (ruler), for example, construction of triangles, drawing the perpendicular bisector of a line segment and much more and also gave their justifications. In this chapter, you will study some advanced constructions by using the concepts learnt in class 9 constructions. This chapter is about the construction of a line segment and a circle, division of a line segment and constructions of tangents to a circle using an analytical approach. Also, students have to write the steps of construction for each answer. In exercise 11.1, we will study how to divide a line segment and in exercise 11.2 we will study the construction of tangents to a circle. NCERT solutions for class 10 maths chapter 11 Constructions is designed to eliminate the wastage of time in searching the solution. NCERT solutions is an important tool to boost your preparation. Here, you will get NCERT solutions from class 6 to 12 for science and maths.
The following constructions that we are going to learn in NCERT class 10 maths chapter 11 Constructions:
To divide a line segment in the given ratio.
To construct a pair of tangents from an external point to a circle.
To construct a triangle similar to a given triangle as per a given scale factor which may be greater than 1 or less than 1.
Construction: To construct the tangents to a circle from a point outside it. Assume a circle with center O and a point P outside it and we have to construct the two tangents from point P to the circle.
Steps of Construction: 1. Join PO and bisect it. Let M is the mid-point of PO. 2. Taking M as a center and MO as radius, draw a circle. Let it intersect the given circle at the points Q and R. 3. Join PR and PQ. Then PR and PQ are the required two tangents. |
NCERT solutions for class 10 maths chapter 11 Constructions Excercise: 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.
Justification:- In the figure, we can see two similar triangles: and
Thus _{ . }
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 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 AB'C' is the required 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 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 AB'C' is the required 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 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 AB'C' is the required triangle.
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 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 AB'C' is the required triangle.
Answer:
Steps of construction:-
(i) Draw a line segment BC.
(ii) Now draw an angle B = 45 ^{ o } and C = 30 ^{ o } and draw rays in these directions.
(iii) Name the intersection of these lines as A.
(iv) Thus 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 AB'C' is the required 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 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 AB'C' is the required triangle.
NCERT solutions for class 10 maths chapter 11 Constructions Excercise: 11.2
Q1 In each of the following, give also the justification of the construction:
Draw a circle of radius 6 cm. From a point 10 cm away from its center, construct the pair of tangents to the circle and measure their lengths.
Answer:
Steps of construction:-
(i) Taking point O as center draw a circle of radius 6 cm.
(ii) Now, name a point P which is 10 cm away from point O. Join OP.
(iii) Draw a perpendicular bisector of OP name the intersection point of bisector and OP as O'.
(iv) Now draw a circle considering O' as center and O'P as the radius.
(v) Name the intersection point of circles as Q and R.
(vi) Join PQ and PR. These are the required tangents.
(vii) Measure lengths of PQ = 8cm and PR = 8cm
Answer:
Steps of constructions:-
(i) Taking point O as a center draw a circle of radius 4 cm.
(ii) Now taking O as center draw a concentric circle of radius 6 cm.
(iii) Taking any point P on the outer circle, join OP.
(iv) Draw a perpendicular bisector of OP.
(v) Name the intersection of bisector and OP as O'.
(vi) Now, draw a circle taking O' as center and O'P as the radius.
(vii) Name the intersection point of two circles as R and Q.
(viii) Join PR and PQ. These are the required tangents.
(ix) Measure the lengths of the tangents. PR = 4.47 cm and PQ = 4.47 cm.
Answer:
Steps of construction:-
(i) Taking O as a center draw a circle of radius 3 cm.
(ii) Now draw a diameter PQ of this circle and extend it.
(iii) Mark two points R and S on the extended diameter such that OR = OS = 7 cm.
(iv) Draw the perpendicular bisector of both the lines and name their mid-points as T and U.
(v) Now, taking T and U as center draw circles of radius TR and QS.
(vi) Name the intersecting points of the circles with the first circles as V, W, X, Y.
(vii) Join the lines. These are the required tangents.
Answer:
Steps of construction:-
(i) Draw a circle with center O and radius 5 cm.
(ii) Now mark a point A on the circumference of the circle. And draw a line AP perpendicular to the radius OA.
(iii) Mark a point B on the circumference of the circle such that AOB = 120 ^{ o } . (As we know, the angle at the center is double that of the angle made by tangents).
(iv) Join B to point P.
(v) AP and BP are the required tangents.
Answer:
Steps of construction:-
(i) Draw a line segment AB having a length of 8 cm.
(ii) Now, taking A as a center draw a circle of radius 4 cm. And taking B as a center draw a circle of radius 3 cm.
(iii) Bisect the line AB and name the mid-point as C.
(iv) Taking C as a center and AC as radius draw a circle.
(v) Name the intersection points of the circle as P, Q, R, S.
(vi) Join the lines and these are our required tangents.
Answer:
Steps of construction:-
(i) Draw a line segment BC of length 8 cm.
(ii) Construct a right angle at point B. Now draw a line of length 6 cm. Name the other point as A.
(iii) Join AC. ABC is the required triangle.
(iv) Now construct a line BD on the line segment AC such that BD is perpendicular to AC.
(v) Now draw a circle taking E as a center (E is the midpoint of line BC) and BE as the radius.
(vi) Join AE. And draw a perpendicular bisector of this line.
(vii) Name the midpoint of AE as F.
(viii) Now, draw a circle with F as center and AF as the radius.
(ix) Name the intersection point of both the circles as G.
(x) Join AG. Thus AB and AG are the required tangents.
Answer:
Steps of construction:-
(i) Draw a circle using a bangle.
(ii) Now draw 2 chords of this circle as QR and ST.
(iii) Take a point P outside the circle.
(iv) Draw perpendicular bisector of both the chords and let them meet at point O.
(v) Joinpoint PO.
(vi) Draw bisector of PO and name the midpoint as U.
(vii) Now, taking U as a center and UP as radius draw a circle.
(viii) Name the intersection point of both the circles as V and W.
(ix) Join PV and PW. These are the required tangents.
NCERT solutions for class 10 maths chapter wise
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables |
Chapter 4 | NCERT solutions for class 10 maths chapter 4 Quadratic Equations |
Chapter 5 | NCERT solutions for class 10 chapter 5 Arithmetic Progressions |
Chapter 6 | |
Chapter 7 | NCERT solutions for class 10 maths chapter 7 Coordinate Geometry |
Chapter 8 | NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry |
Chapter 9 | NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry |
Chapter 10 | |
Chapter 11 | NCERT solutions for class 10 maths chapter 11 Constructions |
Chapter 12 | NCERT solutions for class 10 chapter maths chapter 12 Areas Related to Circles |
Chapter 13 | NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes |
Chapter 14 | |
Chapter 15 |
NCERT solutions of class 10 subject wise
How to use NCERT solutions for class 10 maths chapter 11 Constructions ?
- Before coming to this chapter, please ensure that you have done the construction chapter of class 9.
Learn some basic constructions like angles, triangles, and circles using compass and scale.
After this, go through some examples available in class 10 maths construction.
After covering the above processes, you can come to the practice exercises.
While practicing the question, If you feel some trouble in any problem then take the help of NCERT solutions for class 10 maths chapter 11 Constructions.
Keep working hard & happy learning!
Frequently Asked Question (FAQs) - NCERT Solutions for Class 10 Maths Chapter 11 Constructions
Question: What are the important topics of class 10 maths chapter 11 constructions ?
Answer:
Dividing a line segment in the given ratio, construct a pair of tangents from an external point to a circle, construct a triangle similar to a given triangle as per a given scale factor are the important topics in this chapter.
Question: How does the NCERT solutions are helpful in CBSE board exam?
Answer:
In order to perform well in the CBSE board exam, one must know the NCERT very well. NCERT solutions can help you, if stuck in NCERT problems while solving them. Also, NCERT solutions are provided by the experts who knows how to answer in the board exam in order to get good marks.
Question: Does CBSE provides the solutions of NCERT class 10 ?
Answer:
No, CBSE doesn’t provided NCERT solutions for any class or subject.
Question: Where can I find the complete solutions of NCERT class 10 maths ?
Answer:
Here you will get the detailed NCERT solutions for class 10 maths by clicking on the link
Question: Which is the official website of NCERT ?
Answer:
NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.
Question: What is the weightage of chapter constructions in CBSE class 10 board final exam ?
Answer:
Triangles, circles and constructions combined has 15 marks weightage in class 10 final board exam.