# 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. Solutions of NCERT 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. CBSE 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. In case you want the NCERT solutions for other subjects, classes, and chapters, then you can get these for free by clicking on the above-given link.

The following constructions that we are going to learn in NCERT class 10 maths chapter 11 Constructions:

1. To divide a line segment in the given ratio.

2. To construct a pair of tangents from an external point to a circle.

3. 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.

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 BA13.

(v) Now, from point A5 draw a line parallel to BA13 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  AA1, AA2, AA3.

(vii) Now join A3 to B. Draw a line A2B' parallel to A3B.

(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  AA1, AA2, AA3, AA4, AA5, AA6, AA7,.

(vii) Now join A5 to B. Draw a line A7B' parallel to A5B.

(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  AA1, AA2, AA3.

(vii) Now join A2 to B. Draw a line A3B' parallel to A2B.

(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 60o 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  BB1, BB2, BB3, BB4.

(vi) Now join B4 to C. Draw a line B3C' parallel to B4C.

(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  =  45o   and  $\angle$  C  =  30o  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  BB1, BB2, BB3, BB4.

(vii) Now join B3 to C. Draw a line B4C' parallel to B3C.

(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  AA1, AA2, AA3, AA4, AA5.

(vi) Now join A3 to B. Draw a line A5B' parallel to A3B.

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

Hence $\Delta$ 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.

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

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.

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.

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 $\angle$ AOB  =  120o.   (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.

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.

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. $\Delta$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.

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 CBSE NCERT solutions for class 10 maths chapter 1 Real Numbers Chapter 2 NCERT solutions for class 10 maths chapter 2 Polynomials Chapter 3 Solutions of NCERT class 10 maths chapter 3 Pair of Linear Equations in Two Variables Chapter 4 CBSE NCERT solutions for class 10 maths chapter 4 Quadratic Equations Chapter 5 NCERT solutions for class 10 chapter 5 Arithmetic Progressions Chapter 6 Solutions of NCERT class 10 maths chapter 6 Triangles Chapter 7 CBSE 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 Solutions of NCERT class 10 maths chapter 9 Some Applications of Trigonometry Chapter 10 CBSE NCERT solutions class 10 maths chapter 10 Circles Chapter 11 NCERT solutions for class 10 maths chapter 11 Constructions Chapter 12 Solutions of NCERT class 10 chapter maths chapter 12 Areas Related to Circles Chapter 13 CBSE NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes Chapter 14 NCERT solutions for class 10 maths chapter 14 Statistics Chapter 15 Solutions of NCERT class 10 maths chapter 15 Probability

## 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!