NCERT Exemplar Class 10 Maths Solutions Chapter 10 Constructions
NCERT Exemplar class 10 maths solutions chapter 10 enables students with techniques to draw angles and shapes without actual measurements. We need to use a compass, protractor, scale, and divider. The NCERT exemplar class 10 maths chapter 10 solutions are prepared by our subjectmatter experts keeping in view the practical nature of this chapter so that students can learn the topics of NCERT class 10 maths related to constructions, seamlessly. These NCERT exemplar class 10 maths chapter 10 solutions provide stepbystep solutions to the construction problems to make it easier for students to understand the process flow. CBSE 10th maths syllabus has been followed while preparing these NCERT exemplar class 10 maths solutions chapter 10.
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Question:1
To divide a line segment AB in the ratio , first a ray AX is drawn so that is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is
(A) 8 (B) 10 (C) 11 (D) 12
Solution
Given: is an acute angle.
The required ratio is
Let m = 5, n = 7
Steps of construction
1. Draw any ray AX making an acute angle with AB.
2. Locate 12 points on AX at equal distances. (Because here )
3. Join
4. Through the point draw a line parallel to intersecting AB at the point P.
Then
(By Basic Proportionality theorem)
By construction
Hence the number of points is 12.
Question:2
To divide a line segment AB in the ratio , a ray AX is drawn first such that is an acute angle and then points are located at equal distances on the ray AX and the point B is joined to:
(A) A_{12} (B) A_{11} (C) A_{10} (D) A_{9}
Solution
Given: is an acute angle
The required ratio is
Let m = 4, n = 7
Steps of construction
1. Draw any ray AX making an acute angle with AB.
2. Locate 11 points on AX at equal distances (because m + n = 11)
3. Join
4. Through the point A4 draw a line parallel to intersecting AB at the point P.
Then
Hence point B is joined to A_{11}.
Question:3
To divide a line segment AB in the ratio , draw a ray AX such that is an acute angle, then draw a ray BY parallel to AX and the points and are located at equal distances on ray AX and BY, respectively. Then the points joined are
(A) A_{5} and B_{6} (B) A and B_{5} (C) A_{4} and B_{5} (D) A_{5} and B_{4}
Solution
Given: and both are acute angles and AX parallel to BY
The required ratio is
Let m = 5 , n = 6
Steps of construction
1. Draw any ray AX making an acute angle with AB.
2. Draw a ray .
3. Locate the points on AX at equal distances
4. Locate the points on BY at distance equal to the distance between points on AX line.
5. Join .
Let it intersect AB at a point C in figure.
Then
Here is similar to
Then
by construction
Points joined one A_{5} and B_{6}.
Question:4
To construct a triangle similar to a given with its sides of the corresponding sides of , first draw a ray BX such that CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then locate points B_{1}, B_{2}, B_{3}, ... on BX at equal distances and next step is to join
(A) B_{10} to C (B) B_{3} to C (C) B_{7} to C (D) B_{4} to C
Solution Given: is an acute angle.
Steps of construction
1. Draw any ray BX making angle with BC on the side opposite to vertex A.
2. Locate 7 points on BX in equidistant
3. Now join B_{7 }to C
4. Draw a line through B_{3} parallel to B_{7}C .
Question:5
Answer (B) 8Solution
To construct a triangle similar to a triangle, with its sides of the corresponding sides of given triangle, the minimum number of points to be located at an equal distance is equal to the greater of 8 and 5 in . Here
So, the minimum number of points to be located at equal distance on ray BX is 8.
Question:6
Answer(D) 120°Solution
According to question:
Given :
Let
As we know that angle between tangent and radius of a circle is 90
We know that
[ sum of interior angles of quadrilateral is ]

Question:1
Write True or False and give reasons for your answer in each of the following:
By geometrical construction, it is possible to divide a line segment in the ratio
Answer:
Answer [True]Solution
To divide a line segment in the ratio, we need both positive integer.
So, we can be simplified it by multiply both the terms by .
We obtain
So, the required ratio is
Geometrical construction, is possible to divide a line segment in the ratio
Question:2
Write True or False and give reasons for your answer in each of the following:
To construct a triangle similar to a given with its sides of the corresponding sides of , draw a ray BX making acute angle with BCand X lies on the opposite side of A with respect to BC. The points B_{1}, B_{2}, ....,B_{7} are located at equal distances on BX, B3 is joined to C and then a linesegment is drawn parallel to B_{3}C where lies on BC produced. Finally,line segment is drawn parallel to AC.
Solution
According to question:
To construct a triangle similar to a given with its sides of the corresponding sides of
1. Draw a line segment BC
2. Taking B and C as centres draw two arcs of suitable radii intersecting each other at A.
3. Join BA and CA. ?ABC is the required triangle.
4. From B draw any ray BX downwards making an acute angle CBX.
5. Locate seven points B_{1}, B_{2}, b_{3}, …. B_{7} on BX such that BB_{1} = B_{1}B_{2} = B_{1}B_{3} = B_{3}B_{4} = B_{4}B_{5} = B_{5}B_{6} = B_{6}B_{7}.
6. Join B_{3}C and from B_{7} draw a line B_{7}C’ ? B_{3}C intersecting the extended line segment BC at C’.
7. From point C’ draw C’A’ ? CA intersecting the extended line segment BA at A
But as given if we join B_{3}C and from B_{6} draw a line B_{6}C’ ? B_{3}C intersecting the extended line segment BC at C’.
Hence the sides are not in the ratio of 7:3
So, the required triangle can not be constructed in this way.
Hence the given statement is false.
Question:3
Write True or False and give reasons for your answer in each of the following:
A pair of tangents can be constructed from a point P to a circle of radius 3.5 cm situated at a distance of 3 cm from the center.
Solution
Radius, r = 3.5 cm
Point distance from center = 3 cm
But,
So, the point P lies inside the circle.
So, pair of tangents cannot be drawn to point P to a circle.
Question:4
Answer:
Answer [True]Solution
Tangent :  it is a straight line that touches the curve but not cross it. A pair of tangents can be constructed to a circle inclines at an angle greater than but less than .Here, the inclination angle is Hence it is possible.
Question:1
Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio .
Answer:
SolutionGiven: AB = 7cm
The required ratio is
Let m = 3, n = 5
Steps of constructio
1. Draw a line segment AB = 7cm.
2. Draw a ray AX making acute angle with AB
3. Locate 8 points on AX on equidistant. (because )
4. Join
5. Through the point draw a line parallel to which intersect line AB at P..
Here triangle AA_{3}P is similar to triangle
(by construction)
Therefore
Question:2
Answer:
SolutionGiven BC = 12cm, AB = 5 cm,
Steps of construction
1. Draw a line BC = 12 cm
2. From B draw AB = 5 cm which makes an angle of at B.
3. Join AC
4. Make an acute angle at B as
5. On BX mark 3 point at equal distance at .
6. Join
7. From draw intersect AB at
8. From point draw intersect AB
Now is the required triangle and is also a right triangle.
Question:3
Answer:
SolutionGiven : BC = 6cm, CA = 5cm, AB = 4cm
Scale factor = 5/3
Let m = 5 and n = 3
Steps of construction
 Draw line BC = 6cm
 Taking B and C as centre mark arcs of length 4cm and 5cm respectively which intersecting each other at point A.
 Join BA and CA
 Draw a ray BX making acute angle with BC.
 Locate 5 Points on BX in equidistant as B_{1}, B_{2}, B_{3}, B_{4}, B_{5}.
 Join B_{3} to C and draw a line through B5 parallel to B_{3}C to intersect at BC extended at .
Now is required triangle.
Question:4
Answer:
SolutionGiven : Radius = 4cm
Steps of construction
1. Draw a circle at radius r = 4 cm at point O.
2. Take a point P at a distance 6cm from point O and join PO
3. Draw a perpendicular bisector of line PO M is the midpoint of PO.
4. Taking M as centre draw another circle of radius equal to MO and it intersect the given circle at point Q and R
5. Now, join PQ and PR.
Then PQ and PR one the required two tangents.
Question:1
Answer:
SolutionGiven :
AB = 5 cm and AC = 7 cm
From equation 1
P is any point on B
scale of a line segment AB is
Steps of construction
1. Draw line segment AB = 5 cm
2. Now draw ray AO which makes an angle i.e.,
3. Which A as center and radius equal to 7 cm draw an arc cutting line AO at C
4. Draw ray AP with acute angle BAP
5. Along AP make 4 points with equal distance.
6. Join
7. From draw which is parallel to which meet AB at point P.
Then P is point which divides AB in ratio 3 : 1
AP : PB = 3 : 1
8. Now draw ray AQ, with an acute angle CAQ.
9. Along AQ mark 4 points with equal distance.
10. Join
11. From draw which is parallel to which meet AC at point Q.
Then Q is point which divides AC in ratio 1 : 3
AQ : QC = 1 : 3
12. Finally join PQ and its measurement is 3.25 cm.
Question:2
Answer:
SolutionSteps of construction
1. Draw, A line AB = 3 cm
2 Draw a ray by making
3. Taking B as centre and radius equal to 5 cm. Draw an arc which cut BP at point C
4. Again draw ray AX making
5. With A as centre and radius equal to 5 cm draw an arc which cut AX at point D
6. Join C and D Here ABCD is a parallelogram
7. Join BD , BD is a diagonal of parallelogram ABCD
8. From B draw a ray BQ with any acute angle at point B i.e., is acute angle
9. Locate 4 points on BQ with equal distance.
10. Join and from parallel to which intersect at point
11. From point draw line which is parallel to CD
12. Now draw a line segment parallel to DA
Note : Here and are the extended sides.
13. is a parallelogram in which and and divide it into triangles and by the diagonal
Question:3
Answer:
SolutionSteps of construction
1. Draw two concentric circles with center O and radii 3 cm and 5 cm
2. Taking any point P on outer circle, Join P and O
3. Draw perpendicular bisector of OP let M be the mid point of OP
4. Taking M as centre and OM as radius draw a circle which cuts inner circle at Q and R
5. Join PQ and PR. Thus PQ and PR are required tangents
On measuring PQ and PR we find that PQ = PR = 4 cm
Calculations
[using pythagoras theorem]
Hence the length of both tangents is 4 cm.
Question:4
Answer:
SolutionSteps of construction
1. Draw line BC = 5 cm
2. Taking B and C as centres, draw two arcs of equal radius 6 cm intersecting each other at point A.
3. Join AB and AC DABC is required isosceles triangle
4 From B draw ray with an acute angle
6. draw at with equal distance
7. Join and from draw line , intersect extended segment BC at point D.
8. From point D draw meting BA produced at E.
Then EBD is required triangle. We can name it PQR.
Justification
Question:5
Answer:
SolutionGiven : AB = 5 cm, BC = 6 cm
Steps of construction
 Draw line segment AB = 5 cm
 draw B taking as a centre draw an arc of radius BC=6cm
 Join AC, DABC is the required triangle
 From point A draw any ray with acute angle
 Mark 7 points with equal distance.
 Join and form draw making the angle equal From point X draw intersecting AC at Y. Then, DAMN is the required triangle whose sides are equal to of the corresponding sides of the ..
Now
Also,
Question:6
Answer:
SolutionSteps of construction
1. Draw a circle of radius OA = 4 cm with centre O
2. Produce OA to P such that
3. Draw a perpendicular bisector of OP=8cm
4.Now taking A as Centre draw circle of radius AP = OA = 4 cm
5.Which intersect the circle at x and y
6.Join PX and PY
7.PX and PY is the tangent of the circle
Justification
In we have
(Radius)
(Radius of circle with centre A)
is equilateral triangle
In we have
Hence
Question:7
Answer:
SolutionSteps of construction
 Draw a line segment BC = 6 cm
 Taking B and C as a centres, draw arc of radius AB= 4cm and AC=9cm
 Join AB and AC
 DABC is required triangle From B draw ray BM with acute angle
 Make 3 points on BM with equal distance
 Join and draw intersecting BC at X From point X draw XYCA intersecting the extended line segment BA to Y Then is the required triangle whose sides are equal to of the
Justification :
Here
Also
Here all the three angles are same but three sides are not same.
The two triangles are not congruent because, if two triangles are congruent, then they have same shape and same size.
NCERT Exemplar Solutions Class 10 Maths Chapter 10 Important Topics:
NCERT exemplar class 10 maths solutions chapter 10 takes the student on a gradual learning curve through the topics listed below:
◊ Divide any line segment in a given ratio.
◊ Draw similar triangles with different given conditions.
◊ NCERT Exemplar class 10 maths solutions chapter 10 discusses the method to draw a tangent to the circle.
NCERT Class 10 Exemplar Solutions for Other Subjects:
NCERT Exemplar Class 10 Maths solutions
NCERT Exemplar Class 10 Science solutions
NCERT Class 10 Maths Exemplar Solutions for Other Chapters:
Chapter 
NCERT exemplar solution for class 10 Maths Chapter 1 Real Numbers 
NCERT exemplar solutions for class 10 Maths Chapter 2 Polynomials 
NCERT exemplar solution for class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 
NCERT exemplar solution for class 10 Maths Chapter 4 Quadratic Equations 
NCERT exemplar solution for class 10 Maths Chapter 5 Arithmetic Progressions 
NCERT exemplar solution for class 10 Maths Chapter 6 Triangles 
NCERT exemplar solution for class 10 Maths Chapter 7 Coordinate Geometry 
NCERT exemplar solution for class 10 Maths Chapter 8 Introduction to Trigonometry & Its Equations 
NCERT exemplar solution for class 10 Maths Chapter 9 Circles 
NCERT exemplar solution for class 10 Maths Chapter 10 Constructions 
NCERT exemplar solution for class 10 Maths Chapter 11 Areas related to Circles 
NCERT exemplar solution for class 10 Maths Chapter 12 Surface Areas and Volumes 
NCERT exemplar solution for class 10 Maths Chapter 13 Statistics and Probability 
Features of NCERT exemplar class 10 maths solutions chapter 10:
These class 10 maths NCERT exemplar chapter 10 solutions provide an extension to the learning of constructions studied in class 9.
These techniques used in construction will be beneficial for those students who want to pursue engineering or architecture. The Constructionsbased practice problems attempted by the students from the exemplar will serve as reference material.
These class 10 maths NCERT exemplar solutions chapter 10 Constructions are appropriate to solve other books such as NCERT class 10 maths, A textbook of Mathematics by Monica Kapoor, RD Sharma class 10 maths, RS Aggarwal class 10 maths.
The pdf version for the offline study purposes can be accessed using NCERT exemplar class 10 maths solutions chapter 10 pdf download feature. This will help the student to study NCERT exemplar Class 10 maths chapter 10 in an offline mode too.
Check Chapterwise Solutions of Book Questions
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  
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 
Frequently Asked Question (FAQs)  NCERT Exemplar Class 10 Maths Solutions Chapter 10 Constructions
Question: Can we divide any line segment into equal parts with the help of construction?
Answer:
Yes, we can divide any line segment into any number of parts with any length ratio.
Question: Can we draw a common tangent for two given circles?
Answer:
Yes, we can draw a common tangent for two given circles with the help of construction.
Question: Is the chapter Constructions important for Board examinations?
Answer:
The chapter Constructions is vital for Board examinations as it holds around 23% weightage of the whole paper.
Question: What is the pattern of question type in the chapter of Constructions that can be expected in the board examination?
Answer:
Generally, you can expect to get either a Long answer or a Very Long answer question in the board examination. A thorough study from NCERT Exemplar class 10 maths solutions chapter 10 can help you ace the questions on Constructions.
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Questions related to CBSE Class 10th
I passed 10th compartment 1998 . My Roll Number in that exam was 6379804. The mark sheet certifying my passing of compartment in Mathematics bears sl..No, 98 0469353. I have lost my certificate and Marksheets. How can I get them? Kindly help. Mob.7042522693
See if you have lost your certificate, then you have to atfirst file a complaint at your nearest PS and after that with the receipt copy of complaint you need to visit your school. There you will be provided forms through which you can get a duplicate certificate or you can provided some documents which you have to take to the regional education officer of your district and they will give you some date for the certificate. So you have to follow this and then the else you have to do as per the officials instructs you.
I hope this answer helps. All the very best for your future endeavors!
sir I want class 10 CBSE board compartment exam 2021science paper
Hello,
Class 10 CBSE board compartment exam 2021 will be conducted in September 2021 (Tentative date for science is 23rd September 2021).
You can practice questions from previous year papers. I am sharing with you the link of CBSE website where you will get the previous year papers of all the subjects.
https://www.cbse.gov.in/cbsenew/questionpaper.html
Hope this Helps.
Thank you!!
cut off for cbse class 10th students to get admission in 11th class?
Hey,
Some schools do consider making a merit list before admission into the class 11. However, there is no centralised cutoff released by schools. The merit list or cutoff for the streams are decided by the particular school in which you are seeking admission. There is no as such centralised process announced.
when improvement exam of class 10 held.
As per the recent update CBSE 10th and 12th class Improvement Examinations of this year (2021) are expected to be commenced From Aug 25 and the results of this exams may come out by Sept 30 (tentative).
Hope this is informative
All the best
what to do for admission in good schools if I get 52% marks in cbse class 10th
Dear Student,
If you want to change your school this time then definitely you can get Arts or Commerce stream in some good private schools with this much percentage but not the science group like PCM or PCB. And if you really want a stream like PCM or PCB then you can try some other schools where you can easily get that not the top one.
Good Luck :)