NCERT Exemplar Class 10 Maths Solutions Chapter 10 Constructions

In this chapter, we provide NCERT Exemplar Problems Solutions for Class 10 Maths Solutions Chapter 10 Constructions for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Exemplar Problems Solutions for Class 10 Maths Solutions Chapter 10 Constructions pdf, free NCERT Exemplar Problems Solutions for Class 10 Maths Solutions Chapter 10 Constructions book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 10
Chapter NameConstructions
CategoryNCERT Exemplar

NCERT Exemplar Problems Class 10 Maths Solutions Chapter 10 Constructions

Exercise 10.1 Multiple Choice Questions (MCQs)

Question 1:
To divide a line segment AB in the ratio 5 : 7, first a ray AX is drawn, so that ∠BAX 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:
(d) We know that, to divide a line segment AB in the ratio m: n, first draw a ray AX which makes an acute angle ∠BAX, then marked m + n points at equal distance.
Here,                                                  m = 5, n = 7
So, minimum number of these points = m+n = 5 + 7 = 12.

Question 2:
To divide a line segment AB in the ratio 4 : 7, a ray AX is drawn first such that ∠BAX is an acute angle and then points A1 A2, A3,… are located at equal distances on the ray AY and the point B is joined to
(a) A12                       (b) A11                        (c) A12                         (d) A9
Solution:
(b) Here, minimum 4+7 = 11 points are located at equal distances on the ray AX, and then B is joined to last point is A11

Question 3:
To divide a line segment AB in the ratio 5 : 6, draw a ray AY such that ∠BAX is an acute angle, then draw a ray BY parallel to AY and the points A1, A2, A3,… and B1, B2, B3,… are located to equal distances on ray AY and BY, respectively. Then, the points joined are
(a) A5 and A6                     (b) A6 and B5             (c) A4 and B5              (d) A5 and B4
Solution:
(a) Given a line segment AB and we have to divide it in the ratio 5:6.
ncert-exemplar-problems-class-10-maths-constructions-Ex-10.1-3s
Steps of construction

  1. Draw a ray AX making an acute ∠BAX.
  2. Draw a ray BY parallel to AX by making ∠ABY equal to ∠BAX.
  3. Now, locate the points A1, A 2, A3, A 4 and A5 (m= 5) on AX and B1, B2, B3, B4, B5 and B6 (n = 6) such that all the points are at equal distance from each other.
  4. Join B6A5. Let it intersect AB at a point C.
    Then, AC:BC = 5:6

Question 4:
To construct a triangle similar to a given ΔABC with its sides frac {3}{7} of the corresponding sides of ΔABC, 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 B1, B2, B3,… on BX at equal distances and next step is to join
(a) B10 to C               (b) B13 to C                      (c) B7 to C                   (d)B4 to C
Solution:
(c) Here, we locate points B1, B2, B3, B4, B5, B6 and B7 on BX at equal distance and in next step join the last points is B7 to C.

Question 5:
To construct a triangle similar to a given ΔABC with its sides frac {8}{5} of the corresponding sides of ΔABC draw a ray BX such that ∠ CBX is an acute angle and X is on the opposite side of A with respect to BC. The minimum number of points to be located at equal distances on ray BX is
(a) 5                           (b) 8                           (c)13                          (d) 3
Solution:
(b) To construct a triangle similar to a given triangle, with its sides frac {m}{n} of the corresponding sides of given triangle the minimum number of points to be located at equal distance is equal to the greater of m and n is frac {8}{5}
Hence,                                      frac {m}{n}=frac {8}{5}
So, the minimum number of point to be located at equal distance on ray BX is 8.

MCQ Questions for Class 10 Maths With Answers

Question 6:
To draw a pair of tangents to a circle which are inclined to each other at an angle of 60°, it is required to draw tangents at end points of those two radii of the circle, the angle between them should be
(a) 135°                      (b) 90°                        (c) 60°                         (d) 120°
Solution:
(d) The angle between them should be 120° because in that case the figure formed by the intersection point of pair of tangent, the two end points of those-two radii tangents are drawn) and the centre of the circle is a quadrilateral.
From figure it is quadrilateral,
∠POQ + ∠PRQ = 180° [∴ sum of opposite angles are 180°]
60°+ θ = 180°
θ=120
Hence, the required angle between them is 120°.

Exercise 10.2 Very Short Answer Type Questions

Question 1:
By geometrical construction, it is possible to divide a line segment in the ratio √3 :frac{1}{sqrt{3 }}.
Solution:
True
ncert-exemplar-problems-class-10-maths-constructions-Ex-10.2-1s
Hence, the geometrical constrution is possible to divide a line segment in the ratio 3 : 1

Question 2:
To construct a triangle similar to a given ΔABC with its sides frac {7}{3}  of the corresponding sides of ΔABC, draw a ray BX making acute angle with BC and X lies on the opposite side of A with respect of BC. The points B1, B2, …, B7 are located at equal distances on BX, B3is joined to C and then a line segment B6C’ is drawn parallel to B3C, where C’ lines on BC produced. Finally line segment A’C’ is drawn parallel to AC.
Solution:
False
Steps of construction

  1. Draw a line segment BC with suitable length.
  2. Taking B and C as centres draw two arcs of suitable radii intersecting each other at A
  3. Join BA and CA ΔABC is the required triangle.
  4. From B draw any ray BX downwards making an acute angle CBX.
  5. Locate seven points B1, B2, …, B7 on SX, such that BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7.
  6. Join B3C and from B7 draw a line B7C’|| B3C intersecting the extended line segment BC at C’.
  7. From point C’ draw C’A’|| CA intersecting the extended line segment BA at A’.Then, ΔA’BC’ is the required triangle whose sides are frac {7}{3} of the corresponding sides of ΔABC.

Given that, segment B6C’ is drawn parallel to B3C. But from our construction is never possible that segment B6C’ is parallel to B3C because the similar triangle A’BC’ has its sides frac {7}{3} of the corresponding sides of triangle ABC. So, B7C’ is parallel to B3C.

Question 3:
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 centre.
Solution:
False
Since, the radius of the circle is 3.5 cm i.e., r = 3.5 cm and a point P situated at a distance of 3 cm from the centre i.e.,d= 3 cm
We see that,                                                                r > d
i.e., a point P lies inside the circle. So, no tangent can be drawn to a circle from a point lying inside it. ‘

Question 4:
A pair of tangents can be constructed to a circle inclined at an angle of 170°.
Solution:
ncert-exemplar-problems-class-10-maths-constructions-Ex-10.2-4s

Exercise 10.3 Short Answer Type Questions

Question 1:
Draw a line segment of length 7 cm. Find a point P on it which1 divides it in the ratio 3:5.
Solution:
Steps of construction

  1. Draw a line segment AB = 7 cm.
  2. Draw a ray AX, making an acute ∠BAX
  3. Along AX, mark 3+ 5= 8 points
    A1, A2, A3, A4, A5, A6, A7, A8 such that
    AA1 = A1A2 = A2A3 = A 3A4 = A 4A5 = A 5A6 = A 6A 7 = A 7A8
  4. Join   A8B
  5. From A3, draw A3C || A8B meeting AB at C.
    [by making an angle equal to ∠BA 8A at A 3]

Then, C is the point on AB which divides it in the ratio 3 : 5,
ncert-exemplar-problems-class-10-maths-constructions-Ex-10.3-1s

Question 2:
Draw a right ΔABC in which BC = 12 cm, AB = 5 cm and ∠B = 90°.Construct a triangle similar to it and of scale factor Is the new triangle also a right triangle?
Solution:
Steps of construction

  1. Draw a line segment BC = 12 cm,
  2. From 6 draw a line AB = 5 cm which makes right angle at B.
    ncert-exemplar-problems-class-10-maths-constructions-Ex-10.3-2s
  3. Join AC, ΔABC is the given right triangle.
  4. From B draw an acute ∠CBY downwards.
  5. On ray BY, mark three points B1, B2and B3, such that BB1 = B1B2 = B2B3.
  6. Join B3 C.
  7. From point B2 draw B2N || B3C intersect BC at N.
  8. From point N draw NM || CA intersect BA at M. ΔMBN is the required triangle. ΔMBN is also a right angled triangle at B.

Question 3:
Draw a ΔABC in which BC = 6 cm, CA = 5 cm and AB = 4 cm. Construct a triangle similar to it and of scale factor frac {5}{3}
Solution:
Steps of construction

  1. Draw a line segment BC = 6 cm.
  2. Taking Sand C as centres, draw two arcs of radii 4 cm and 5 cm intersecting each other at A.
  3. Join BA and CA. ΔABC is the required triangle.
  4. From B, draw any ray BX downwards making at acute angle.
  5. Mark five points B1, B2,B3, B4 and B5 on BX, such that
    BB, = B,B2 = B2B3 = B3B4 = B4B5.
  6. Join B3C and from B5 draw B5M || B3C intersecting the extended line segment BC at
  7. From point M draw MN || CA intersecting the extended line segment BA at N.
    Then, ΔNBM is the required triangle whose sides is equal to frac {5}{3} of the corresponding
    sides of the ΔABC.
    Hence, ΔNBM is the required triangle.

Question 4:
Construct a tangent to a circle of radius 4 cm from a point which is at a distance of 6 cm from its centre.
Solution:
Given, a point M’ is at a distance of 6 cm from the centre of a circle of radius 4 cm.
ncert-exemplar-problems-class-10-maths-constructions-Ex-10.3-4s

Exercise 10.4 Long Answer Type Questions

Question 1:
Two line segments AB and AC include an angle of 60°, where AB = 5 cm and AC = 7 cm. Locate points P and Q on AB and AC, respectively such that AP = frac {3}{4}AB and AQ = frac {1}{4}AC. Join P and Q and measure the length PQ.
Solution:
Given that, AB = 5 cm and AC = 7 cm
ncert-exemplar-problems-class-10-maths-constructions-Ex-10.4-1s-1
ncert-exemplar-problems-class-10-maths-constructions-Ex-10.4-1s-2
Steps of construction

  1. Draw a line segment AB = 5 cm.
  2. Now draw a ray AZ making an acute ∠BAZ = 60°.
  3. With A as centre and radius equal to 7 cm draw an arc cutting the line AZ at C.
  4. Draw a ray AX, making an acute ∠BAX
  5. Along AX, mark 1+3 = 4 points A1, A2, A3, and A4 Such that AA1 = A1A2 = A2A3 = A3A4
  6. Join A4B
  7. From A3 draw A3P || A4B meeting AB at  P                   [by making an angle equal to ∠AA4B]
    Then, Pis the point on AB which divides it in the ratio 3:1.
    So,                                        AP: PB = 3:1
  8. Draw a ray AY, making an acute  ∠CAY
ncert-exemplar-problems-class-10-maths-constructions-Ex-10.4-1s-3
  1. Along AY, mark 3+1 = 4 points B1, B2, B3 and B4.
    Such that AB1 = B1B2 = B2B3 =B3B4
  2. Join   B4C
  3. From B1 draw B1Q || B4C meeting AC atQ. [by making an angle equal to ∠AB4C]
    Then, Q is the point on AC which divides it in the ratio 1 : 3.
    So  AQ:OC = 1:3
  4. Finally, join PQ and its measurment is 3.25 cm.

Question 2:
Draw a parallelogram ABCD in which BC = 5 cm, AB = 3 cm and ∠ABC = 60°, divide it into triangles BCD and ABD by the diagonal BD. Construct the triangles BD’C’ similar to ΔBDC with Scale factor frac {1}{4}. Draw the line segment D’A’ parallel to DA, where A’ lies on extended side BA. Is A’BC’D’ a parallelogram?
Solution:
Steps of construction

  1. Draw a line segment AB = 3 cm.
  2. Now, draw a ray BY making an acute ∠ABY = 60°.
  3. With B as centre and radius equal to 5 cm draw an arc cut the point C on
  4. Again draw a ray AZ making an acute ∠ZAX’ = 60°.                [∴ BY || AZ, ∴ ∠YBX’ = TAX’ = 60°]
  5. With A as centre and radius equal to 5 cm draw an arc cut the point D on AZ.
ncert-exemplar-problems-class-10-maths-constructions-Ex-10.4-2s
  1. Now, join CD and finally make a parallelogram ABCD
  2. Join BD, which is a diagonal of parallelogram ABCD
  3. From B draw any ray BX downwards making an acute ∠CBX.
  4. Locate 4 points B1, B2, B3, B4 on BX, such that BB1 = B1B2 = B2B3 = B3B4,
  5. Join B4C and from B3C draw a line B4C’ || B3C intersecting the extended line segment BC at C’.
  6. From point C’ draw C’D’|| CD intersecting the extended line segment BD at D’. Then, AD’BC’ is the required triangle whose sides are frac {4}{3} of the corresponding sides of ΔDBC
  7. Now draw a line segment D’A’ parallel to DA, where A’ lies on extended side BA i.e ray BX’.
  8. Finally, we observe that A’BCD’ is a parallelogram in which A’D’ = 6.5,cm A’B = 4 cm and ∠A’BD’ = 60° divide it into triangles BCD’ and A’BD’ by the diagonal BD.

Question 3:
Draw two concentric circles of radii 3 cm and 5 cm. Taking a point on outer circle construct the pair of tangents to the other. Measure the length of a tangent and verify it by actual calculation.
Solution:
Given, two concentric circles of radii 3 cm and 5 cm with centre 0. We have to draw pair of
tangents from point P on outer circle to the other.
ncert-exemplar-problems-class-10-maths-constructions-Ex-10.4-3s -1
Steps of construction

  1. Draw two concentric circles with centre 0 and radii 3 cm and 5 cm.
  2. Taking any point P on outer circle. Join OP.
  3. Bisect OP, let M’ be the mid-point of .
    Taking M’ as centre and OM’ as radius draw a circle dotted which cuts the inner circle at M and P’.
  4. Join P M and PP’. Thus, PM and PP’ are the required tangents.
  5. On measuring PM and PP’, we find that PM = PP’ = 4 cm.
ncert-exemplar-problems-class-10-maths-constructions-Ex-10.4-3s -2

Question 4:
Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to AABC in which PQ = 8 cm. Also justify the construction.
Solution:
Let ΔPQR and ΔABC are similar triangles, then its scale factor between the corresponding sides is frac { PQ}{AB} =frac{8}{6} =frac{4}{3}
Steps of construction

  1. Draw a line segment BC = 5 cm.
  2. Construct OQ the perpendicular bisector of line segment BC meeting BC at P’.
  3. Taking B and C as centres draw two arcs of equal radius 6 cm intersecting each other at A
  4. Join BA and CA. So, ΔABC is the required isosceles triangle.
    ncert-exemplar-problems-class-10-maths-constructions-Ex-10.4-4s-1
  5. From B, draw any ray BX making an acute ∠CBX
  6. Locate four points B1, B2, B3 and B4 on BX such that BB1 = B1B2 = B2B3 = B3B4
  7. Join B3C and from B4 draw a line B4R || B3C intersecting the extended line segment BC at R.
  8. From point R, draw RP||CA meeting BA produced at P
    Then, ΔPBR is the required triangle.
ncert-exemplar-problems-class-10-maths-constructions-Ex-10.4-4s-2

Question 5:
Draw a ΔABC in which AB = 5 cm, BC = 6 cm and ∠ABC = 60°. Construct a triangle similar to ABC with scale factor  frac {5}{7} Justify the construction.
Solution:
ncert-exemplar-problems-class-10-maths-constructions-Ex-10.4-5s

Question 6:
Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is 60°. Also justify the  construction.  Measure the distance between the centre of the circle and the point of intersection of tangents.
Solution:
In order to draw the pair of tangents, we follow the following steps
Steps of construction

  1. Take a point 0 on the plane of the paper and draw a circle of radius OA = 4 cm.
  2. Produce OA to B such that OA = AB = 4 cm.
  3. Taking A as the centre draw a circle of radius AO = AB = 4 cm.
    Suppose it cuts the circle drawn in step 1 at P and Q.
  4. Join BP and BQ to get desired tangents.
ncert-exemplar-problems-class-10-maths-constructions-Ex-10.4-6s-1
ncert-exemplar-problems-class-10-maths-constructions-Ex-10.4-6s-3

Question 7:
Draw a ΔABC in which AB = 4 cm, SC = 6 cm and AC = 9 cm. Construct a triangle similar to ΔABC with scale factor  frac {1}{4} Justify the construction. Are the two triangles congruent? Note that, all the three angls and two sides of the two triangles are equal.
Solution:
Steps of construction

  1. Draw a line segment BC = 6 cm.
  2. Taking B and C as centres, draw two arcs of radii 4 cm and 9 cm intersecting each other at A.
  3. Join BA and CA, ΔABC is the required triangle.
  4. From B, draw any ray BX downwards making an acute angle.
  5. Mark three points B1, B2, B3 on BX, such that BB1 = B1B2 = B2B3.
ncert-exemplar-problems-class-10-maths-constructions-Ex-10.4-7s-1
  1. Join B2C and from B3 draw B3M || B2C intersecting the extended line segment BC at
  2. From point M, draw MN||CA intersecting the extended line segment BA to N.
    Then,ΔNBM is the required triangle whose sides are equals to   of the corresponding sides of the ΔABC

The two triangles are not congruent because, if two triangles are congruent, then they have same shape and same size. Here, all the three angles are same but three sides are not same one side is different.
ncert-exemplar-problems-class-10-maths-constructions-Ex-10.4-7s-2

All Chapter NCERT Exemplar Problems Solutions For Class 10 maths

—————————————————————————–

All Subject NCERT Exemplar Problems Solutions For Class 10

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

Leave a Comment

Your email address will not be published. Required fields are marked *