According to the CBSE Syllabus 202324, this chapter has been removed fromÂ NCERT Class 10 Maths textbook.
CBSE Class 10 Maths Chapter 11 Constructions Objective Questions include topics such as Division of a Line Segment, Constructions of Tangents to a Circle, Line Segment Bisector, and so on. Class 10, Chapter 11 Constructions is one of the scoring chapters that come under the unit Geometry. Based on the latest modification of the exam pattern, we have compiled here the CBSE Class 10 Maths Chapter 11 – Construction Objective Questions for the students to practise and score well in the board examination.
Find the subtopics covered in the chapter, based on which the CBSE Class 10 Maths objective questions are formed.
Subtopics Covered in Chapter 11
11.1 Constructing Similar Triangles (8 MCQs from the Topic)
11.2 Construction of Tangents to a Circle (3 MCQs from the Topic)
11.3 Drawing Tangents to a Circle (3 MCQs from the Topic)
11.4 Dividing a Line Segment (6 MCQs from the Topic)
Find below the link to access the list of MCQs organised topicwise from Chapter 11 Constructions of CBSE Class 10 Maths.
Download CBSE Class 10 Maths Chapter 11 – Construction Objective Questions
Constructing Similar Triangles

 The ratio of corresponding sides for the pair of triangles whose construction is given as follows: Triangle ABC of dimensions AB=4cm, BC= 5 cm, and âˆ B= 60^{o}. A ray BX is drawn from B, making an acute angle with AB. 5 points B1, B2, B3, B4 and B5 are located on the ray, such that BB1=B1B2=B2B3=B3B4=B4B5. B4 is joined to A, and a line parallel to B4A is drawn through B5 to intersect the extended line AB at A’. Another line is drawn through A’ parallel to AC, intersecting the extended line BC at C’. Find the ratio of the corresponding sides of Î”ABC and Î”Aâ€²BCâ€².
 1:5
 1:4
 4:5
 4:1
 The ratio of corresponding sides for the pair of triangles whose construction is given as follows: Triangle ABC of dimensions AB=4cm, BC= 5 cm, and âˆ B= 60^{o}. A ray BX is drawn from B, making an acute angle with AB. 5 points B1, B2, B3, B4 and B5 are located on the ray, such that BB1=B1B2=B2B3=B3B4=B4B5. B4 is joined to A, and a line parallel to B4A is drawn through B5 to intersect the extended line AB at A’. Another line is drawn through A’ parallel to AC, intersecting the extended line BC at C’. Find the ratio of the corresponding sides of Î”ABC and Î”Aâ€²BCâ€².
Answer: (C) 4:5
Solution:
According to the construction,Â Î”BB_{4}Aâˆ¼Î”BB_{5}Aâ€²
And, for similar triangles, the ratio of corresponding sides is AB/Aâ€²B.
Hence, the ratio of the corresponding sides is 4:5.

 If I ask you to construct â–³PQR ~ â–³ABC exactly (when we say exactly, we mean the exact relative positions of the triangles) as given in the figure (Assuming I give you the dimensions of â–³ABC and the scale factor for â–³PQR), what additional information would you ask for?


 The information given is sufficient.
 We cannot construct the triangle because there is no connection between the two triangles.
 The perpendicular distance between AC and Point P and the angle between AC and PR.
 Dimensions of PQR.

Answer: (C) The perpendicular distance between AC and Point P and the angle between AC and PR.
Solution: Consider that we need to draw a triangle â–³PQR ~ â–³ABC. But â–³PQR is at an isolated position w.r.t. â–³ABC. Therefore, we need to know the perpendicular distance between AC and point PÂ and theÂ angle between AC and PR.
Now let us reframe the question with the required information.
Construct a triangle â–³PQR similar to â–³ABC, such that point P is at a perpendicular distance of 5 cm from the line AC, PR makes an angle of 60^{ o} and which is 23rd of â–³ABC. In triangle, â–³ABC, AB = 6 cm, AC = 7 cm, and âˆ ABC = 30^{o}.
Steps of construction:
1. Draw a straight line AC = 7 cm.
2. Measure an angle ofÂ 30^{ o} w.r.t AC and draw a straight line.
3. Set the compass at 6 cm in length, and with A as the centre, mark the length on the line, and name the point of intersection as B.
Reason: This is done to draw the line AB, i.e. we are marking the length of 6 cm on the line 30^{o} inclined to AC.
4. Connect the points to BC to complete the triangle.
5. Draw a line perpendicular to AC at A.
Reason: We are drawing this step to identify the line on which point P might lie.
6. For the given length of AP, cut the perpendicular at P.
Reason: We are drawing the arc to identify the exact position of point P on the line perpendicular to AC.
7. Extrapolate line segment AC.
8. Draw a Ray XYÂ 60^{ o}Â to AC, such that the line passes through the point P.
9. Measure the distance AC with a compass and draw an arc with point P as the centre and radius equal to AC, and the point of intersection is R’.
10. Measure z and BAC, and draw a similar angle on PR’.
11. Measure AB and mark PZ as PQ. Join QR’.
12. Draw a ray PM, mark 3 pointsÂ P_{1}P_{2}P_{3}Â such thatÂ PP_{1}=P_{1}P_{2}=P_{2}P_{3}
13. Join Râ€²P_{3}
14. Through P2, draw a line parallel to P3Râ€², and let the point of intersection of the line with PR’ is R.
15. Through point R, draw a line parallel to R’Q’. Let the point of intersection of the line with line PO is Q. The triangle â–³PQR is similar to â–³ABC, such that the angle between PR and AC is 120^{ o} and the perpendicular distance of point P from AC is 5 cm.

 If the perpendicular distance between AP is given, which vertices of a similar triangle would you find first?


 R
 Q
 P
 A

Answer: (C) P
Solution: If the perpendicular distance of AP is given, then we would start the construction of a similar triangle by finding the position of the point P.

 If you need to construct a triangle with point P as oneÂ of its vertices, which is the angle that you need to construct a side of the triangle?


 âˆ QPR
 âˆ RQP
 âˆ PRQ
 Angle PR makes with AC

Answer: (D) Angle PR makes with AC.
Solution: Once we know the position of Point P, we need to find the orientation of the â–³PQR w.r.t â–³ABC. For that purpose, we need to know the angle of any one side that makes the original triangle. Therefore, among the given options, we will be able to determine the orientation of the â–³PQR if we know the angle at which side PR makes with side AC.

 Match the following based on the construction of similar triangles, if the scale factor (m/n) is


 I – c, II – a, III â€“ b
 I – b, II – a, III â€“ c
 I – a, II – c, III â€“ b
 I – a, II – b, III â€“ c

Answer: (A) I – c, II – a III â€“ b
Solution: The scale factor basically defines the ratio between the sides of the constructed triangle to that of the original triangle.
So, when we see the scale factor (m/n)>1, it means the sides of the constructed triangle are larger than the original triangle, i.e. the triangle constructed is larger than the original triangle.
Similarly, if scale factor (m/n) <1, then the sides of the constructed triangle are smaller than the original triangle, i.e. the constructed triangle is smaller than the original triangle.
When we have scale factor (m/n) = 1, then the sides of both the constructed triangle and the original triangle are equal.
When a pair of similar triangles have equal corresponding sides, then the pair of similar triangles can be called congruent because the triangles will have equal corresponding sides and equal corresponding angles.

 The image of the construction of A,’ C’B, a similar triangle of Î”ACB, is given below. Choose the correct option:


 âˆ BAâ€²Câ€² =âˆ BAC
 âˆ CAB=âˆ ACB
 âˆ Aâ€²BCâ€²â‰ âˆ CBA
 BAâ€²/Aâ€²Câ€²=BC/BC

Answer: (A) âˆ BAâ€²Câ€² =âˆ BAC
Solution: AsÂ Î”ABCâˆ¼Î”Aâ€²BCâ€²
âˆ BAâ€²Câ€²=âˆ BACÂ (corresponding angles of similar triangles)

 If a triangle similar to the given Î”ABC with sides equal to 3/4 of the sides of Î”ABC is to be constructed, then the number of points to be marked on ray BX is __.


 3
 4
 7
 6

Answer: (B) 4
Solution: In the ratio between sidesÂ 3/4,Â 4>3
â‡’Â The number of points to be marked on BX to construct similar triangles is 4.

 The construction of similar polygons is similar to that of the construction of similar triangles. If you are asked to construct a parallelogram similar to a given parallelogram with a given scale factor, which of the given steps will help you construct a similar parallelogram?
 Find a point on the larger side that divides it in the ratio of the given scale factor, and use the smaller side as the other parallel side to construct the parallelogram.
 Find two points, one on the larger side and the other on the smaller side, using the given scale factor and using these scaled lengths to construct a similar parallelogram.
 With one of the vertices as the centre and radius (scale factor multiplied by the length of the larger side), draw two arcs on the larger and smaller sides. Use these 2 points to construct the parallelogram.
 None of these helps in constructing a similar parallelogram.
 The construction of similar polygons is similar to that of the construction of similar triangles. If you are asked to construct a parallelogram similar to a given parallelogram with a given scale factor, which of the given steps will help you construct a similar parallelogram?
Answer: (B) Find two points, one on the larger side and the other on the smaller side, using the given scale factor and using these scaled lengths to construct a similar parallelogram.
Solution:
The following steps will give you information on how to construct a parallelogram similar to ABCD.
Step 1: Find points E and F on longer and smaller sides, respectively, using the given scale factor.
Step 2: Draw a line from E parallel to the smaller side AD.
Step 3: Taking the length of AF and E as the centre, cut an arc on the line parallel to AD, and let this new point be G.
Step 4: Join EG and FG.
Step 5: AEFG is the required parallelogram.
Now, we have constructed the parallelogram AEFG âˆ¼ ABCD.
Construction of Tangents to a Circle

 You are given a circle with radius â€˜râ€™ and centre O. You are asked to draw a pair of tangents that are inclined at an angle of 60âˆ˜ with each other. Refer to the figure and select the option which would lead us to the required construction. d is the distance OE.


 Using trigonometry, arrive at d= âˆš5 and mark E.
 Using trigonometry, arrive atÂ d= âˆš3 and mark E.
 Mark M and N on the circle such thatÂ âˆ MOE =Â 60 Â°Â andÂ âˆ NOE =Â 60 Â°
 Construct theÂ â–³MNO as it is equilateral

Answer: (C) Mark M and N on the circle, such that âˆ MOE = 60 Â°Â andÂ âˆ NOE =Â 60 Â°
Solution: The angle between the tangents is 60Â°, and OE bisects âˆ MEN, âˆ MEO = 30Â°.
Since â–³OME is rightangled at M, we realise that the âˆ MOE = 60Â°. SinceÂ âˆ MOE =Â 60Â°, we must haveÂ âˆ NOE =60Â° and âˆ MON = 120Â°. Hence, â–³MNO is NOT equilateral.
Next, since inÂ â–³OME,Â sin30Â°=Â 1/2Â =Â OM/OEÂ =Â r/d, we have d = 2r.
Recalling thatÂ âˆ MOE =Â 60Â°, the following are the steps of construction:
1. Draw a ray from the centre of O.
2. With O as centre, constructÂ âˆ MOE =Â 60Â°Â [constructing angleÂ 60Â°Â is easy]
3. Now, extend OM, and from M, draw a line perpendicular to OM. This intersects the ray at E. This is the point from where the tangents should be drawn; EM is one tangent.
4. Similarly, EN is another tangent.

 In the above scenario, after drawing the circle with radius R, what is the next thing to be constructed?
 The point B
 The point O
 Circle with radius â€˜r.â€™
 Tangent PO
 In the above scenario, after drawing the circle with radius R, what is the next thing to be constructed?
Answer: (B) The point O
Solution: Since we need to draw a circle with a radius â€˜râ€™, we need the following points:
i) Centre of the smaller circle
ii) Radius of the smaller circle.
We have the radius â€˜râ€™, but we need to adjust the centre on the line OA.
Realising that the tangents are common to both the circles, the radius of each circle at their point of contact being perpendicular to the common tangent, we can say that the radii are parallel. So, we also have the ratio PQ:QO because we have a pair of similar triangles.
But, before we can do all this, first, we need to have the line AO and, even before that, the point O. Only then we can draw tangents and then the inner circle. So, the next step would be determining the point ‘Oâ€™.

 In reference to the above question, what would be the first thing to determine?
 The radius of the circle â€˜C.â€™
 The radius of the circle â€˜D.â€™
 Centre of the circle â€˜C.â€™
 None of these
 In reference to the above question, what would be the first thing to determine?
Answer: (C) Centre of the circle â€˜C.â€™
Solution: Since we need to finally construct a circle of radius â€˜Râ€™ concentric to the previous circle, we need to determine the centre of these circles first, before proceeding with anything else.
The radius of the circle â€˜Dâ€™ can be figured out after we get the radius of the first circle using the centre.
The radius of the circle â€˜Câ€™ can be found after finding the centre of this circle.
Drawing Tangents to a Circle

 Which of the following is notÂ true for a point P on the circle?
 Only 1 tangent can be drawn from point P
 There are 2 tangents to the circle from point P
 Perpendicular to the tangent passes through the centre
 None of these
 Which of the following is notÂ true for a point P on the circle?
Answer: (B) There are 2 tangents to the circle from point P
Solution: Only one tangent can be drawn from a point on the circle, and the tangent is always perpendicular to the radius.

 There is a circle with centre O. P is a point from there only one tangent can be drawn to this circle. What can we say about P?
 O and P are coincident points
 P is on the circle
 P is inside the circle
 P is outside the circle
 There is a circle with centre O. P is a point from there only one tangent can be drawn to this circle. What can we say about P?
Answer: (B) P is on the circle.
Solution: Since only one tangent can be drawn, this point P should be present on the circle.
Any point on the circle is at a distance equal to the radius of the circle. So, OP is equal to the radius of the circle.

 A circle of radius rÂ has a centre O. What is the first step to construct a tangent from a generic point P which is at a distance rÂ from O?
 With P as centre and radius > r, draw a circle and then join OP.
 With P as centre and radius < r, draw a circle and then join OP.
 With P as centre and radius = r, draw a circle and then join OP.
 Join OP.
 A circle of radius rÂ has a centre O. What is the first step to construct a tangent from a generic point P which is at a distance rÂ from O?
Answer: (D) Join OP
Solution: P is a point on the circle. We know that only one tangent can be drawn, and it is perpendicular to the line joining the centre of the circle O and the point of contact P.
So, the first step would be to join OP.
Dividing a Line Segment

 A point C divides a line segment AB in the ratio 5:6. The ratio of lengths AB:BC is:


 11:5
 11:6
 6:11
 5:11

Answer: (B) 11:6
Solution: GivenÂ AC/BCÂ =Â 5/6
AB/BC=Â (AC+BC)/BCÂ =Â (AC/BC) +1Â =Â 5/6+1Â =Â 11/6
So, the ratio is 11:6.

 Point W divides the line XY in the ratio m:n. Then, the ratio of lengths of the line segments XY:WX, is
 m+n:m
 m+n:n
 m:m+n
 m:n
 Point W divides the line XY in the ratio m:n. Then, the ratio of lengths of the line segments XY:WX, is
Answer: (A) m+n:m
Solution:
XY/ XW= (XW/XW) + (WY/XW) = 1 + (n/m) = (m+n)/m = m+n: m

 What is the ratio AC/BC for line segment AB following the construction method below? Step 1. A ray is extended from A, and 30 arcs of equal lengths are cut, cutting the ray at A_{1}, A_{2},…A_{30}Step 2. A line is drawn fromÂ A_{30} to B, and a line parallel to A_{30}B is drawn, passing through point A_{17} and meeting AB at C.
 13:30
 13:17
 17:13
 17:30
 What is the ratio AC/BC for line segment AB following the construction method below? Step 1. A ray is extended from A, and 30 arcs of equal lengths are cut, cutting the ray at A_{1}, A_{2},…A_{30}Step 2. A line is drawn fromÂ A_{30} to B, and a line parallel to A_{30}B is drawn, passing through point A_{17} and meeting AB at C.
Answer: (C) 17:13
Solution: Here, the total number of arcs is equal to m+n in the ratio m:n.
The trianglesÂ â–³Â AA_{17}CÂ andÂ â–³Â AA_{30}BÂ are similar.
Hence, AC/ABÂ =Â AA_{17}/AA_{30}Â =Â 17/30
BC/ABÂ = (ABâˆ’AC)/AB
BC/ABÂ = 1 –Â 17/30Â =Â 13/30
Hence,Â AC/BCÂ =Â 17/13Â =Â 17:13.

 What is the ratio AC/BC for the following construction: A Â line segment AB is drawn. A single ray is extended from A, and 12 arcs of equal lengths are cut, cutting the ray at A1, A2â€¦ A12. A line is drawn from A12 to B, and a line parallel to A12B is drawn, passing through point A6 and cutting AB at C.
 1:2
 1:1
 2:1
 3:1
 What is the ratio AC/BC for the following construction: A Â line segment AB is drawn. A single ray is extended from A, and 12 arcs of equal lengths are cut, cutting the ray at A1, A2â€¦ A12. A line is drawn from A12 to B, and a line parallel to A12B is drawn, passing through point A6 and cutting AB at C.
Answer: (B) 1:1
Solution:
In the construction process given, trianglesÂ â–³AA_{12}BÂ Â Â â–³AA_{6}CÂ are similar.
Hence, we getÂ AC/ABÂ =Â 6/12Â =Â 1/2.
By construction, BC/AB = 6/12 = 1/2.
AC/BCÂ =Â (AC/AB)/ (BC/AB)
=Â (Â½)/ (Â½)Â = 1.

 The basic principle used in dividing a line segment is:
 Tangent to a circle
 Congruency of triangles
 Similarity of triangles
 None of these
 The basic principle used in dividing a line segment is:
Answer: (C) Similarity of triangles
Solution:
The similarity of triangles is the basic principle used in dividing a line segment.
In this case, similar trianglesÂ ACA_{3}Â andÂ ABA_{5}Â have been constructed to divide the line segment AB.

 To divide a line segment, the ratio of division must be:
 Negative and rational
 Greater than 1
 Less than 1
 Positive and rational
 To divide a line segment, the ratio of division must be:
Answer: (D) Positive and rational
Solution: The ratio of division must always be positive and rational. It can be greater than or less than 1.
Furthermore, Class 10 Maths Chapter 11 deals with various concepts of Construction, such as the Construction of Line Segments, Division of a Line Segment and Construction of a Circle, Construction of Tangents to a Circle using an analytical approach, and more.
Also, we have compiled below some extra questions for the students to solve:
CBSE Class 10 Maths Chapter 11 Extra Questions
1. To divide a line segment PQ in the ratio 5:7, first, a ray PX is drawn so that âˆ QPX is an acute angle, and then at equal distances, points are marked on the ray PX. If so, what will be the minimum number of these points?
(a) 5
(b) 7
(c) 12
(d) 10
2. If a line segment is divided in the ratio 2:3, calculate the number of parts it is divided into.Â
(a)Â 2/3
(b) 2
(c) 3
(d) 5
3.Â If a pair of tangents have to be constructed from a point P to a circle of radius 3.5 cm. Then at what distance from the centre should the point be?
(a) 5 cm
(b) 2 cm
(c) 3 cm
(d) 3.5 cm
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