NCERT Solutions for Class 10 Maths Chapter 11 Constructions
Here, we have provided NCERT Solutions for Class 10 Maths Chapter 11. Students can view these NCERT Solutions for Class 10 Maths Chapter 11 Constructions before exams for better understanding of the chapter.
Neha Tanna22 Mar, 2024
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NCERT Solutions for Class 10 Maths Chapter 11
NCERT Solutions for Class 10 Maths Chapter 11: Students may obtain a step-by-step solution to every question for quick revisions by using the comprehensive NCERT Solutions for Class 10 Maths Chapter 11 Constructions. To help students prepare for their board exams, subject matter experts have created solutions for the 11th chapter of NCERT Class 10 Maths, following NCERT criteria.
To speed up your preparation for the CBSE exam, get free NCERT Solutions for Class 10 Maths, Chapter 11 - Constructions here. Every NCERT exercise question has a diagram with a step-by-step construction process to help with the solution. Students can improve their concepts and dispel uncertainties with the aid of NCERT Solutions .NCERT Solutions for Class 10 Maths Chapter 11 PDF
Chapter 11 of Class 10 Maths focuses on building various geometric forms. Teaching the art of drawing division points for line segments and closed figures is the main goal of this chapter. To properly prepare for this foundational chapter, the entire study material is required. The answers to the Class 10 Maths Constructions exercises should be in this resource. Here we have provided NCERT Solutions for Class 10 Maths Chapter 11 pdf -NCERT Solutions for Class 10 Maths Chapter 11 PDF
NCERT Solutions for Class 10 Maths Chapter 11 Constructions
1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts .
Construction Procedure A line segment with a measure of 7.6 cm length is divided in the ratio of 5:8 as follows. 1. Draw line segment AB with a length measure of 7.6 cm. 2. Draw a ray AX that makes an acute angle with line segment AB. 3. Locate the points, i.e.,13 (= 5+8) points, such as A1, A2, A3, A4 …….. A13, on the ray AX, such that it becomes AA1 = A1A2 = A2A3 and so on. 4. Join the line segment and the ray, BA13. 5. Through the point A5, draw a line parallel to BA13 which makes an angle equal to ∠AA13B. 6. Point A5, which intersects line AB at point C. 7. C is the point that divides line segment AB of 7.6 cm in the required ratio of 5:8. 8. Now, measure the lengths of the lines AC and CB. It becomes the measure of 2.9 cm and 4.7 cm, respectively.
Justification:
The construction of the given problem can be justified by proving that
AC/CB = 5/ 8
By construction, we have A5C || A13B. From the Basic proportionality theorem for the triangle AA13B, we get
AC/CB =AA
5
/A
5
A
13
….. (1)
From the figure constructed, it is observed that AA5 and A5A13 contain 5 and 8 equal divisions of line segments, respectively.
Therefore, it becomes
AA
5
/A
5
A
13
=5/8… (2)
Compare the equations (1) and (2), we obtain
AC/CB = 5/ 8
Hence, justified.
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2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of
the corresponding sides of the first triangle.
Construction Procedure 1. Draw a line segment AB which measures 4 cm, i.e., AB = 4 cm. 2. Take point A as the centre, and draw an arc of radius 5 cm. 3. Similarly, take point B as its centre, and draw an arc of radius 6 cm. 4. The arcs drawn will intersect each other at point C. 5. Now, we have obtained AC = 5 cm and BC = 6 cm, and therefore, ΔABC is the required triangle. 6. Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C. 7. Locate 3 points such as A1, A2, and A3 (as 3 is greater between 2 and 3) on line AX such that it becomes AA1= A1A2 = A2A3. 8. Join point BA3 and draw a line through A2, which is parallel to the line BA3 that intersects AB at point B’. 9. Through point B’, draw a line parallel to line BC that intersects the line AC at C’. 10. Therefore, ΔAB’C’ is the required triangle.
Justification
The construction of the given problem can be justified by proving that
AB’ = (2/3)AB
B’C’ = (2/3)BC
AC’= (2/3)AC
From the construction, we get B’C’ || BC
∴ ∠AB’C’ = ∠ABC (Corresponding angles)
In ΔAB’C’ and ΔABC,
∠ABC = ∠AB’C (Proved above)
∠BAC = ∠B’AC’ (Common)
∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion)
Therefore, AB’/AB = B’C’/BC= AC’/AC …. (1)
In ΔAAB’ and ΔAAB,
∠A
2
AB’ =∠A
3
AB (Common)
From the corresponding angles, we get
∠AA
2
B’ =∠AA
3
B
Therefore, from the AA similarity criterion, we obtain
ΔAA
2
B’ and AA
3
B
So, AB’/AB = AA
2
/AA
3
Therefore, AB’/AB = 2/3 ……. (2)
From equations (1) and (2), we get
AB’/AB=B’C’/BC = AC’/ AC = 2/3
This can be written as
AB’ = (2/3)AB
B’C’ = (2/3)BC
AC’= (2/3)AC
Hence, justified.
3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle
Construction Procedure 1. Draw a line segment AB =5 cm. 2. Take A and B as the centre, and draw the arcs of radius 6 cm and 7 cm, respectively. 3. These arcs will intersect each other at point C, and therefore, ΔABC is the required triangle with the length of sides as 5 cm, 6 cm, and 7 cm, respectively. 4. Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C. 5. Locate the 7 points, such as A 1 , A 2 , A 3 , A 4 , A 5 , A 6 , A 7 (as 7 is greater between 5 and 7), on line AX such that it becomes AA 1 = A 1 A 2 = A 2 A 3 = A 3 A 4 = A 4 A 5 = A 5 A 6 = A 6 A 7 6. Join the points BA 5 and draw a line from A 7 to BA 5, which is parallel to the line BA 5 where it intersects the extended line segment AB at point B’. 7. Now, draw a line from B’ to the extended line segment AC at C’, which is parallel to the line BC, and it intersects to make a triangle. 8. Therefore, ΔAB’C’ is the required triangle.
Justification
The construction of the given problem can be justified by proving that
AB’ = (7/5)AB
B’C’ = (7/5)BC
AC’= (7/5)AC
From the construction, we get B’C’ || BC
∴ ∠AB’C’ = ∠ABC (Corresponding angles)
In ΔAB’C’ and ΔABC,
∠ABC = ∠AB’C (Proved above)
∠BAC = ∠B’AC’ (Common)
∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion)
Therefore, AB’/AB = B’C’/BC= AC’/AC …. (1)
In ΔAA
7
B’ and ΔAA
5
B,
∠A
7
AB’=∠A
5
AB (Common)
From the corresponding angles, we get
∠A A
7
B’=∠A A
5
B
Therefore, from the AA similarity criterion, we obtain
ΔA A
2
B’ and A A
3
B
So, AB’/AB = AA
5
/AA
7
Therefore, AB /AB’ = 5/7 ……. (2)
From equations (1) and (2), we get
AB’/AB = B’C’/BC = AC’/ AC = 7/5
This can be written as
AB’ = (7/5)AB
B’C’ = (7/5)BC
AC’= (7/5)AC
Hence, justified.
Construction Procedure:
1. Draw a line segment BC with a measure of 8 cm.
2. Now, draw the perpendicular bisector of the line segment BC and intersect at point D.
3. Take the point D as the centre and draw an arc with a radius of 4 cm, which intersects the perpendicular bisector at the point A.
4. Now, join the lines AB and AC, and the triangle is the required triangle.
5. Draw a ray BX which makes an acute angle with the line BC on the side opposite to the vertex A.
6. Locate the 3 points B
1
, B
2
and B
3
on the ray BX such that BB
1
= B
1
B
2
= B
2
B
3
7. Join the points B
2
C and draw a line from B
3,
which is parallel to the line B
2
C where it intersects the extended line segment BC at point C’.
8. Now, draw a line from C’ to the extended line segment AC at A’, which is parallel to the line AC, and it intersects to make a triangle.
9. Therefore, ΔA’BC’ is the required triangle.
Justification
The construction of the given problem can be justified by proving that
A’B = (3/2)AB
BC’ = (3/2)BC
A’C’= (3/2)AC
From the construction, we get A’C’ || AC
∴ ∠ A’C’B = ∠ACB (Corresponding angles)
In ΔA’BC’ and ΔABC,
∠B = ∠B (Common)
∠A’BC’ = ∠ACB
∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)
Therefore, A’B/AB = BC’/BC= A’C’/AC
Since the corresponding sides of the similar triangle are in the same ratio, it becomes
A’B/AB = BC’/BC= A’C’/AC = 3/2
Hence, justified.
5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.
Construction Procedure 1. Draw a ΔABC with base side BC = 6 cm, and AB = 5 cm and ∠ABC = 60°. 2. Draw a ray BX which makes an acute angle with BC on the opposite side of vertex A. 3. Locate 4 points (as 4 is greater in 3 and 4), such as B1, B2, B3, and B4, on line segment BX. 4. Join the points B4C and also draw a line through B3, parallel to B4C intersecting the line segment BC at C’. 5. Draw a line through C’ parallel to the line AC, which intersects the line AB at A’. 6. Therefore, ΔA’BC’ is the required triangle.6. Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ∆ ABC.
To find ∠C: Given: ∠B = 45°, ∠A = 105° We know that, The sum of all interior angles in a triangle is 180°. ∠A+∠B +∠C = 180° 105°+45°+∠C = 180° ∠C = 180° − 150° ∠C = 30° So, from the property of the triangle, we get ∠C = 30° Construction Procedure The required triangle can be drawn as follows. 1. Draw a ΔABC with side measures of base BC = 7 cm, ∠B = 45°, and ∠C = 30°. 2. Draw a ray BX that makes an acute angle with BC on the opposite side of vertex A. 3. Locate 4 points (as 4 is greater in 4 and 3), such as B1, B2, B3, and B4, on the ray BX. 4. Join the points B3C. 5. Draw a line through B4 parallel to B3C, which intersects the extended line BC at C’. 6. Through C’, draw a line parallel to the line AC that intersects the extended line segment at C’. 7. Therefore, ΔA’BC’ is the required triangle.
Justification
The construction of the given problem can be justified by proving that
Since the scale factor is 4/3, we need to prove
A’B = (4/3)AB
BC’ = (4/3)BC
A’C’= (4/3)AC
From the construction, we get A’C’ || AC
In ΔA’BC’ and ΔABC,
∴ ∠A’C’B = ∠ACB (Corresponding angles)
∠B = ∠B (Common)
∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)
Since the corresponding sides of the similar triangle are in the same ratio, it becomes
Therefore, A’B/AB = BC’/BC= A’C’/AC
So, it becomes A’B/AB = BC’/BC= A’C’/AC = 4/3
Hence, justified.
7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.
Given: The sides other than the hypotenuse are of lengths 4cm and 3cm. It defines that the sides are perpendicular to each other Construction Procedure The required triangle can be drawn as follows. 1. Draw a line segment BC =3 cm. 2. Now, measure and draw an angle of 90° 3. Take B as the centre draw an arc with a radius of 4 cm, and intersect the ray at point B. 4. Now, join the lines AC, and the triangle ABC is the required triangle. 5. Draw a ray BX that makes an acute angle with BC on the opposite side of vertex A. 6. Locate 5 such as B1, B2, B3, B4, on the ray BX, such that BB 1 = B 1 B 2 = B 2 B 3 = B 3 B 4 = B 4 B 5 7. Join the points B3C. 8. Draw a line through B5 parallel to B3C, which intersects the extended line BC at C’. 9. Through C’, draw a line parallel to the line AC that intersects the extended line AB at A’. 10. Therefore, ΔA’BC’ is the required triangle.
Justification
The construction of the given problem can be justified by proving that
Since the scale factor is 5/3, we need to prove
A’B = (5/3)AB
BC’ = (5/3)BC
A’C’= (5/3)AC
From the construction, we get A’C’ || AC
In ΔA’BC’ and ΔABC,
∴ ∠ A’C’B = ∠ACB (Corresponding angles)
∠B = ∠B (Common)
∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)
Since the corresponding sides of the similar triangle are in the same ratio, it becomes
Therefore, A’B/AB = BC’/BC= A’C’/AC
So, it becomes A’B/AB = BC’/BC= A’C’/AC = 5/3
Hence, justified.
NCERT Solutions for Class 10 Maths Chapter 11 FAQs
Is maths class 10 tough?
The paper was of easy to moderate level in difficulty with NCERT questions. The CBSE Class 10 Mathematics board exam, held on March 11, 2024, was considered reasonably easy though a little lengthy.
How many exercises are there in ch 11 maths class 10?
Class 10 Maths Chapter 11 Constructions includes the two exercises having variety of questions: This chapter will make use of the basics taught in class 9 such as the construction of a line segment, bisecting an angle, constructing a perpendicular bisector of a line segment, construction of triangles etc.
Which is the easiest math chapter?
Real numbers are considered as one of the easiest chapters.
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