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Chapter 11

Constructions

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NCERT Solution of Class 10th Maths Chapter 11

The NCERT Solutions for Class 10 Maths Chapter 11 Constructions are comprehensive, providing step-by-step answers to all questions for quick review. In this chapter, you will learn how to find a point by internally dividing a line segment by a ratio, how to make construct triangles, how to make a tangent to a circle, how to make a pair of tangents, and how to make a pair of tangents that are at an angle to each other. The following are some general reference notes to help you solve NCERT Chapter 11 Solutions.

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For each of the following, give the justification of the constructions made:

Question 1. Draw a line segment of length 7.6 cm and divide it in the ratio of 5:8. Measure the two parts.

Answer.

Procedures for construction: 

A line segment with a length of 7.6 cm is divided in the ratio of 5:8. 

1. Create a line segment AB with a length of 7.6 cm. 

2. Create an acute angle between ray AX and line segment AB by drawing a ray AX. 

3. Locate the points on the ray AX, i.e., 13 (= 5+8) points, such as A1, A2, A3, A4........ A13, so that AA1 = A1A2 = A2A3 and so on.

4. Create a connection between the line segment and the ray, BA13. 

5. Parallel to BA13, draw a line through point A5 that produces an angle equal to AA13B. 

6. The location A5, where the line AB intersects at point C. 

7. C is the point that divides the 7.6-cm-long line segment AB in the appropriate 5:8 ratio. 

8. Now, determine the lengths of the AC and CB lines. It comes to 2.9 cm and 4.7 cm in length.

Justification:

The given problem's structure can be justified by demonstrating that 

AC/CB = 5/8

A5C || A13B is our construction. We obtain the following from the fundamental proportionality theorem for the triangle AA13B:

AC/CB = AA5/A5A13.... (1) 

AA5 and A5A13 contain five and eight equal divisions of line segments, respectively, as seen in the figure. 

As a result, it becomes 

AA5/A5A13=5/8… (2) 

By comparing equations (1) and (2), we obtain at 

AC/CB = 5/8 

Hence proved.

 

Question 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.

Answer.

Procedure for Construction: 

1. Draw a 4-centimetre line segment AB, i.e. AB = 4 centimetres. 

2. Draw a 5 cm radius arc with point A at the centre. 

3. Draw an arc with a radius of 6 cm with point B as the centre. 

4. At point C, the sketched arcs will cross. 

5. With AC = 5 cm and BC = 6 cm, the needed triangle is ΔABC. 

6. On the opposite side of vertex C, draw a ray AX that forms an acute angle with the line segment AB. 

7. Draw a line AX with three points A1, A2, and A3 (since 3 is higher than 2 and 3) so that AA1= A1A2 = A2A3. 

8. Draw a line through A2 parallel to BA3 and intersecting AB at point B'. 

9. Draw a line parallel to the line BC from point B' to point C', intersecting the line AC. 

10. Therefore, ΔAB’C’ is the required triangle.

Justification:

It is possible to justify the problem's construction by demonstrating that 

(2/3)AB = AB' 

(2/3)BC = B'C'

AC' equals (2/3)AC 

We get B'C' || BC from the construction. 

∴ ∠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,

∠A2AB’ =∠A3AB (Common) 

We get, from the corresponding angles, 

∠AA2B’ =∠AA3B 

As a result of applying the AA similarity criterion, we obtain: 

ΔAA2B’ and AA3B

So, 

AB’/AB = AA2/AA3 

AB'/AB = 2/3 as a result..... (2) 

We get (1) and (2) from the equations. 

AB’/AB=B’ 

C’/BC = AC’/ AC = 2/3 

This can be expressed as 

(2/3)AB = AB' 

(2/3)BC = B'C' 

AC' = (2/3)AC

Thus the result is justified.

Question 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.

Answer.

Construction Procedures: 

1. Draw a line segment with the length AB = 5 cm. 

2. Using A and B as the centres, draw arcs with radii of 6 and 7 cm, respectively. 

3. Because these arcs will cross at point C, ΔABC is the requisite triangle with sides of 5 cm, 6 cm, and 7 cm, respectively. 

4. Draw a ray AX on the opposite side of vertex C that forms an acute angle with the line segment AB. 

5. On the line AX, locate the seven points A1, A2, A3, A4, A5, A6, and A7 (since 7 is greater than 5 and 7), such that AA1 = A1A2 = A2A3 = A3A4 = A3A5 = A4A6 = A6A7 

6. Connect the points BA5 and create a parallel line from A7 to BA5 that intersects the extended line segment AB at point B'. 

7. Now, draw a line from B' to C' that is parallel to the extended line segment AC, intersecting the line BC to form a triangle. 

8. So, the needed triangle is ΔAB’C’.

Justification: 

The given problem's structure can be justified by demonstrating that 

AB' = (7/5)AB 

B’C’ = (7/5)BC 

AC’= (7/5)AC

We get B'C' || BC from the construction. 

∴ ∠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 ΔAA7B’ and ΔAA5B, 

∠A7AB’=∠A5AB (Common) 

We obtain the following from the corresponding angles: 

∠A A7B’=∠A A5B 

As a result of the AA similarity criterion, we arrive at 

ΔA A2B’ and A A3B

Thus, AB'/AB = AA5/AA7. 

As a result, AB /AB' = 5/7... (2) 

We get from equations (1) and (2) 

AB'/AB = B'C'/BC = AC'/AC = 7/5 

This can also be expressed as 

AB' = (7/5)AB 

B’C’ = (7/5)BC 

AC’= (7/5)AC 

Hence justified.

Question 4. Construct an isosceles triangle whose base is 8 cm and altitude is 4cm, then another triangle whose sides are 112 times the corresponding sides of the isosceles triangle.

Answer.

Procedure for Construction: 

1. Draw a line segment BC with an 8 cm length. 

2. Now, create a perpendicular bisector of the line segment BC that intersects at point D. 

3. Using D as the centre, draw an arc with a radius of 4 cm that intersects the perpendicular bisector at A.

4. Now, connect the lines AB and AC to form the needed triangle. 

5. On the side opposite the vertex A, draw a ray BX that forms an acute angle with the line BC. 

6. On the ray BX, locate the three points B1, B2, and B3 such that BB1 = B1B2 = B2B3. 

7. Connect the points B2C and construct a line parallel to the line B2C from B3 to point C' where it intersects the extended line segment BC. 

8. Now, draw a line from C' to A' so it is parallel to the extended line segment AC and intersects the line AC to form a triangle. 

9. As a result, the needed triangle is ΔA’BC’'.

Justification: 

The given problem's structure can be justified by demonstrating that 

A'B = (3/2)AB 

BC' = (3/2)BC 

A’C’= (3/2)AC

We get A'C' || AC from the construction.

∴ ∠ A’C’B = ∠ACB (Corresponding angles)

In ΔA’BC’ and ΔABC,

∠B = ∠B (common)

∠A’BC’ = ∠ACB

∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

As a result, A'B/AB = BC'/BC = A'C'/AC.

Because the similar triangle's corresponding sides have the same ratio, it becomes

A'B/AB = BC'/BC = A'C'/AC equals 3/2

As a result, justified.

Points to remember

  • A point that moves in the same direction as two other points has a locus that is normal to the line linking them. 

  • A bisector divides a line segment in half, whereas perpendicular or normal refers to straight angles. 

  • Using a pair of compasses and a ruler, you can make a variety of forms. 

  • Examine the situation and offer a clear solution.

NCERT solutions can also help students with homework and exam preparation for CBSE term-based and competitive exams. The learning experience is enhanced by using visuals to illustrate each question. Subject experts have supplied solutions for the 11th chapter of NCERT Class 10 mathematics in accordance with NCERT rules to assist students in preparation for their second term exams.

 To help students prepare for the second term exams, msvgo provides free NCERT Solutions for Class 10 Maths, Chapter 11: Constructions. All NCERT exercises are done by constructing diagrams in a step-by-step manner. An answer to a question on the NCERT test helps students understand concepts well and clear up any questions they may have.

If you want to gain access to 15,000+ videos explaining NCERT concepts and our 10,000+ questions bank, download the msvgo learning app and become a msvgo champ now!

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