NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

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NCERT Solutions for Class 10 Maths Chapter 11 Constructions (Ex 11.1) Exercise 11.1 

Mathematics is the foundation of all subjects and helps to develop brainpower.Numbers make up the structure of the universe. Science, Data Management, Engineering, Technology, Space Exploration, and other fields have all relied on Mathematics. The use of Mathematics is essential in daily life. Therefore, understanding Mathematics is necessary in order to see and understand the universe. Mathematics gives life structure and minimises uncertainty. Mathematics enhances students’ capacity for conceptual and visual thinking, creativity, critical analysis, problem-solving, and even effective communication. Students are unfamiliar with the various concepts and formulas used in Mathematics. It is also regarded as a challenging subject, thus students frequently have difficulties with it.Students can use the Extramarks website to access various study materials, such as NCERT Solutions for various Mathematics chapters. For Class 10 Maths Chapter 11 Exercise 11.1, the NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.1 are beneficial for getting good examination results.

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Geometrical Construction involves properly drawing lines, line segments, forms, circles, and other figures with a ruler, compass, or protractor. A straightedge (ruler), a compass, and a pencil are some of the geometric instruments used in the process of building geometric forms. Given that there are no numbers involved, this is the most basic kind of Geometric Construction. The straightedge (ruler) is used to draw the line segments and measure their lengths, while the compass is used to form arcs and circles and mark off equal lengths. Students study several constructions in Geometric Construction basics, including the Construction of Line Segments, Copy of Line Segments, Construction of Angle, and Angle Bisector. For CBSE students preparing for examinations, using the NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.1 provided by the Extramarks’ website is thought to be the best alternative. There are numerous exercises in this chapter. On Extramarks’ website, students can access the NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.1 in PDF format. Students can study the NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.1 straight from the website or mobile application, or can download it as needed.

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Access NCERT Solutions for Class 10 Maths Chapter 11 – Construction

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NCERT Solutions for Class 10 Maths Chapter 11 Constructions (Ex 11.1) Exercise 11.1

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The NCERT solutions are made up of all the chapter-by-chapter answers to the questions found in the NCERT book for Class 10 Mathematics. All of the solutions were created in a very precise and orderly manner while preserving the objectives of the textbooks. The NCERT solutions may be used by the students as additional resources and study tools. It is clear that practising NCERT solutions will help students prepare for their examinations. The NCERT Solutions are made to help students prepare for examinations.  Students who have registered on Extramarks’ website can access the NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.1 to help them in their educational careers. Experts at Extramarks developed the NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.1 to help students get ready for the Class 10 board examination. To help students quickly complete the NCERT questions, specialists created the NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.1. Along with ensuring that students can understand the NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.1 quickly, they also take into account how well the students understand the NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.1.  Along with NCERT solutions, all of the examples from the NCERT textbooks are available on Extramarks’ website. Every NCERT solution for Class 10 is given in a practical and complete manner in order to help students grasp the concepts covered in Class 10 in-depth. Students might feel more confident that they will perform well in their exams because the majority of the questions from the NCERT books appear often in the examination. The purpose of the NCERT answers for Class 10 is to help students understand the principles of each topic and thus improve their core knowledge.

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

Ans.

Following are the steps to divide a line segment of length7.6 cm in the ratio of 5:8.Step 1: Draw line segment AB of 7.6 cm and draw a ray AX making an acute angle with line segment AB.Step 2: Locate 13(=5+8) points A1, A2,  A3,  ...,  A13 on AX                such that AA1=A1A2=A2A3=...=A12A13.Step 3: Join the points B and A13.Step 4: Through the point A5, draw a line parallel to BA13 at                A5 intersecting AB at point C.C is the point which divides line segment AB of length 7.6 cmin the ratio 5:8.On measuring lengths of AC and BC, we get AC=2.9  cm andBC=4.7 cm.

Q.2

Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 23 of thecorresponding sides of the first triangle.

Ans.

Step 1: Draw a line segment AB=4  cm. Taking point A as centre, draw an arc of 5 cm radius. Again, taking point B as centre, draw an arc of 6 cm. These arcs​​​​​​​​​​​​​​​​​​​​ intersect each other at point C. So, we have AC=5 cm and BC=6  cm. Δ ABC is the required triangle.Step 2: Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.Step 3: Locate 3 points A1, A2,  A3 on AX such that                AA1=A1A2=A2A3.Step 4: Join the points B and A3.Step 5: Through the point A2, draw a line parallel to BA3                intersecting AB at point B‘.Step 6: Draw a line through B‘ parallel to the line BC to intersect AC at C‘.The required triangle is ΔAB‘C‘.

Q.3

Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 75 of thecorresponding sides of the first triangle.

Ans.

Step 1: Draw a line segment AB=5  cm. Taking point A as centre, draw an arc of 6 cm radius. Again, taking point B as centre, draw an arc of 7 cm. These arcs​​​​​​​​​​​​​​​​​​​​ intersect each other at point C. So, we have AC=6 cm and BC=7  cm. Δ ABC is the required triangle.Step 2: Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.Step 3: Locate 7 points A1, A2,  A3,...,A7 on AX such that                AA1=A1A2=A2A3=...=A6A7.Step 4: Join the points B and A5.Step 5: Through the point A7, draw a line parallel to BA5                intersecting extended line segment AB at point B’.Step 6: Draw a line through B’ parallel to the line BC to intersect extended line segment AC at C’.The required triangle is ΔAB ’C’.

Q.4

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

Ans.

Steps of Construction:Step 1: Draw a line segment AB=8  cm. Draw arcs of same                 radius on both sides of the line segment while taking point A and B as its centre. Let these arcs intersect                each other at O and O’. Join OO’. Let OO’ intersect  AB at D.Step 2: Taking D as centre, draw an arc of 4 cm radius which cuts the extended line segment OO’ at point C. An isosceles Δ ABC is formed havind CD as 4 cm and AB as 8 cm.​​​​​​​​​​​​​​​​​Step 3: Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.Step 4: Locate 3 points A1, A2,  A3 on AX such that                AA1=A1A2=A2A3.Step 5: Join the points B and A2.Step 6: Through the point A3, draw a line parallel to BA2                intersecting extended line segment AB at point B’.Step 7: Draw a line through B’ parallel to the line BC to intersect extended line segment AC at C’.The required triangle is ΔAB ’C’.

Q.5

Draw a triangle ABC with side BC=6 cm, AB=5 cm and ∠ABC=60°. Then construct a triangle whose sides are34  of the corresponding sides of the triangle ABC.

Ans.

Steps of Construction:Step 1: Draw a Δ ABC with sides AB=5  cm, BC=6 cm and ∠ABC=60°. ​​​​​​​​​​​​​​​​​Step 2: Draw a ray BX making an acute angle with line BC on the opposite side of vertex A.Step 3: Locate 4 points B1, B2,  B3,  B4 on BX such that                BB1=B1B2=B2B3=B3B4.Step 4: Join the points C and B4.Step 5: Through the point B3, draw a line parallel to CB4                intersecting line segment BC at point C’.Step 6: Draw a line through C’ parallel to the line AC to intersect line segment AB at A’.The required triangle is ΔA ’BC’.

Q.6

Draw a triangle ABC with sides BC=7 cm, ∠=°,∠=°, then construct a triangle whose sides are 43 times the corresponding sides of Δ ABC.

Ans.

It is given that ∠B=45°, ∠A=105°.∴ In Δ ABC, we have ∠C=180°−105°−45°=30°StepsofConstruction:Step 1: Draw a Δ ABC with side BC=7 cm and ∠B=45°, ∠C=30°. ​​​​​​​​​​​​​​​​​Step 2: Draw a ray BX making an acute angle with line BC on the opposite side of vertex A.Step 3: Locate 4 points B1, B2,  B3,  B4 on BX such that                BB1=B1B2=B2B3=B3B4.Step 4: Join the points C and B3.Step 5: Through the point B4, draw a line parallel to CB3                intersecting extended line segment BC at point C’.Step 6: Draw a line through C’ parallel to the line AC to intersect extended line segment BA at A’.The required triangle is ΔA ’BC’.

Q.7

Draw a right triangle in which the sides (other than hypotenuse) are of length 4 cm and 3 cm. Thenconstruct another triangle whose sides are 53 times the corresponding sides of the given triangle.

Ans.

Steps of Construction:Step 1: Draw a right angle Δ ABC with base AB=4 cm, AC=3 cm and ∠A=90°. ​​​​​​​​​​​​​​​​​Step 2: Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.Step 3: Locate 5 points A1, A2,  A3,  A4, A5 on AX such that                AA1=A1A2=A2A3=A3A4=A4A5.Step 4: Join the points B and A3.Step 5: Through the point A5, draw a line parallel to BA3                intersecting extended line segment AB at point B’.Step 6: Draw a line through B’ parallel to the line BC to intersect extended line segment AC at C’.The required triangle is ΔAB’ C‘.

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