NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry (EX 10.3) Exercise 10.3

Mathematics has been a part of the academic careers of students since the beginning of their studies. It can be the most scoring subject for the learners of Class 7, but some students may find it challenging. Therefore, Extramarks offers them  Class 7 Maths Chapter 10 Exercise 10.3 to help learners gain a better understanding of the chapters of the Class 7 Mathematics curriculum. The Extramarks website also provides students with revision notes, important questions, extra questions, sample papers, the Class 7 Maths Chapter 10 Exercise 10.3 Solutions and much more. This allows them to prepare for Mathematics and various other subjects very efficiently. It has always been Extramarks’ goal to help students achieve academic excellence. It provides them with all the study material they need to succeed in their examinations. In addition to the NCERT Solutions Class 7 Maths Chapter 10 Exercise 10.3, the Extramarks website also provides students with CBSE past years’ papers, important questions, revision notes, and much more. Students should thoroughly review the past years’ papers, in order to prepare for the CBSE examination, as they provide them with a glimpse of the Class 7 Mathematics examinations. They are one of the most important resources to enable students to prepare for examinations.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry (EX 10.3) Exercise 10.3 

Class 7 Mathematics examination may not be easy, and it may not contain straightforward questions. Furthermore, it is imperative that students develop a steady pace to complete the examination on time, as the question paper may be lengthy. In order to familiarise themselves with the examination pattern, students can refer to Extramarks’ NCERT Solutions Class 7 Maths Chapter 10 Exercise 10.3. The solutions provided by Extramarks are precise and well-detailed, so students can easily understand them. These solutions are available for download on any browser from the Extramarks website for free. In these solutions, every question is answered in a way that helps students understand how the questions are actually answered. As a result, Extramarks recommends that students learn and practise NCERT Solutions Class 7 Maths Chapter 10 Exercise 10.3 in order to score better in their examinations. These solutions are written in simple and straightforward language. This helps students find answers to all the difficult problems in the exercise questions with the help of NCERT Solutions Class 7 Maths Chapter 10 Exercise 10.3.Students can enhance their academic development on an individual basis with the assistance of Extramarks. It is difficult for Class 7 students to keep  track of all the topics of their curriculum due to the extensive syllabus for each subject. As a result, they can not always complete their entire academic curriculum on time. To ensure that students do not miss any important topics of the curriculum, Extramarks offers them the NCERT Solutions Class 7 Maths Chapter 10 Exercise 10.3. As part of its initiative, Extramarks provides students with all the learning tools they need to succeed academically. These tools can help them stand out in any examination and have a bright academic future.

Access NCERT Solutions for Class 7 Chapter 10 – Practical Geometry

Class 7 is a very essential academic session for the students as it introduces them to new concepts that are useful in their higher classes and are very important for their academic careers. Due to the wide variety of chapters in Class 7 Mathematics, students may face challenges. Therefore, Extramarks provides them with the NCERT Solutions Class 7 Maths Chapter 10 Exercise 10.3 in order to help them deeply understand the concepts of the chapter and score higher grades. Students must practise these solutions in order to identify their errors and correct them. Furthermore, the NCERT Solutions Class 7 Maths Chapter 10 Exercise 10.3 assists students in identifying the types of questions that may appear in their examinations. They can use these solutions to score higher marks in examinations. The Extramarks website provides students with various tools such as K12 Live Classes, Curriculum Mapping, Complete Syllabus Coverage, Live Doubt Solving Classes and much more, along with the NCERT Solutions Class 7 Maths Chapter 10 Exercise 10.3. Extramarks offers a curriculum-mapped learning experience that can be used to improve performance in class. With Extramarks’ In-depth Performance Reports, students can track their academic progress and preparation level. Students can also build their fundamentals through the NCERT Solutions Class 7 Maths Chapter 10 Exercise 10.3 in order to answer any question accurately in their examinations.

Exercise 10.3

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NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.3

The NCERT Solutions Class 7 Maths Chapter 10 Exercise 10.3 provided by Extramarks are compiled by expert teachers of Extramarks. Students can easily understand the concepts of the chapter because the solutions are well-structured and written in simple language. The one-step solution to the problems of students is Extramarks. They can find credible and authentic solutions on the Extramarks website without having to look elsewhere. Students can prepare for their examinations more effectively by going through these solutions. As the NCERT Solutions Class 7 Maths Chapter 10 Exercise 10.3 are available in PDF format, students can download them very easily and access them anytime or anywhere.

Q.1

Construct ΔDEF such that DE=5 cm, DF=3 cm and m EDF=90°.

Ans.

The steps of construction are as follows: (i) Draw a line segment DE of length 5 cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A79F@

(ii) At point D, draw a ray DX making an angle of 90° with DE. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8A92@

(iii) Taking D as centre, draw an arc of 3 cm radius. It will intersect DX at point F MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A6EE@

(iv) Join F to E MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqG2bGDcqqGPaqkcqqGGaaicqqGkbGscqqGVbWBcqqGPbqAcqqGUbGBcqqGGaaicqqGgbGrcqqGGaaicqqG0baDcqqGVbWBcqqGGaaicqqGfbqraaa@53F9@

Thus, DEF is the required triangle. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGubavcqqGObaAcqqG1bqDcqqGZbWCcqqGSaalcqqGGaaicqqGebarcqqGfbqrcqqGgbGrcqqGGaaicqqGPbqAcqqGZbWCcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGYbGCcqqGLbqzcqqGXbqCcqqG1bqDcqqGPbqAcqqGYbGCcqqGLbqzcqqGKbazcqqGGaaicqqG0baDcqqGYbGCcqqGPbqAcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqqGUaGlaaa@6D24@

Q.2

Construct an isosceles triangle in which the lengths ofeach of its equal sides is 6.5 cm and the angle betweenthem is 110°.

Ans.

An isosceles triangle PQR has to be constructed with PQ=QR=6.5 cm. The steps of construction are as follows: (i) Draw a line segment QR of length 6.5 cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@F827@

(ii) At point Q, draw a ray QX making an angle 110°with QR. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqGPbqAcqqGPaqkcqqGGaaicqqGbbqqcqqG0baDcqqGGaaicqqGWbaCcqqGVbWBcqqGPbqAcqqGUbGBcqqG0baDcqqGGaaicqqGrbqucqqGSaalcqqGGaaicqqGKbazcqqGYbGCcqqGHbqycqqG3bWDcqqGGaaicqqGHbqycqqGGaaicqqGYbGCcqqGHbqycqqG5bqEcqqGGaaicqqGrbqucqqGybawcqqGGaaicqqGTbqBcqqGHbqycqqGRbWAcqqGPbqAcqqGUbGBcqqGNbWzcqqGGaaicqqGHbqycqqGUbGBcqqGGaaicqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqqGGaaicqqGXaqmcqqGXaqmcqqGWaamcqGHWcaScqqG3bWDcqqGPbqAcqqG0baDcqqGObaAcqqGGaaicqqGrbqucqqGsbGucqqGUaGlaaa@878D@

(iii) Taking Q as centre, draw an arc of 6.5 cm radius. It intersects QX at point P. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A5D4@

(iv) Join P to R to obtain the required triangle PQR.

Q.3

Construct ΔABC with BC=7.5 cm, AC=5 cm and mC=60°.

Ans.

The steps of constructions are as follows: (i) Draw a line segment BC of length 7.5 cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@AADC@

(ii) At point C, draw a ray CX making 60 with BC. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqGPbqAcqqGPaqkcqqGGaaicqqGbbqqcqqG0baDcqqGGaaicqqGWbaCcqqGVbWBcqqGPbqAcqqGUbGBcqqG0baDcqqGGaaicqqGdbWqcqqGSaalcqqGGaaicqqGKbazcqqGYbGCcqqGHbqycqqG3bWDcqqGGaaicqqGHbqycqqGGaaicqqGYbGCcqqGHbqycqqG5bqEcqqGGaaicqqGdbWqcqqGybawcqqGGaaicqqGTbqBcqqGHbqycqqGRbWAcqqGPbqAcqqGUbGBcqqGNbWzcqqGGaaicqqG2aGncqqGWaamcqqGGaaicqqG3bWDcqqGPbqAcqqG0baDcqqGObaAcqqGGaaicqqGcbGqcqqGdbWqcqqGUaGlaaa@7A2A@

(iii) Taking C as centre, draw an arc of 5 cm radius. It intersect CX at point A. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A23B@

(iv) Join A to B to obtain the required triangle ABC. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqG2bGDcqqGPaqkcqqGGaaicqqGkbGscqqGVbWBcqqGPbqAcqqGUbGBcqqGGaaicqqGbbqqcqqGGaaicqqG0baDcqqGVbWBcqqGGaaicqqGcbGqcqqGGaaicqqG0baDcqqGVbWBcqqGGaaicqqGVbWBcqqGIbGycqqG0baDcqqGHbqycqqGPbqAcqqGUbGBcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGYbGCcqqGLbqzcqqGXbqCcqqG1bqDcqqGPbqAcqqGYbGCcqqGLbqzcqqGKbazcqqGGaaicqqG0baDcqqGYbGCcqqGPbqAcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqqGGaaicqqGbbqqcqqGcbGqcqqGdbWqcqqGUaGlaaa@8184@

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1. Why is it essential to go through all the NCERT Solutions Class 7 Maths Chapter 10 Exercise 10.3?

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