Class 10 Exam > Class 10 Notes > Mathematics (Maths) Class 10 > RD Sharma Solutions: Triangles (Exercise 6A)
Triangles (Exercise 6A) RD Sharma Solutions | Mathematics (Maths) Class 10 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1 1. (i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC. (ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD. (iii) If = and AC = 6.6cm, find AE. (iv) If = and EC = 3.5cm, find AE. Sol: (i) BC. AD = 3.6 cm , AB = 10 cm, AE = 4.5cm DB = 10 3.6 = 6.4cm Or, Or, EC = Or, EC= 8 cm Thus, AC = AE + EC = 4.5 + 8 = 12.5 cm (ii) t is given that DE || BC. Adding 1 to both sides, we get : +1 (iii) BC. Adding 1 to both the sides, we get : EC = = 4.2 cm Therefore, Page 2 1. (i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC. (ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD. (iii) If = and AC = 6.6cm, find AE. (iv) If = and EC = 3.5cm, find AE. Sol: (i) BC. AD = 3.6 cm , AB = 10 cm, AE = 4.5cm DB = 10 3.6 = 6.4cm Or, Or, EC = Or, EC= 8 cm Thus, AC = AE + EC = 4.5 + 8 = 12.5 cm (ii) t is given that DE || BC. Adding 1 to both sides, we get : +1 (iii) BC. Adding 1 to both the sides, we get : EC = = 4.2 cm Therefore, AE = AC EC = 6.6 4.2 = 2.4 cm (iv) 8AE + 28 = 15AE 7AE = 28 AE = 4cm 2. D and E are the value of x, when (i) AD = x cm, DB = (x 2) cm, AE = (x + 2) cm and EC = (x 1) cm. (ii) AD = 4cm, DB = (x 4) cm, AE = 8cm and EC = (3x 19) cm. (iii) AD = (7x 4) cm, AE = (5x 2) cm, DB = (3x + 4) cm and EC = 3x cm. Sol: (i) || BC. : (ii) : = 4 (3x-19) = 8 (x-4) 12x 76 = 8x 32 4x = 44 x = 11 cm (iii) || BC. 3x (7x-4) = (5x-2) (3x+4) Page 3 1. (i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC. (ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD. (iii) If = and AC = 6.6cm, find AE. (iv) If = and EC = 3.5cm, find AE. Sol: (i) BC. AD = 3.6 cm , AB = 10 cm, AE = 4.5cm DB = 10 3.6 = 6.4cm Or, Or, EC = Or, EC= 8 cm Thus, AC = AE + EC = 4.5 + 8 = 12.5 cm (ii) t is given that DE || BC. Adding 1 to both sides, we get : +1 (iii) BC. Adding 1 to both the sides, we get : EC = = 4.2 cm Therefore, AE = AC EC = 6.6 4.2 = 2.4 cm (iv) 8AE + 28 = 15AE 7AE = 28 AE = 4cm 2. D and E are the value of x, when (i) AD = x cm, DB = (x 2) cm, AE = (x + 2) cm and EC = (x 1) cm. (ii) AD = 4cm, DB = (x 4) cm, AE = 8cm and EC = (3x 19) cm. (iii) AD = (7x 4) cm, AE = (5x 2) cm, DB = (3x + 4) cm and EC = 3x cm. Sol: (i) || BC. : (ii) : = 4 (3x-19) = 8 (x-4) 12x 76 = 8x 32 4x = 44 x = 11 cm (iii) || BC. 3x (7x-4) = (5x-2) (3x+4) 21 12x = 15 + 14x-8 6 26x + 8 = 0 (x-4) (6x-2) =0 x = 4, (as if x = x = 4 cm 3. D and E are points on the sides AB and AC (i) AD = 5.7cm, DB = 9.5cm, AE = 4.8cm and EC = 8cm. (ii) AB = 11.7cm, AC = 11.2cm, BD = 6.5cm and AE = 4.2cm. (iii) AB = 10.8cm, AD = 6.3cm, AC = 9.6cm and EC = 4cm. (iv) AD = 7.2cm, AE = 6.4cm, AB = 12cm and AC = 10cm. Sol: (i) We have: Hence, We conclude that DE || BC. (ii) We have: AB = 11.7 cm, DB = 6.5 cm Therefore, AD = 11.7 -6.5 = 5.2 cm Similarly, AC = 11.2 cm, AE = 4.2 cm Therefore, EC = 11.2 4.2 = 7 cm Now, Thus, We conclude that DE is not parallel to BC. (iii) We have: AB = 10.8 cm, AD = 6.3 cm Therefore, Page 4 1. (i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC. (ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD. (iii) If = and AC = 6.6cm, find AE. (iv) If = and EC = 3.5cm, find AE. Sol: (i) BC. AD = 3.6 cm , AB = 10 cm, AE = 4.5cm DB = 10 3.6 = 6.4cm Or, Or, EC = Or, EC= 8 cm Thus, AC = AE + EC = 4.5 + 8 = 12.5 cm (ii) t is given that DE || BC. Adding 1 to both sides, we get : +1 (iii) BC. Adding 1 to both the sides, we get : EC = = 4.2 cm Therefore, AE = AC EC = 6.6 4.2 = 2.4 cm (iv) 8AE + 28 = 15AE 7AE = 28 AE = 4cm 2. D and E are the value of x, when (i) AD = x cm, DB = (x 2) cm, AE = (x + 2) cm and EC = (x 1) cm. (ii) AD = 4cm, DB = (x 4) cm, AE = 8cm and EC = (3x 19) cm. (iii) AD = (7x 4) cm, AE = (5x 2) cm, DB = (3x + 4) cm and EC = 3x cm. Sol: (i) || BC. : (ii) : = 4 (3x-19) = 8 (x-4) 12x 76 = 8x 32 4x = 44 x = 11 cm (iii) || BC. 3x (7x-4) = (5x-2) (3x+4) 21 12x = 15 + 14x-8 6 26x + 8 = 0 (x-4) (6x-2) =0 x = 4, (as if x = x = 4 cm 3. D and E are points on the sides AB and AC (i) AD = 5.7cm, DB = 9.5cm, AE = 4.8cm and EC = 8cm. (ii) AB = 11.7cm, AC = 11.2cm, BD = 6.5cm and AE = 4.2cm. (iii) AB = 10.8cm, AD = 6.3cm, AC = 9.6cm and EC = 4cm. (iv) AD = 7.2cm, AE = 6.4cm, AB = 12cm and AC = 10cm. Sol: (i) We have: Hence, We conclude that DE || BC. (ii) We have: AB = 11.7 cm, DB = 6.5 cm Therefore, AD = 11.7 -6.5 = 5.2 cm Similarly, AC = 11.2 cm, AE = 4.2 cm Therefore, EC = 11.2 4.2 = 7 cm Now, Thus, We conclude that DE is not parallel to BC. (iii) We have: AB = 10.8 cm, AD = 6.3 cm Therefore, DB = 10.8 6.3 = 4.5 cm Similarly, AC = 9.6 cm, EC = 4cm Therefore, AE = 9.6 4 = 5.6 cm Now, (iv) We have : AD = 7.2 cm, AB = 12 cm Therefore, DB = 12 7.2 = 4.8 cm Similarly, AE = 6.4 cm, AC = 10 cm Therefore, EC = 10 6.4 = 3.6 cm Now, This, heorem, We conclude that DE is not parallel to BC. 4. In a ABC, AD is the bisector of A. (i) If AB = 6.4cm, AC = 8cm and BD = 5.6cm, find DC. (ii) If AB = 10cm, AC = 14cm and BC = 6cm, find BD and DC. (iii) If AB = 5.6cm, BD = 3.2cm and BC = 6cm, find AC. (iv) If AB = 5.6cm, AC = 4cm and DC = 3cm, find BC. Sol: (i) It is give that AD bisects A. Applying angle Page 5 1. (i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC. (ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD. (iii) If = and AC = 6.6cm, find AE. (iv) If = and EC = 3.5cm, find AE. Sol: (i) BC. AD = 3.6 cm , AB = 10 cm, AE = 4.5cm DB = 10 3.6 = 6.4cm Or, Or, EC = Or, EC= 8 cm Thus, AC = AE + EC = 4.5 + 8 = 12.5 cm (ii) t is given that DE || BC. Adding 1 to both sides, we get : +1 (iii) BC. Adding 1 to both the sides, we get : EC = = 4.2 cm Therefore, AE = AC EC = 6.6 4.2 = 2.4 cm (iv) 8AE + 28 = 15AE 7AE = 28 AE = 4cm 2. D and E are the value of x, when (i) AD = x cm, DB = (x 2) cm, AE = (x + 2) cm and EC = (x 1) cm. (ii) AD = 4cm, DB = (x 4) cm, AE = 8cm and EC = (3x 19) cm. (iii) AD = (7x 4) cm, AE = (5x 2) cm, DB = (3x + 4) cm and EC = 3x cm. Sol: (i) || BC. : (ii) : = 4 (3x-19) = 8 (x-4) 12x 76 = 8x 32 4x = 44 x = 11 cm (iii) || BC. 3x (7x-4) = (5x-2) (3x+4) 21 12x = 15 + 14x-8 6 26x + 8 = 0 (x-4) (6x-2) =0 x = 4, (as if x = x = 4 cm 3. D and E are points on the sides AB and AC (i) AD = 5.7cm, DB = 9.5cm, AE = 4.8cm and EC = 8cm. (ii) AB = 11.7cm, AC = 11.2cm, BD = 6.5cm and AE = 4.2cm. (iii) AB = 10.8cm, AD = 6.3cm, AC = 9.6cm and EC = 4cm. (iv) AD = 7.2cm, AE = 6.4cm, AB = 12cm and AC = 10cm. Sol: (i) We have: Hence, We conclude that DE || BC. (ii) We have: AB = 11.7 cm, DB = 6.5 cm Therefore, AD = 11.7 -6.5 = 5.2 cm Similarly, AC = 11.2 cm, AE = 4.2 cm Therefore, EC = 11.2 4.2 = 7 cm Now, Thus, We conclude that DE is not parallel to BC. (iii) We have: AB = 10.8 cm, AD = 6.3 cm Therefore, DB = 10.8 6.3 = 4.5 cm Similarly, AC = 9.6 cm, EC = 4cm Therefore, AE = 9.6 4 = 5.6 cm Now, (iv) We have : AD = 7.2 cm, AB = 12 cm Therefore, DB = 12 7.2 = 4.8 cm Similarly, AE = 6.4 cm, AC = 10 cm Therefore, EC = 10 6.4 = 3.6 cm Now, This, heorem, We conclude that DE is not parallel to BC. 4. In a ABC, AD is the bisector of A. (i) If AB = 6.4cm, AC = 8cm and BD = 5.6cm, find DC. (ii) If AB = 10cm, AC = 14cm and BC = 6cm, find BD and DC. (iii) If AB = 5.6cm, BD = 3.2cm and BC = 6cm, find AC. (iv) If AB = 5.6cm, AC = 4cm and DC = 3cm, find BC. Sol: (i) It is give that AD bisects A. Applying angle (ii) It is given that AD bisects . Applying angle bisector Let BD be x cm. Therefore, DC = (6- x) cm (iii) It is given that AD bisector Applying angle get: BD = 3.2 cm, BC = 6 cm Therefore, DC = 6- 3.2 = 2.8 cm (iv) It is given that AD bisects Applying angle 5. M is a point on the side BC of a parallelogram ABCD. DM when produced meets AB produced at N. Prove that (i) D M D C M N B N (ii) D N A N D M D C Sol: (i) Given: ABCD is a parallelogramRead More
|
126 videos|457 docs|75 tests
|
Related Exams
About this Document
|
4.89/5 Rating |
|
Dec 26, 2024 Last updated |
Document Description: RD Sharma Solutions: Triangles (Exercise 6A) for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation.
The notes and questions for RD Sharma Solutions: Triangles (Exercise 6A) have been prepared according to the Class 10 exam syllabus. Information about RD Sharma Solutions: Triangles (Exercise 6A) covers topics
like and RD Sharma Solutions: Triangles (Exercise 6A) Example, for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Triangles (Exercise 6A).
Introduction of RD Sharma Solutions: Triangles (Exercise 6A) in English is available as part of our Mathematics (Maths) Class 10
for Class 10 & RD Sharma Solutions: Triangles (Exercise 6A) in Hindi for Mathematics (Maths) Class 10 course.
Download more important topics related with notes, lectures and mock test series for Class 10
Exam by signing up for free. Class 10: Triangles (Exercise 6A) RD Sharma Solutions | Mathematics (Maths) Class 10
Description
Full syllabus notes, lecture & questions for Triangles (Exercise 6A) RD Sharma Solutions | Mathematics (Maths) Class 10 - Class 10 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 10 | Best notes, free PDF download
Information about RD Sharma Solutions: Triangles (Exercise 6A)
In this doc you can find the meaning of RD Sharma Solutions: Triangles (Exercise 6A) defined & explained in the simplest way possible. Besides explaining types of
RD Sharma Solutions: Triangles (Exercise 6A) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Triangles (Exercise 6A) tests, examples and also practice Class 10
tests
|
Explore Courses for Class 10 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
practice quizzes
,Previous Year Questions with Solutions
,MCQs
,Triangles (Exercise 6A) RD Sharma Solutions | Mathematics (Maths) Class 10
,Triangles (Exercise 6A) RD Sharma Solutions | Mathematics (Maths) Class 10
,Objective type Questions
,ppt
,Free
,Summary
,mock tests for examination
,Exam
,Sample Paper
,Semester Notes
,Triangles (Exercise 6A) RD Sharma Solutions | Mathematics (Maths) Class 10
,study material
,Important questions
,Viva Questions
,past year papers
,video lectures
,shortcuts and tricks
,Extra Questions
;
Additional Information about RD Sharma Solutions: Triangles (Exercise 6A) for Class 10 Preparation
RD Sharma Solutions: Triangles (Exercise 6A) Free PDF Download
The RD Sharma Solutions: Triangles (Exercise 6A) is an invaluable resource that delves deep into the core of the Class 10 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the RD Sharma Solutions: Triangles (Exercise 6A) now and kickstart your journey towards success in the Class 10 exam.
Importance of RD Sharma Solutions: Triangles (Exercise 6A)
The importance of RD Sharma Solutions: Triangles (Exercise 6A) cannot be overstated, especially for Class 10 aspirants.
This document holds the key to success in the Class 10 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
RD Sharma Solutions: Triangles (Exercise 6A) Notes
RD Sharma Solutions: Triangles (Exercise 6A) Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to RD Sharma Solutions: Triangles (Exercise 6A).
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, RD Sharma Solutions: Triangles (Exercise 6A) Notes on EduRev are your ultimate resource for success.
RD Sharma Solutions: Triangles (Exercise 6A) Class 10 Questions
The "RD Sharma Solutions: Triangles (Exercise 6A) Class 10 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 10 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study RD Sharma Solutions: Triangles (Exercise 6A) on the App
Students of Class 10 can study RD Sharma Solutions: Triangles (Exercise 6A) alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RD Sharma Solutions: Triangles (Exercise 6A),
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of RD Sharma Solutions: Triangles (Exercise 6A) is prepared as per the latest Class 10 syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup