Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1
Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 | Mathematics (Maths) Class 8 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
EXERCISE 17. 1 2 . 7 1 : O N S E G A P 1
1. Given below is a parallelogram ABCD. Complete each statement along with the
definition or property used.
(i) AD =
(ii) ?DCB =
(iii) OC =
(iv) ?DAB + ?CDA =
Solution:
(i) AD = BC. Because, diagonals bisect each other in a parallelogram.
(ii) ?DCB = ?BAD. Because, alternate interior angles are equal.
(iii) OC = OA. Because, diagonals bisect each other in a parallelogram.
(iv) ?DAB+ ?CDA = 180°. Because sum of adjacent angles in a parallelogram is 180°.
2. The following figures are parallelograms. Find the degree values of the
unknowns x, y, z.
Page 2
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
EXERCISE 17. 1 2 . 7 1 : O N S E G A P 1
1. Given below is a parallelogram ABCD. Complete each statement along with the
definition or property used.
(i) AD =
(ii) ?DCB =
(iii) OC =
(iv) ?DAB + ?CDA =
Solution:
(i) AD = BC. Because, diagonals bisect each other in a parallelogram.
(ii) ?DCB = ?BAD. Because, alternate interior angles are equal.
(iii) OC = OA. Because, diagonals bisect each other in a parallelogram.
(iv) ?DAB+ ?CDA = 180°. Because sum of adjacent angles in a parallelogram is 180°.
2. The following figures are parallelograms. Find the degree values of the
unknowns x, y, z.
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
Solution:
(i) ?ABC = ?y = 100
o
(opposite angles are equal in a parallelogram)
?x + ?y = 180
o
(sum of adjacent angles is = 180
o
in a parallelogram)
?x + 100
o
= 180
o
?x = 180
o
- 100
o
= 80
o
? ?x = 80
o
?y = 100
o
?z = 80
o
(opposite angles are equal in a parallelogram)
(ii) ?RSP + ?y = 180
o
(sum of adjacent angles is = 180
o
in a parallelogram)
?y + 50
o
= 180
o
?y = 180
o
- 50
o
= 130
o
? ?x = ?y = 130
o
(opposite angles are equal in a parallelogram)
?RSP = ?RQP = 50
o
(opposite angles are equal in a parallelogram)
?RQP + ?z = 180
o
(linear pair)
50
o
+ ?z = 180
o
?z = 180
o
– 50
o
= 130
o
? ?x = 130
o
?y = 130
o
?z = 130
o
(iii) In ?PMN
?NPM + ?NMP + ?MNP = 180° [Sum of all the angles of a triangle is 180°]
Page 3
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
EXERCISE 17. 1 2 . 7 1 : O N S E G A P 1
1. Given below is a parallelogram ABCD. Complete each statement along with the
definition or property used.
(i) AD =
(ii) ?DCB =
(iii) OC =
(iv) ?DAB + ?CDA =
Solution:
(i) AD = BC. Because, diagonals bisect each other in a parallelogram.
(ii) ?DCB = ?BAD. Because, alternate interior angles are equal.
(iii) OC = OA. Because, diagonals bisect each other in a parallelogram.
(iv) ?DAB+ ?CDA = 180°. Because sum of adjacent angles in a parallelogram is 180°.
2. The following figures are parallelograms. Find the degree values of the
unknowns x, y, z.
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
Solution:
(i) ?ABC = ?y = 100
o
(opposite angles are equal in a parallelogram)
?x + ?y = 180
o
(sum of adjacent angles is = 180
o
in a parallelogram)
?x + 100
o
= 180
o
?x = 180
o
- 100
o
= 80
o
? ?x = 80
o
?y = 100
o
?z = 80
o
(opposite angles are equal in a parallelogram)
(ii) ?RSP + ?y = 180
o
(sum of adjacent angles is = 180
o
in a parallelogram)
?y + 50
o
= 180
o
?y = 180
o
- 50
o
= 130
o
? ?x = ?y = 130
o
(opposite angles are equal in a parallelogram)
?RSP = ?RQP = 50
o
(opposite angles are equal in a parallelogram)
?RQP + ?z = 180
o
(linear pair)
50
o
+ ?z = 180
o
?z = 180
o
– 50
o
= 130
o
? ?x = 130
o
?y = 130
o
?z = 130
o
(iii) In ?PMN
?NPM + ?NMP + ?MNP = 180° [Sum of all the angles of a triangle is 180°]
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
30° + 90° + ?z = 180°
?z = 180°-120°
= 60°
?y = ?z = 60° [opposite angles are equal in a parallelogram]
?z = 180°-120° [sum of the adjacent angles is equal to 180° in a parallelogram]
?z = 60°
?z + ?LMN = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]
60° + 90°+ ?x = 180°
?x = 180°-150°
?x = 30°
? ?x = 30
o
?y = 60
o
?z = 60
o
(iv) ?x = 90° [vertically opposite angles are equal]
In ?DOC
?x + ?y + 30° = 180° [Sum of all the angles of a triangle is 180°]
90° + 30° + ?y = 180°
?y = 180°-120°
?y = 60°
?y = ?z = 60° [alternate interior angles are equal]
? ?x = 90
o
?y = 60
o
?z = 60
o
(v) ?x + ?POR = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]
?x + 80° = 180°
?x = 180°-80°
?x = 100°
?y = 80° [opposite angles are equal in a parallelogram]
?SRQ = ?x = 100°
?SRQ + ?z = 180° [Linear pair]
100° + ?z = 180°
?z = 180°-100°
?z = 80°
? ?x = 100
o
?y = 80
o
?z = 80
o
(vi) ?y = 112° [In a parallelogram opposite angles are equal]
?y + ?VUT = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]
?z + 40° + 112° = 180°
?z = 180°-152°
?z = 28°
?z = ?x = 28° [alternate interior angles are equal]
Page 4
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
EXERCISE 17. 1 2 . 7 1 : O N S E G A P 1
1. Given below is a parallelogram ABCD. Complete each statement along with the
definition or property used.
(i) AD =
(ii) ?DCB =
(iii) OC =
(iv) ?DAB + ?CDA =
Solution:
(i) AD = BC. Because, diagonals bisect each other in a parallelogram.
(ii) ?DCB = ?BAD. Because, alternate interior angles are equal.
(iii) OC = OA. Because, diagonals bisect each other in a parallelogram.
(iv) ?DAB+ ?CDA = 180°. Because sum of adjacent angles in a parallelogram is 180°.
2. The following figures are parallelograms. Find the degree values of the
unknowns x, y, z.
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
Solution:
(i) ?ABC = ?y = 100
o
(opposite angles are equal in a parallelogram)
?x + ?y = 180
o
(sum of adjacent angles is = 180
o
in a parallelogram)
?x + 100
o
= 180
o
?x = 180
o
- 100
o
= 80
o
? ?x = 80
o
?y = 100
o
?z = 80
o
(opposite angles are equal in a parallelogram)
(ii) ?RSP + ?y = 180
o
(sum of adjacent angles is = 180
o
in a parallelogram)
?y + 50
o
= 180
o
?y = 180
o
- 50
o
= 130
o
? ?x = ?y = 130
o
(opposite angles are equal in a parallelogram)
?RSP = ?RQP = 50
o
(opposite angles are equal in a parallelogram)
?RQP + ?z = 180
o
(linear pair)
50
o
+ ?z = 180
o
?z = 180
o
– 50
o
= 130
o
? ?x = 130
o
?y = 130
o
?z = 130
o
(iii) In ?PMN
?NPM + ?NMP + ?MNP = 180° [Sum of all the angles of a triangle is 180°]
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
30° + 90° + ?z = 180°
?z = 180°-120°
= 60°
?y = ?z = 60° [opposite angles are equal in a parallelogram]
?z = 180°-120° [sum of the adjacent angles is equal to 180° in a parallelogram]
?z = 60°
?z + ?LMN = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]
60° + 90°+ ?x = 180°
?x = 180°-150°
?x = 30°
? ?x = 30
o
?y = 60
o
?z = 60
o
(iv) ?x = 90° [vertically opposite angles are equal]
In ?DOC
?x + ?y + 30° = 180° [Sum of all the angles of a triangle is 180°]
90° + 30° + ?y = 180°
?y = 180°-120°
?y = 60°
?y = ?z = 60° [alternate interior angles are equal]
? ?x = 90
o
?y = 60
o
?z = 60
o
(v) ?x + ?POR = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]
?x + 80° = 180°
?x = 180°-80°
?x = 100°
?y = 80° [opposite angles are equal in a parallelogram]
?SRQ = ?x = 100°
?SRQ + ?z = 180° [Linear pair]
100° + ?z = 180°
?z = 180°-100°
?z = 80°
? ?x = 100
o
?y = 80
o
?z = 80
o
(vi) ?y = 112° [In a parallelogram opposite angles are equal]
?y + ?VUT = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]
?z + 40° + 112° = 180°
?z = 180°-152°
?z = 28°
?z = ?x = 28° [alternate interior angles are equal]
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
? ?x = 28
o
?y = 112
o
?z = 28
o
3. Can the following figures be parallelograms? Justify your answer.
Solution:
(i) No, opposite angles are equal in a parallelogram.
(ii) Yes, opposite sides are equal and parallel in a parallelogram.
(iii) No, diagonals bisect each other in a parallelogram.
4. In the adjacent figure HOPE is a parallelogram. Find the angle measures x,
y and z. State the geometrical truths you use to find them.
Solution:
We know that
?POH + 70° = 180° [Linear pair]
?POH = 180°-70°
?POH = 110°
Page 5
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
EXERCISE 17. 1 2 . 7 1 : O N S E G A P 1
1. Given below is a parallelogram ABCD. Complete each statement along with the
definition or property used.
(i) AD =
(ii) ?DCB =
(iii) OC =
(iv) ?DAB + ?CDA =
Solution:
(i) AD = BC. Because, diagonals bisect each other in a parallelogram.
(ii) ?DCB = ?BAD. Because, alternate interior angles are equal.
(iii) OC = OA. Because, diagonals bisect each other in a parallelogram.
(iv) ?DAB+ ?CDA = 180°. Because sum of adjacent angles in a parallelogram is 180°.
2. The following figures are parallelograms. Find the degree values of the
unknowns x, y, z.
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
Solution:
(i) ?ABC = ?y = 100
o
(opposite angles are equal in a parallelogram)
?x + ?y = 180
o
(sum of adjacent angles is = 180
o
in a parallelogram)
?x + 100
o
= 180
o
?x = 180
o
- 100
o
= 80
o
? ?x = 80
o
?y = 100
o
?z = 80
o
(opposite angles are equal in a parallelogram)
(ii) ?RSP + ?y = 180
o
(sum of adjacent angles is = 180
o
in a parallelogram)
?y + 50
o
= 180
o
?y = 180
o
- 50
o
= 130
o
? ?x = ?y = 130
o
(opposite angles are equal in a parallelogram)
?RSP = ?RQP = 50
o
(opposite angles are equal in a parallelogram)
?RQP + ?z = 180
o
(linear pair)
50
o
+ ?z = 180
o
?z = 180
o
– 50
o
= 130
o
? ?x = 130
o
?y = 130
o
?z = 130
o
(iii) In ?PMN
?NPM + ?NMP + ?MNP = 180° [Sum of all the angles of a triangle is 180°]
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
30° + 90° + ?z = 180°
?z = 180°-120°
= 60°
?y = ?z = 60° [opposite angles are equal in a parallelogram]
?z = 180°-120° [sum of the adjacent angles is equal to 180° in a parallelogram]
?z = 60°
?z + ?LMN = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]
60° + 90°+ ?x = 180°
?x = 180°-150°
?x = 30°
? ?x = 30
o
?y = 60
o
?z = 60
o
(iv) ?x = 90° [vertically opposite angles are equal]
In ?DOC
?x + ?y + 30° = 180° [Sum of all the angles of a triangle is 180°]
90° + 30° + ?y = 180°
?y = 180°-120°
?y = 60°
?y = ?z = 60° [alternate interior angles are equal]
? ?x = 90
o
?y = 60
o
?z = 60
o
(v) ?x + ?POR = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]
?x + 80° = 180°
?x = 180°-80°
?x = 100°
?y = 80° [opposite angles are equal in a parallelogram]
?SRQ = ?x = 100°
?SRQ + ?z = 180° [Linear pair]
100° + ?z = 180°
?z = 180°-100°
?z = 80°
? ?x = 100
o
?y = 80
o
?z = 80
o
(vi) ?y = 112° [In a parallelogram opposite angles are equal]
?y + ?VUT = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]
?z + 40° + 112° = 180°
?z = 180°-152°
?z = 28°
?z = ?x = 28° [alternate interior angles are equal]
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
? ?x = 28
o
?y = 112
o
?z = 28
o
3. Can the following figures be parallelograms? Justify your answer.
Solution:
(i) No, opposite angles are equal in a parallelogram.
(ii) Yes, opposite sides are equal and parallel in a parallelogram.
(iii) No, diagonals bisect each other in a parallelogram.
4. In the adjacent figure HOPE is a parallelogram. Find the angle measures x,
y and z. State the geometrical truths you use to find them.
Solution:
We know that
?POH + 70° = 180° [Linear pair]
?POH = 180°-70°
?POH = 110°
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
?POH = ?x = 110° [opposite angles are equal in a parallelogram]
?x + ?z + 40° = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]
110° + ?z + 40° = 180°
?z = 180° - 150°
?z = 30°
?z + ?y = 70°
?y + 30° = 70°
?y = 70°- 30°
?y = 40°
5. In the following figures GUNS and RUNS are parallelograms. Find x and y.
Solution:
(i) 3y – 1 = 26 [opposite sides are of equal length in a parallelogram]
3y = 26 + 1
y = 27/3
y = 9
3x = 18 [opposite sides are of equal length in a parallelogram]
x = 18/3
x = 6
? x = 6 and y = 9
(ii) y – 7 = 20 [diagonals bisect each other in a parallelogram]
y = 20 + 7
y = 27
x - y = 16 [diagonals bisect each other in a parallelogram]
x -27 = 16
Read More
|
79 videos|408 docs|31 tests
|
Related Exams
About this Document
|
4.67/5 Rating |
|
Dec 22, 2024 Last updated |
Document Description: Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 for Class 8 2024 is part of Mathematics (Maths) Class 8 preparation.
The notes and questions for Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 have been prepared according to the Class 8 exam syllabus. Information about Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 covers topics
like and Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 Example, for Class 8 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1.
Introduction of Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 in English is available as part of our Mathematics (Maths) Class 8
for Class 8 & Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 in Hindi for Mathematics (Maths) Class 8 course.
Download more important topics related with notes, lectures and mock test series for Class 8
Exam by signing up for free. Class 8: Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 | Mathematics (Maths) Class 8
Description
Full syllabus notes, lecture & questions for Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 | Mathematics (Maths) Class 8 - Class 8 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 8 | Best notes, free PDF download
Information about Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1
In this doc you can find the meaning of Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 defined & explained in the simplest way possible. Besides explaining types of
Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 theory, EduRev gives you an ample number of questions to practice Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 tests, examples and also practice Class 8
tests
|
Explore Courses for Class 8 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Important questions
,MCQs
,Extra Questions
,mock tests for examination
,study material
,Sample Paper
,Summary
,shortcuts and tricks
,ppt
,Objective type Questions
,video lectures
,Previous Year Questions with Solutions
,Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 | Mathematics (Maths) Class 8
,Viva Questions
,Free
,Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 | Mathematics (Maths) Class 8
,Semester Notes
,past year papers
,Exam
,practice quizzes
,Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 | Mathematics (Maths) Class 8
;
Additional Information about Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 for Class 8 Preparation
Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 Free PDF Download
The Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 is an invaluable resource that delves deep into the core of the Class 8 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 Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 now and kickstart your journey towards success in the Class 8 exam.
Importance of Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1
The importance of Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 cannot be overstated, especially for Class 8 aspirants.
This document holds the key to success in the Class 8 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.
Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 Notes
Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1.
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, Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 Notes on EduRev are your ultimate resource for success.
Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 Class 8 Questions
The "Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 Class 8 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 8 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 Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 on the App
Students of Class 8 can study Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1,
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 Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1 is prepared as per the latest Class 8 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