Open App

Class 8 Exam > Class 8 Notes > Mathematics Class 8 ICSE > Selina Textbook Solutions: Special Types of Quadrilaterals

Selina Textbook Solutions: Special Types of Quadrilaterals | Mathematics Class 8 ICSE PDF Download

Download, print and study this document offline

Please wait while the PDF view is loading

 Page 1


17. Special Types of Quadrilaterals 
EXERCISE 17 
Question 1. 
In parallelogram ABCD, ?A = 3 times ?B. Find all the angles of the parallelogram. In the 
same parallelogram, if AB = 5x – 7 and CD = 3x +1 ; find the length of CD. 
Solution: 
 
Let ?B = x 
?A = 3 ?B = 3x 
AD||BC 
?A + ?B = 180° 
3x + x = 180° 
? 4x = 180° 
? x = 45° 
?B = 45° 
?A = 3x = 3 x 45 = 135° 
and ?B = ?D = 45° 
opposite angles of || gm are equal. 
?A = ?C = 135° 
opposite sides of //gm are equal. 
AB = CD 
5x – 7 = 3x + 1 
? 5x – 3x = 1+7 
? 2x = 8 
? x = 4 
CD = 3 x 4+1 = 13 
Hence 135°, 45°, 135° and 45° ; 13 
Question 2. 
In parallelogram PQRS, ?Q = (4x – 5)° and ?S = (3x + 10)°. Calculate : ?Q and ?R. 
Solution: 
In parallelogram PQRS, 
?Q = (4x – 5)° and ?S = (3x + 10)° 
Page 2


17. Special Types of Quadrilaterals 
EXERCISE 17 
Question 1. 
In parallelogram ABCD, ?A = 3 times ?B. Find all the angles of the parallelogram. In the 
same parallelogram, if AB = 5x – 7 and CD = 3x +1 ; find the length of CD. 
Solution: 
 
Let ?B = x 
?A = 3 ?B = 3x 
AD||BC 
?A + ?B = 180° 
3x + x = 180° 
? 4x = 180° 
? x = 45° 
?B = 45° 
?A = 3x = 3 x 45 = 135° 
and ?B = ?D = 45° 
opposite angles of || gm are equal. 
?A = ?C = 135° 
opposite sides of //gm are equal. 
AB = CD 
5x – 7 = 3x + 1 
? 5x – 3x = 1+7 
? 2x = 8 
? x = 4 
CD = 3 x 4+1 = 13 
Hence 135°, 45°, 135° and 45° ; 13 
Question 2. 
In parallelogram PQRS, ?Q = (4x – 5)° and ?S = (3x + 10)°. Calculate : ?Q and ?R. 
Solution: 
In parallelogram PQRS, 
?Q = (4x – 5)° and ?S = (3x + 10)° 
 
opposite ?s of //gm are equal. 
?Q = ?S 
4x – 5 = 3x + 10 
4x – 3x = 10+5 
x = 15 
?Q = 4x – 5 =4 x 15 – 5 = 55° 
Also ?Q + ?R = 180° 
55° + ?R = 180° 
?R = 180°-55° = 125° 
?Q = 55° ; ?R = 125° 
Question 3. 
In rhombus ABCD ; 
(i) if ?A = 74° ; find ?B and ?C. 
(ii) if AD = 7.5 cm ; find BC and CD. 
Solution: 
AD || BC 
?A + ?B = 180° 
74° + ?B = 180° 
?B =180° – 74°= 106° 
 
opposite angles of Rhombus are equal. 
?A = ?C = 74° 
Sides of Rhombus are equal. 
BC = CD = AD = 7.5 cm 
(i) ?B = 106° ; ?C = 74° 
(ii) BC = 7.5 cm and CD = 7.5 cm Ans. 
Question 4. 
In square PQRS : 
Page 3


17. Special Types of Quadrilaterals 
EXERCISE 17 
Question 1. 
In parallelogram ABCD, ?A = 3 times ?B. Find all the angles of the parallelogram. In the 
same parallelogram, if AB = 5x – 7 and CD = 3x +1 ; find the length of CD. 
Solution: 
 
Let ?B = x 
?A = 3 ?B = 3x 
AD||BC 
?A + ?B = 180° 
3x + x = 180° 
? 4x = 180° 
? x = 45° 
?B = 45° 
?A = 3x = 3 x 45 = 135° 
and ?B = ?D = 45° 
opposite angles of || gm are equal. 
?A = ?C = 135° 
opposite sides of //gm are equal. 
AB = CD 
5x – 7 = 3x + 1 
? 5x – 3x = 1+7 
? 2x = 8 
? x = 4 
CD = 3 x 4+1 = 13 
Hence 135°, 45°, 135° and 45° ; 13 
Question 2. 
In parallelogram PQRS, ?Q = (4x – 5)° and ?S = (3x + 10)°. Calculate : ?Q and ?R. 
Solution: 
In parallelogram PQRS, 
?Q = (4x – 5)° and ?S = (3x + 10)° 
 
opposite ?s of //gm are equal. 
?Q = ?S 
4x – 5 = 3x + 10 
4x – 3x = 10+5 
x = 15 
?Q = 4x – 5 =4 x 15 – 5 = 55° 
Also ?Q + ?R = 180° 
55° + ?R = 180° 
?R = 180°-55° = 125° 
?Q = 55° ; ?R = 125° 
Question 3. 
In rhombus ABCD ; 
(i) if ?A = 74° ; find ?B and ?C. 
(ii) if AD = 7.5 cm ; find BC and CD. 
Solution: 
AD || BC 
?A + ?B = 180° 
74° + ?B = 180° 
?B =180° – 74°= 106° 
 
opposite angles of Rhombus are equal. 
?A = ?C = 74° 
Sides of Rhombus are equal. 
BC = CD = AD = 7.5 cm 
(i) ?B = 106° ; ?C = 74° 
(ii) BC = 7.5 cm and CD = 7.5 cm Ans. 
Question 4. 
In square PQRS : 
(i) if PQ = 3x – 7 and QR = x + 3 ; find PS 
(ii) if PR = 5x and QR = 9x – 8. Find QS 
Solution: 
(i) sides of square are equal. 
 
PQ = QR 
=> 3x – 7 = x + 3 
=> 3x – x = 3 + 7 
=> 2x = 10 
x = 5 
PS=PQ = 3x – 7 = 3 x 5 – 7 =8 
(ii) PR = 5x and QS = 9x – 8 
 
As diagonals of square are equal. 
PR = QS 
5x = 9x – 8 
=> 5x – 9x = -8 
=> -4x = -8 
=> x = 2 
QS = 9x – 8 = 9 x 2 – 8 =10 
Question 5. 
ABCD is a rectangle, if ?BPC = 124° 
Calculate : (i) ?BAP (ii) ?ADP 
Page 4


17. Special Types of Quadrilaterals 
EXERCISE 17 
Question 1. 
In parallelogram ABCD, ?A = 3 times ?B. Find all the angles of the parallelogram. In the 
same parallelogram, if AB = 5x – 7 and CD = 3x +1 ; find the length of CD. 
Solution: 
 
Let ?B = x 
?A = 3 ?B = 3x 
AD||BC 
?A + ?B = 180° 
3x + x = 180° 
? 4x = 180° 
? x = 45° 
?B = 45° 
?A = 3x = 3 x 45 = 135° 
and ?B = ?D = 45° 
opposite angles of || gm are equal. 
?A = ?C = 135° 
opposite sides of //gm are equal. 
AB = CD 
5x – 7 = 3x + 1 
? 5x – 3x = 1+7 
? 2x = 8 
? x = 4 
CD = 3 x 4+1 = 13 
Hence 135°, 45°, 135° and 45° ; 13 
Question 2. 
In parallelogram PQRS, ?Q = (4x – 5)° and ?S = (3x + 10)°. Calculate : ?Q and ?R. 
Solution: 
In parallelogram PQRS, 
?Q = (4x – 5)° and ?S = (3x + 10)° 
 
opposite ?s of //gm are equal. 
?Q = ?S 
4x – 5 = 3x + 10 
4x – 3x = 10+5 
x = 15 
?Q = 4x – 5 =4 x 15 – 5 = 55° 
Also ?Q + ?R = 180° 
55° + ?R = 180° 
?R = 180°-55° = 125° 
?Q = 55° ; ?R = 125° 
Question 3. 
In rhombus ABCD ; 
(i) if ?A = 74° ; find ?B and ?C. 
(ii) if AD = 7.5 cm ; find BC and CD. 
Solution: 
AD || BC 
?A + ?B = 180° 
74° + ?B = 180° 
?B =180° – 74°= 106° 
 
opposite angles of Rhombus are equal. 
?A = ?C = 74° 
Sides of Rhombus are equal. 
BC = CD = AD = 7.5 cm 
(i) ?B = 106° ; ?C = 74° 
(ii) BC = 7.5 cm and CD = 7.5 cm Ans. 
Question 4. 
In square PQRS : 
(i) if PQ = 3x – 7 and QR = x + 3 ; find PS 
(ii) if PR = 5x and QR = 9x – 8. Find QS 
Solution: 
(i) sides of square are equal. 
 
PQ = QR 
=> 3x – 7 = x + 3 
=> 3x – x = 3 + 7 
=> 2x = 10 
x = 5 
PS=PQ = 3x – 7 = 3 x 5 – 7 =8 
(ii) PR = 5x and QS = 9x – 8 
 
As diagonals of square are equal. 
PR = QS 
5x = 9x – 8 
=> 5x – 9x = -8 
=> -4x = -8 
=> x = 2 
QS = 9x – 8 = 9 x 2 – 8 =10 
Question 5. 
ABCD is a rectangle, if ?BPC = 124° 
Calculate : (i) ?BAP (ii) ?ADP 
 
Solution: 
Diagonals of rectangle are equal and bisect each other. 
?PBC = ?PCB = x (say) 
But ?BPC + ?PBC + ?PCB = 180° 
124° + x + x = 180° 
2x = 180° – 124° 
2x = 56° 
=> x = 28° 
?PBC = 28° 
But ?PBC = ?ADP [Alternate ?s] 
?ADP = 28° 
Again ?APB = 180° – 124° = 56° 
Also PA = PB 
?BAP =  (180° – ?APB) 
=  x (180°- 56°) =  x 124° = 62° 
Hence (i) ?BAP = 62° (ii) ?ADP =28° 
Question 6. 
ABCD is a rhombus. If ?BAC = 38°, find : 
(i) ?ACB 
(ii) ?DAC 
(iii) ?ADC. 
 
Solution: 
ABCD is Rhombus (Given) 
AB = BC 
?BAC = ?ACB ( ?s opp. to equal sides) 
But ?BAC = 38° (Given) 
?ACB = 38° 
Page 5


17. Special Types of Quadrilaterals 
EXERCISE 17 
Question 1. 
In parallelogram ABCD, ?A = 3 times ?B. Find all the angles of the parallelogram. In the 
same parallelogram, if AB = 5x – 7 and CD = 3x +1 ; find the length of CD. 
Solution: 
 
Let ?B = x 
?A = 3 ?B = 3x 
AD||BC 
?A + ?B = 180° 
3x + x = 180° 
? 4x = 180° 
? x = 45° 
?B = 45° 
?A = 3x = 3 x 45 = 135° 
and ?B = ?D = 45° 
opposite angles of || gm are equal. 
?A = ?C = 135° 
opposite sides of //gm are equal. 
AB = CD 
5x – 7 = 3x + 1 
? 5x – 3x = 1+7 
? 2x = 8 
? x = 4 
CD = 3 x 4+1 = 13 
Hence 135°, 45°, 135° and 45° ; 13 
Question 2. 
In parallelogram PQRS, ?Q = (4x – 5)° and ?S = (3x + 10)°. Calculate : ?Q and ?R. 
Solution: 
In parallelogram PQRS, 
?Q = (4x – 5)° and ?S = (3x + 10)° 
 
opposite ?s of //gm are equal. 
?Q = ?S 
4x – 5 = 3x + 10 
4x – 3x = 10+5 
x = 15 
?Q = 4x – 5 =4 x 15 – 5 = 55° 
Also ?Q + ?R = 180° 
55° + ?R = 180° 
?R = 180°-55° = 125° 
?Q = 55° ; ?R = 125° 
Question 3. 
In rhombus ABCD ; 
(i) if ?A = 74° ; find ?B and ?C. 
(ii) if AD = 7.5 cm ; find BC and CD. 
Solution: 
AD || BC 
?A + ?B = 180° 
74° + ?B = 180° 
?B =180° – 74°= 106° 
 
opposite angles of Rhombus are equal. 
?A = ?C = 74° 
Sides of Rhombus are equal. 
BC = CD = AD = 7.5 cm 
(i) ?B = 106° ; ?C = 74° 
(ii) BC = 7.5 cm and CD = 7.5 cm Ans. 
Question 4. 
In square PQRS : 
(i) if PQ = 3x – 7 and QR = x + 3 ; find PS 
(ii) if PR = 5x and QR = 9x – 8. Find QS 
Solution: 
(i) sides of square are equal. 
 
PQ = QR 
=> 3x – 7 = x + 3 
=> 3x – x = 3 + 7 
=> 2x = 10 
x = 5 
PS=PQ = 3x – 7 = 3 x 5 – 7 =8 
(ii) PR = 5x and QS = 9x – 8 
 
As diagonals of square are equal. 
PR = QS 
5x = 9x – 8 
=> 5x – 9x = -8 
=> -4x = -8 
=> x = 2 
QS = 9x – 8 = 9 x 2 – 8 =10 
Question 5. 
ABCD is a rectangle, if ?BPC = 124° 
Calculate : (i) ?BAP (ii) ?ADP 
 
Solution: 
Diagonals of rectangle are equal and bisect each other. 
?PBC = ?PCB = x (say) 
But ?BPC + ?PBC + ?PCB = 180° 
124° + x + x = 180° 
2x = 180° – 124° 
2x = 56° 
=> x = 28° 
?PBC = 28° 
But ?PBC = ?ADP [Alternate ?s] 
?ADP = 28° 
Again ?APB = 180° – 124° = 56° 
Also PA = PB 
?BAP =  (180° – ?APB) 
=  x (180°- 56°) =  x 124° = 62° 
Hence (i) ?BAP = 62° (ii) ?ADP =28° 
Question 6. 
ABCD is a rhombus. If ?BAC = 38°, find : 
(i) ?ACB 
(ii) ?DAC 
(iii) ?ADC. 
 
Solution: 
ABCD is Rhombus (Given) 
AB = BC 
?BAC = ?ACB ( ?s opp. to equal sides) 
But ?BAC = 38° (Given) 
?ACB = 38° 
In ?ABC, 
?ABC + ?BAC + ?ACB = 180° 
?ABC + 38°+ 38° = 180° 
?ABC = 180° – 76° = 104° 
But ?ABC = ?ADC (opp. ?s of rhombus) 
?ADC = 104° 
?DAC = ?DCA ( AD = CD) 
?DAC =  [180° – 104°] 
?DAC =  x 76° = 38° 
Hence (i) ?ACB = 38° (ii) ?DAC = 38° (iii) ?ADC = 104° Ans. 
Question 7. 
ABCD is a rhombus. If ?BCA = 35°. find ?ADC. 
Solution: 
Given : Rhombus ABCD in which ?BCA = 35° 
 
To find : ?ADC 
Proof : AD || BC 
?DAC = ?BCA (Alternate ?s) 
But ?BCA = 35° (Given) 
?DAC = 35° 
But ?DAC = ?ACD ( AD = CD) & ?DAC + ?ACD + ?ADC = 180° 
35°+ 35° + ?ADC = 180° 
?ADC = 180° – 70° = 110° 
Hence ?ADC = 110° 
Question 8. 
PQRS is a parallelogram whose diagonals intersect at M. 
If ?PMS = 54°, ?QSR = 25° and ?SQR = 30° ; find : 
(i) ?RPS 
(ii) ?PRS 
(iii) ?PSR. 
Solution: 
Given : ||gm PQRS in which diagonals PR & QS intersect at M. 
?PMS = 54° ; ?QSR = 25° and ?SQR=30°

	Mathematics Class 8 ICSE 23 videos\|98 docs\|14 tests

Mathematics Class 8 ICSE

23 videos|98 docs|14 tests

Join Course for Free

About this Document

Oct 06, 2025 Last updated

Related Exams

Class 8

Document Description: Selina Textbook Solutions: Special Types of Quadrilaterals for Class 8 2025 is part of Mathematics Class 8 ICSE preparation. The notes and questions for Selina Textbook Solutions: Special Types of Quadrilaterals have been prepared according to the Class 8 exam syllabus. Information about Selina Textbook Solutions: Special Types of Quadrilaterals covers topics like and Selina Textbook Solutions: Special Types of Quadrilaterals Example, for Class 8 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Selina Textbook Solutions: Special Types of Quadrilaterals.

Introduction of Selina Textbook Solutions: Special Types of Quadrilaterals in English is available as part of our Mathematics Class 8 ICSE for Class 8 & Selina Textbook Solutions: Special Types of Quadrilaterals in Hindi for Mathematics Class 8 ICSE course. Download more important topics related with notes, lectures and mock test series for Class 8 Exam by signing up for free. Class 8: Selina Textbook Solutions: Special Types of Quadrilaterals | Mathematics Class 8 ICSE

Description

Full syllabus notes, lecture & questions for Selina Textbook Solutions: Special Types of Quadrilaterals | Mathematics Class 8 ICSE - Class 8 | Plus excerises question with solution to help you revise complete syllabus for Mathematics Class 8 ICSE | Best notes, free PDF download

Information about Selina Textbook Solutions: Special Types of Quadrilaterals

In this doc you can find the meaning of Selina Textbook Solutions: Special Types of Quadrilaterals defined & explained in the simplest way possible. Besides explaining types of Selina Textbook Solutions: Special Types of Quadrilaterals theory, EduRev gives you an ample number of questions to practice Selina Textbook Solutions: Special Types of Quadrilaterals tests, examples and also practice Class 8 tests

	Mathematics Class 8 ICSE 23 videos\|98 docs\|14 tests

Mathematics Class 8 ICSE

23 videos|98 docs|14 tests

Join Course for Free

Download as PDF

Explore Courses for Class 8 exam

Free

Viva Questions

video lectures

Selina Textbook Solutions: Special Types of Quadrilaterals | Mathematics Class 8 ICSE

study material

Exam

Extra Questions

ppt

pdf

practice quizzes

Objective type Questions

MCQs

Semester Notes

Sample Paper

Previous Year Questions with Solutions

Important questions

shortcuts and tricks

Selina Textbook Solutions: Special Types of Quadrilaterals | Mathematics Class 8 ICSE

mock tests for examination

Selina Textbook Solutions: Special Types of Quadrilaterals | Mathematics Class 8 ICSE

Summary

past year papers

;

Additional Information about Selina Textbook Solutions: Special Types of Quadrilaterals for Class 8 Preparation

Selina Textbook Solutions: Special Types of Quadrilaterals Free PDF Download

The Selina Textbook Solutions: Special Types of Quadrilaterals 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 Selina Textbook Solutions: Special Types of Quadrilaterals now and kickstart your journey towards success in the Class 8 exam.

Importance of Selina Textbook Solutions: Special Types of Quadrilaterals

The importance of Selina Textbook Solutions: Special Types of Quadrilaterals 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.

Selina Textbook Solutions: Special Types of Quadrilaterals Notes

Selina Textbook Solutions: Special Types of Quadrilaterals Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to Selina Textbook Solutions: Special Types of Quadrilaterals. 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, Selina Textbook Solutions: Special Types of Quadrilaterals Notes on EduRev are your ultimate resource for success.

Selina Textbook Solutions: Special Types of Quadrilaterals Class 8 Questions

The "Selina Textbook Solutions: Special Types of Quadrilaterals 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 Selina Textbook Solutions: Special Types of Quadrilaterals on the App

Students of Class 8 can study Selina Textbook Solutions: Special Types of Quadrilaterals alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the Selina Textbook Solutions: Special Types of Quadrilaterals, 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 Selina Textbook Solutions: Special Types of Quadrilaterals is prepared as per the latest Class 8 syllabus.

Education Revolution