NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals (EX 3.2) Exercise 3.2
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The National Council of Educational Research and Training (NCERT) creates and provides textbooks for students in elementary and high schools. The easiest way to explain the significance of the NCERT books for CBSE board exam preparation is that they put a strong emphasis on the basics to aid students in understanding the important ideas. Since NCERT textbooks are comprehensive and efficient in and of themselves, CBSE seldom ever requires students to read any other materials apart from NCERT books. The Department of Education in Science and Mathematics (DESM) and NCERT have created some excellent exercises in Science and Mathematics for classes 912 that are referred to as “Exemplar Problems” in order to improve students’ learning abilities and assess their comprehension, analytical thinking, and problemsolving skills. For a deeper understanding of Class 8 Maths Chapter 3 Exercise 3.2 Solutions, students may consult NCERT Solutions For Class 8 Maths Chapter 3 Exercise 3.2.
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NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals (EX 3.2) Exercise 3.2
The NCERT Solutions For Class 8 Maths Chapter 3 Exercise 3.2 are available for download from the Extramarks website and mobile application, making it simple for CBSE Class 8 Maths Chapter 3.2 students to access and review the solutions even when they are not online. The NCERT Class 8 Maths Chapter 3 Exercise 3.2 will aid in their comprehension of the problems.
For Class 8 Chapter 3: Understanding Quadrilaterals, the NCERT Solutions For Class 8 Maths Chapter 3 Exercise 3.2 are useful. A student’s experience in Class 8 is crucial since it serves as a transitional year after Class 7. Students will not feel lost or puzzled about a question if they apply and follow the methods in the NCERT Solutions For Class 8 Maths Chapter 3 Exercise 3.2. Extramarks follow the format used in the NCERT Mathematics book. Eighth grade is a pivotal year for students’ academic goals.
Students who carefully study from the NCERT textbooks will do well in their examinations and have a solid conceptual understanding. As a result, NCERT Solutions For Class 8 Maths Chapter 3 Exercise 3.2 are helpful for times when students encounter challenges when attempting to answer the questions, and it may get complicated when they get stuck in a particular subject. Extramarks has been a great help to students in overcoming all of these problems and has improved their conceptual understanding. When students answer the questions, they will quickly comprehend the answers since the NCERT Solutions For Class 8 Maths Chapter 3 Exercise 3.2 is written in an approachable and understandable manner.
Access NCERT Solutions for Maths Chapter 3 – Understanding Quadrilaterals
If students carefully read NCERT textbooks, they will do well on examinations and have a solid conceptual understanding. NCERT Solutions for Class 8 Maths Chapter 3 Exercise 3.2 can be useful when students are having difficulty responding to questions. Things can get complicated when students are unable to move on from their issues. Extramarks has helped students significantly in resolving all of these problems and improving their conceptual knowledge. Students will comprehend the solutions right away after answering the questions since the NCERT Solutions For Class 8 Maths Chapter 3 Exercise 3.2 are written in a straightforward and accessible manner.
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Students who get extra credit become more talented and skilled. Students will have access to additional study tools in addition to the NCERT Solutions For Class 8 Maths Chapter 3 Exercise 3.2, such as live courses, practice exams, past years’ papers, and the whole curriculum. After entering Class 8, some students begin to fear Mathematics, but the NCERT Solutions For Class 8 Maths Chapter 3 Exercise 3.2 have made things simple for them and left an impression on their brains that has caused them to start studying more effectively than ever. The way professionals attempt to answer the problems with the help of NCERT Solutions For Class 8 Maths Chapter 3 Exercise 3.2, teaches students a lot.
NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.2
The NCERT Solutions For Class 8 Maths Chapter 3 Exercise 3.2, are solely dependent on the exterior angle property, which asserts that the sum of all an object’s outside angles would be 360 degrees. There are six easy sums in NCERT Solutions for Class 8 Maths Chapter 3 Exercise 3.1. Although they are each framed differently, they are all based on the angle sum characteristic. For instance, whereas other questions would ask for the number of sides given the measurement of an external angle, others will require the student to apply the formula directly.
The student’s ability to calculate accurately is required to tackle all of the tasks in this exercise successfully. Additionally, they should be able to read the questions and comprehend how the angle sum theorem is used. The NCERT Solutions For Class 8 Maths Chapter 3 Exercise 3.2, is a fantastic resource that can be viewed by accessing Extramarks’ website.
NCERT Solutions for Class 8
Through NCERT Solutions for Class 8, students get access to many different study materials. The Extramarks website and mobile application both offer access to all course materials. As a result, it is encouraged that students look up any pertinent questions they may have about the chapter on the Extramarks website. The NCERT Solutions For Class 8 Maths Chapter 3 Exercise 3.2, are also available to students. For NCERT Solutions For Class 8 Maths Chapter 3 Exercise 3.2, they may locate notes, significant issues, and example questions. The NCERT Solutions For Class 8 Maths Chapter 3 Exercise 3.2, provides students with access to problems from prior years and assistance with their answers.
The NCERT Solutions For Class 8 Maths Chapter 3 Exercise 3.2, are dependent on one extremely significant formula: the exterior angle measure of a Polygon. Exterior Angle of a Polygon: Any Polygon, regardless of the number of sides, has a total exterior angle measurement of 360°. The NCERT Solutions For Class 8 Maths Chapter 3 Exercise 3.2 guides how to appropriately use this formula and understand the outcomes.
All the other chapters’ NCERT solutions for Class 8 are available on Extramarks’ website and mobile application.
Chapter 1 Rational Numbers
Chapter 2 Linear Equation
Chapter 3 Understanding Quadrilaterals
Chapter 4 Applied Practical Geometry
Chapter 5 About Data Handling
Chapter 6 Square Square Roots
Chapter 7 Cube And Cube Roots
Chapter 8 Comparing Quantities
Chapter 9 Algebraic Expressions And Identities
Chapter 10 Visualizing Solid Shapes
Chapter 11 Mensuration
Chapter 12 Exponents And Powers
Chapter 13 Direct Inverse Proportions
Chapter 14 Factorisation
Chapter 15 Introduction To Graphs
Chapter 16 Playing With Number
Q.1 Find x in the following figures.
Ans
$\begin{array}{l}\text{(a)}\\ \text{We know that the sum of all exterior angles of any}\\ \text{polygon is 36}{0}^{\circ}.\\ \therefore \text{125}\mathrm{\xb0}+\text{125}\mathrm{\xb0}+\mathrm{x}=\text{36}0\mathrm{\xb0}\\ \Rightarrow \text{25}0\mathrm{\xb0}+\mathrm{x}=\text{36}0\mathrm{\xb0}\\ \Rightarrow \mathrm{x}=\text{11}0\mathrm{\xb0}\end{array}$
$\begin{array}{l}\text{(b)}\\ \text{We know that the sum of all exterior angles of any}\\ \text{polygon is 36}{0}^{\circ}.\\ \therefore \text{6}0\xb0+\text{9}0\xb0+\text{7}0\xb0+x+\text{9}0\xb0=\text{36}0\xb0\\ \Rightarrow \text{31}0\xb0+x=\text{36}0\xb0\\ \Rightarrow x=\text{5}0\xb0\end{array}$
Q.2 Find the measure of each exterior angle of a regular polygon of
(i) 9 sides
(ii) 15 sides.
Ans
$\begin{array}{l}\text{(i)}\\ \text{We know that, the}\mathrm{sum}\mathrm{of}\mathrm{}\mathrm{all}\mathrm{}\mathrm{exterior}\mathrm{}\mathrm{angles}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\\ \mathrm{given}\text{}\mathrm{polygon}={360}^{\circ}\\ \\ \mathrm{Thus},\mathrm{}\mathrm{measure}\mathrm{}\mathrm{of}\mathrm{}\mathrm{each}\mathrm{}\mathrm{exterior}\mathrm{}\mathrm{angle}\mathrm{}\mathrm{of}\mathrm{}\mathrm{a}\mathrm{}\mathrm{regular}\mathrm{}\\ \mathrm{polygon}\text{}\mathrm{of}\text{}9\text{}\mathrm{sides}=\frac{{360}^{\circ}}{9}={40}^{\circ}\\ \end{array}$
$\begin{array}{l}\left(\mathrm{ii}\right)\\ \mathrm{We}\mathrm{know}\mathrm{that},\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{}\mathrm{all}\mathrm{}\mathrm{exterior}\mathrm{}\mathrm{angles}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\\ \mathrm{given}\text{}\mathrm{polygon}={360}^{\circ}\\ \\ \mathrm{Thus},\mathrm{}\mathrm{measure}\mathrm{}\mathrm{of}\mathrm{}\mathrm{each}\mathrm{}\mathrm{exterior}\mathrm{}\mathrm{angle}\mathrm{}\mathrm{of}\mathrm{}\mathrm{a}\mathrm{}\mathrm{regular}\mathrm{}\\ \mathrm{polygon}\mathrm{}\mathrm{of}\mathrm{}15\mathrm{}\mathrm{sides}=\frac{{360}^{\circ}}{15}={24}^{\circ}\end{array}$
Q.3 How many sides does a regular polygon have if the measure of an exterior angle is 24°?
Ans
\begin{array}{l}\text{We know that, the sum of all exterior angles of the given}\\ \text{polygon}=\text{36}0\xb0\\ \\ \text{It is given that the measure of each exterior angle}=\text{24}\xb0\\ \\ \text{Thus},\text{number of sides of the regular polygon=}\frac{{360}^{\circ}}{24\xb0}=15\end{array}
Q.4 How many sides does a regular polygon have if each of its interior angles is 165°?
Ans
$\begin{array}{l}\text{Given,}\\ \text{Measure of each interior angle}=\text{165}\xb0\\ \\ \therefore \text{Measure of each exterior angle}=\text{18}0\xb0\text{165}\xb0=\text{15}\xb0\\ \\ \text{Also, the sum of all exterior angles of any polygon is 36}0\xba.\\ \\ \mathrm{Therefore},\text{number of sides of the polygon =}\frac{{360}^{\circ}}{{15}^{\circ}}=24\end{array}$
Q.5 (a) Is it possible to have a regular polygon with measure of each exterior angle as 22°?
(b) Can it be an interior angle of a regular polygon? Why?
Ans
$\begin{array}{l}\text{(a) Given:}\\ \text{Exterior angle}={22}^{\circ}\\ \\ \text{We know that,}\\ \text{the sum of all the exterior angles of a polygon is 3}{60}^{\circ}.\end{array}$
$\begin{array}{l}\mathrm{Also},\text{all the exterior angles are of equal measure.That}\\ {\text{means 22}}^{\circ}\text{should be exact divisor of 3}{60}^{\circ}.\\ \\ \mathrm{Since},{\text{\hspace{0.17em}22}}^{\circ}\text{\hspace{0.17em}is not an exact divisor of 3}{60}^{\circ},\mathrm{therefore},\text{such}\\ \text{a polygon is not possible.}\\ \end{array}$
\begin{array}{l}\text{(b) Given:}\\ \text{Interior angle}=\text{22}\xb0\\ \\ \therefore \text{Exterior angle}=\text{18}0\xb0\text{22}\xb0=\text{158}\xb0\\ \\ \text{As 36}0\xb0\text{is not a perfect multiple of 158}\xb0,so\text{such a polygon}\\ \text{is not possible}\text{.}\end{array}
Q.6 (a) What is the minimum interior angle possible for a regular polygon? Why?
(b) What is the maximum exterior angle possible for a regular polygon?
Ans
For a regular polygon, as the number of sides decreases, interior angle also decreases and exterior angle increases.
Therefore, polygon with maximum exterior angle and minimum interior angle is an equilateral triangle as it has the minimum number of sides possible for a polygon.
Consider an equilateral triangle. The exterior angle of this triangle will be the maximum exterior angle possible for any regular polygon.
Therefore,
$\mathrm{Exterior}\text{}\mathrm{angle}\text{}\mathrm{of}\text{}\mathrm{an}\text{}\mathrm{equilateral}\text{}\mathrm{triangle}=\frac{{360}^{\circ}}{3}=120\mathrm{\xb0}$
$\begin{array}{l}\mathrm{Hence},\text{}\mathrm{maximum}\text{}\mathrm{possible}\text{}\mathrm{measure}\text{}\mathrm{of}\text{}\mathrm{exterior}\text{}\mathrm{angle}\text{}\mathrm{for}\text{}\mathrm{any}\\ \mathrm{polygon}\text{}\mathrm{is}\text{}120\mathrm{\xb0}.\text{}\\ \\ \mathrm{Also},\text{}\mathrm{we}\text{}\mathrm{know}\text{}\mathrm{that}\text{}\mathrm{an}\text{}\mathrm{exterior}\text{}\mathrm{angle}\text{}\mathrm{and}\text{}\mathrm{an}\text{}\mathrm{interior}\text{}\mathrm{angle}\\ \mathrm{are}\text{}\mathrm{always}\text{}\mathrm{in}\text{}\mathrm{a}\text{}\mathrm{linear}\text{}\mathrm{pair}.\\ \mathrm{Hence},\text{}\mathrm{minimum}\text{}\mathrm{interior}\text{}\mathrm{angle}=180\mathrm{\xb0}120\mathrm{\xb0}=60\mathrm{\xb0}\end{array}$
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