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Important Questions Class 8 Maths Chapter 3
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Important Questions Class 8 Mathematics Chapter 3 – Understanding Quadrilaterals
Mathematics deals with numbers of various forms, shapes, logic, quantity and arrangements. Mathematics also teaches us to solve problems based on numerical calculations and find solutions.
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Chapter 3 of Class 8 Mathematics is called ‘Understanding Quadrilaterals’. A quadrilateral is a closed shape and also a type of polygon that has four sides, four vertices and four angles. It is formed by joining four noncollinear points. The sum of all the interior angles of a quadrilateral is always equal to 360 degrees. In a quadrilateral, the sides are straight lines and are twodimensional. Square, rectangle, rhombus, parallelogram, etc., are examples of quadrilaterals. The formula for the angle sum of a polygon = (n – 2) × 180°.
Extramarks is the best study buddy for students and helps them with comprehensive online study solutions from Class 1 to Class 12. Our team of expert Mathematics teachers have prepared a variety of NCERT solutions to help students in their studies and exam preparation. Students can refer to our Important Questions Class 8 Mathematics Chapter 3 to practise examoriented questions. We have collated questions from various sources such as NCERT textbooks and exemplars, CBSE sample papers, CBSE past year question papers, etc. Students can prepare well for their exams and tests by solving a variety of chapter questions from our Important Questions Class 8 Mathematics Chapter 3.
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Important Questions Class 8 Mathematics Chapter 3 – With Solution
Mentioned below are some sets of questions and their answers from our Chapter 3 Class 8 Mathematics important questions.
Question 1: A quadrilateral has three acute angles, each measuring 80°. What is the measure of the fourth angle of the quadrilateral?
Answer 1: – Let x be the measure of the fourth angle of a quadrilateral.
The sum of all the angles of a quadrilateral + 360°
80° + 80° + 80° + x = 360° …………(since the measure of all the three acute angles = 80°)
240° + x = 360°
x = 360° – 240°
Hence, the fourth angle made by the quadrilateral is 120°.
Question 2: Find the measure of all the exterior angles of a regular polygon with
(i) 9 sides and (ii) 15 sides.
Answer 2 : (i) Total measure of all exterior angles = 360°
Each exterior angle =sum of exterior angle = 360° = 40°
number of sides 9
Each exterior angle = 40°
(ii) Total measure of all exterior angles = 360°
Each exterior angle = sum of exterior angle =360° = 24°
number of sides 15
Each exterior angle = 24°
Answer 3: a) The sum of all the angles of the triangle = 180°
One side of a triangle
= 180° (90° + 30°) = 60°
In a linear pair, the sum of two adjacent angles altogether measures up to 180°
x + 90° = 180°
x = 180° – 90°
= 90°
Similarly,
y + 60° = 180°
y = 180° – 60°
= 120°
similarly,
z + 30° = 180°
z = 180° – 30°
= 150°
Hence,x + y + z
= 90° + 120° + 150°
= 360°
Thus, the sum of the angles x, y, and z is altogether 360°
 b) Sum of all angles of quadrilateral = 360°
One side of quadrilateral = 360° (60° + 80° + 120°) = 360° – 260° = 100°
x + 120° = 180°
x = 180° – 120°
= 60°
y + 80° = 180°
y = 180° – 80°
= 100°
z + 60° = 180°
z = 180° – 60°
= 120°
w + 100° = 180°
w = 180° – 100° = 80°
x + y + z + w = 60° + 100° + 120° + 80° = 360°
Question 4: Adjacent sides of a rectangle are in the ratio 5: 12; if the perimeter of the given rectangle is 34 cm, find the length of the diagonal.
Answer 4: The ratio of the adjacent sides of the rectangle is 5: 12
Let 5x and 12x be adjacent sides.
The perimeter is the sum of all the given sides of a rectangle.
5x + 12x + 5x + 12x = 34 cm ……(since the opposite sides of the rectangle are the
same)
34x = 34
x = 34/34
x = 1 cm
Therefore, the adjacent sides of the rectangle are 5 cm and 12 cm, respectively.
That is,
Length =12 cm
Breadth = 5 cm
Length of the diagonal = √( l2 + b2)
= √( 122 + 52)
= √(144 + 25)
= √169
= 13 cm
Hence, the length of the diagonal of a rectangle is 13 cm.
Question 5: How many sides do regular polygons consist of if each interior angle is 165 ° ?
Answer 5: A regular polygon with an interior angle of 165°
We need to find the sides of the given regular polygon:
The sum of all exterior angles of any given polygon is 360°.
Formula Used: Number of sides = 360∘ /Exterior angle
Exterior angle=180∘−Interior angle
Thus,
Each interior angle =165°
Hence, the measure of every exterior angle will be
=180°−165°
=15°
Therefore, the number of sides of the given polygon will be
=360°/15°
=24°
Question 6: Find x in the following figure.
Answer 6: The two interior angles in the given figures are right angles = 90°
70° + m = 180°
m = 180° – 70°
(In a linear pair, the sum of two adjacent angles altogether measures up to 180°)
60° + n = 180°
n = 180° – 60°
= 120°
The given figure has five sides, and it is a pentagon.
Thus, the sum of the angles of the pentagon = 540°
90° + 90° + 110° + 120° + y = 540°
410° + y = 540°
y = 540° – 410° = 130°
x + y = 180°….. (Linear pair)
x + 130° = 180°
x = 180° – 130°
Question 7: ABCD is a parallelogram with ∠A = 80°. The internal bisectors of ∠B and ∠C meet each other at O. Find the measure of the three angles of ΔBCO.
Answer 7: The measure of angle A = 80°.
In a parallelogram, the opposite angles are the same.
Hence,
∠A = ∠C = 80°
And
∠OCB = (1/2) × ∠C
= (1/2) × 80°
= 40°
∠B = 180° – ∠A (the sum of interior angles situated on the same side of the transversal is supplementary)
= 180° – 80°
= 100°
Also,
∠CBO = (1/2) × ∠B
∠CBO= (1/2) × 100°
∠CBO= 50°.
By the property of the sum of the angle BCO, we get,
∠BOC + ∠OBC + ∠CBO = 180°
∠BOC = 180° – (∠OBC + CBO)
= 180° – (40° + 50°)
= 180° – 90°
= 90°
Hence, the measure of all the angles of triangle BCO is 40°, 50° and 90°.
Question 8: The measure of the two adjacent angles of the given parallelogram is the ratio of 3:2. Then, find the measure of each angle of the parallelogram.
Answer 8: A parallelogram with adjacent angles in the ratio of 3:2
To find: The measure of each of the angles of the parallelogram.
Let the measure of angle A be 3x
Let the measure of angle B be 2x
Since the sum of the measures of adjacent angles is 180° for a parallelogram,
3x+2x=180°
∠A=∠C =3x=108°
∠B=∠D =2x=72° (Opposite angles of a parallelogram are equal).
Hence, the angles of a parallelogram are 108°, 72°,108°and 72°
Question 9: Is it ever possible to have a regular polygon, each of whose interior angles is 100?
Answer 9: The sum of all the exterior angles of a regular polygon is 360°
As we also know, the sum of interior and exterior angles are 180°
Exterior angle + interior angle = 180100=80°
When we divide the exterior angle, we will get the number of exterior angles
since it is a regular polygon means the number of exterior angles equals the number of sides.
Therefore n=360/ 80=4.5
And we know that 4.5 is not an integer, so having a regular polygon is impossible.
Whose exterior angle is 100°
Question 10: ABCD is a parallelogram in which ∠A=110 ° . Find the measure of the angles B, C and D, respectively.
Answer 10: The measure of angle A=110°
the sum of all adjacent angles of a parallelogram is 180°
∠A + ∠B = 180
110°+ ∠B = 180°
∠B = 180° 110°
= 70°.
Also ∠B + ∠C = 180° [Since ∠B and ∠C are adjacent angles]
70°+ ∠C = 180°
∠C = 180° 70°
= 110°.
Now ∠C + ∠D = 180° [Since ∠C and ∠D are adjacent angles]
110o+ ∠D = 180°
∠D = 180° 110°
= 70°
Question11: A diagonal and a side of a rhombus are of equal length. Find the measure of the angles of the rhombus.
Answer 11: Let ABCD be the rhombus.
All the sides of a rhombus are the same.
Thus, AB = BC = CD = DA.
The side and diagonal of a rhombus are equal.
AB = BD
Therefore, AB = BC = CD = DA = BD
Consider triangle ABD,
Each side of a triangle ABD is congruent.
Hence, ΔABD is an equilateral triangle.
Similarly,
ΔBCD is also an equilateral triangle.
Thus, ∠BAD = ∠ABD = ∠ADB = ∠DBC = ∠BCD = ∠CDB = 60°
∠ABC = ∠ABD + ∠DBC = 60° + 60° = 120°
And
∠ADC = ∠ADB + ∠CDB = 60° + 60° = 120°
Hence, all angles of the given rhombus are 60°, 120°, 60° and 120°, respectively.
Question 12: The two adjacent angles of a parallelogram are the same. Find the measure of each and every angle of the parallelogram.
Answer 12: A parallelogram with two equal adjacent angles.
To find: the measure of each of the angles of the parallelogram.
The sum of all the adjacent angles of a parallelogram is supplementary.
∠B = ∠A = 90°
In a parallelogram, the opposite sides are the same.
Hence, each angle of the parallelogram measures 90°.
Question 13: The measures of the two adjacent angles of a parallelogram are in the given ratio 3: 2. Find the measure of every angle of the parallelogram.
Answer 13: Let the measures of two adjacent angles ∠A and ∠B be 3x and 2x, respectively, in parallelogram ABCD.
∠A + ∠B = 180°
⇒ 3x + 2x = 180°
⇒ 5x = 180°
The opposite sides of a parallelogram are the same.
∠A = ∠C = 3x = 3 × 36° = 108°
∠B = ∠D = 2x = 2 × 36° = 72°
Question 14: State whether true or false.
(a) All the rectangles are squares.
(b) All the rhombuses are parallelograms.
(c) All the squares are rhombuses and also rectangles.
(d) All the squares are not parallelograms.
(e) All the kites are rhombuses.
(f) All the rhombuses are kites.
(g) All the parallelograms are trapeziums.
(h) All the squares are trapeziums.
Answer 14: (a) This statement is false.
Since all squares are rectangles, all rectangles are not squares.
(b) This statement is true.
(c) This statement is true.
(d) This statement is false.
Since all squares are parallelograms, the opposite sides are parallel, and opposite angles are
congruent.
(e) This statement is false.
Since, for example, the length of the sides of a kite is not the same length.
(f) This statement is true.
(g) This statement is true.
(h) This statement is true.
Question 15: Two adjacent angles of a parallelogram are equal. What is the measure of each of these angles?
Answer 15: Let ∠A and ∠B be two adjacent angles.
But we know that the sum of adjacent angles of a parallelogram is 180o
But given that ∠A = ∠B
Now substituting, we get
∠A + ∠A = 180°
∠A=180/2 = 90°
Question 16:Triangle ABC is a rightangled triangle, and O is the midpoint of the side opposite to the right angle. State why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you).
Answer 16: AD and DC are drawn in such a way that AD is parallel to BC
and AB is parallel to DC
AD = BC and AB = DC
ABCD is a rectangle since the opposite sides are equal and parallel to each other, and the measure of all the interior angles is altogether 90°.
In a rectangle, all the diagonals bisect each other and are of equal length.
Therefore, AO = OC = BO = OD
Hence, O is equidistant from A, B and C.
Question 17: Is the quadrilateral ABCD a parallelogram if
(i) the measure of angle D + the measure of angle B = 180°?
(ii) AB = DC = 8 cm , the length of AD = 4 cm and the length of BC = 4.4 cm?
(iii)The measure of angle A = 70° and the measure of angle C = 65°?
Answer 17: (i) Yes, the quadrilateral ABCD can be a parallelogram if ∠D + ∠B = 180° but it should also fulfil certain conditions, which are as follows:
(a) The sum of all the adjacent angles should be 180°.
(b) Opposite angles of a parallelogram must be equal.
(ii) No, opposite sides should be of the same length. Here, AD ≠ BC
(iii) No, opposite angles should be of the same measures. ∠A ≠ ∠C
Question 18: Find the measure of angles P and S if SP and RQ are parallel.
Answer 18: ∠P + ∠Q = 180° (angles on the same side of transversal)
∠P + 130° = 180°
∠P = 180° – 130° = 50°
also, ∠R + ∠S = 180° (angles on the same side of transversal)
⇒ 90° + ∠S = 180°
⇒ ∠S = 180° – 90° = 90°
Thus, ∠P = 50° and ∠S = 90°
Yes, there is more than one method to find m∠P.
PQRS is a quadrilateral. The sum of measures of all angles is 360°.
Since we know the measurement of ∠Q, ∠R and ∠S.
∠Q = 130°, ∠R = 90° and ∠S = 90°
∠P + 130° + 90° + 90° = 360°
⇒ ∠P + 310° = 360°
⇒ ∠P = 360° – 310° = 50°
Question 19: The opposite angles of a parallelogram are (3x + 5)° and (61 – x)°. Find the measure of four angles.
Answer 19: (3x + 5)° and (61 – x)° are the opposite angles of a parallelogram.
The opposite angles of a parallelogram are the same.
Therefore, (3x + 5)° = (61 – x)°
3x + x = 61° – 5°
4x = 56°
x = 56°/4
x = 14°
The first angle of the parallelogram =3x + 5
= 3(14) + 5
= 42 + 5 = 47°
The second angle of the parallelogram=61 – x
= 61 – 14 = 47°
The measure of angles adjacent to the given angles = 180° – 47° = 133°
Hence, the measure of the four angles of the parallelogram is 47°, 133°, 47°, and 133°.
Question 20: What is the maximum exterior angle possible for a regular polygon?
Answer 20: To find: The maximum exterior angle possible for a regular polygon.
A polygon with minimum sides is an equilateral triangle.
So, the number of sides =3
The sum of all exterior angles of a polygon is 360°
Exterior angle =360°/Number of sides
Therefore, the maximum exterior angle possible will be
Benefits of Solving Important Questions Class 8 Mathematics Chapter 3
Mathematics demands a lot of practice. Class 8, 9 and 10 are very important for students to develop a strong fundamental knowledge of Algebra as well as Geometry. We recommend that students get access to Extramarks comprehensive set of Important Questions Class 8 Mathematics Chapter 3. By regularly solving questions and going through our answer solutions, students will gain good confidence to solve complex problems from the Understanding Quadrilaterals chapter.
Below are a few benefits of frequently solving NCERT textbook and NCERT exemplar questions for Class 8.
 Our experienced Mathematics subject teachers have carefully assembled the most important questions Class 8 Mathematics Chapter 3 by analysing many past exam questions.
 Mathematics Class 8 Chapter 3 important questions provide information about the types of questions that may be asked in exams, which helps to minimise stress and exam anxiety.
 The questions and solutions provided are based on the latest CBSE syllabus and as per CBSE guidelines. So the students can completely count on it.
 Studying something just once may help you grasp the concepts, but going over it again will help you remember it. You enhance your chances of getting good grades on exams while you solve more important questions Class 8 Mathematics Chapter 3.
 The questions covered in our set of Important Questions Class 8 Mathematics Chapter 3 are based on various topics covered in the Quadrilaterals chapter. So while solving these questions, students will be able to revise the chapter and also clear any doubts they have.
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Mathematics requires a lot of practice. To score, one must have a strong conceptual understanding of the chapter, be good with numbers and calculations, practice Important Questions Class 8 Mathematics Chapter 3 regularly, give practice tests from time to time, get feedback and avoid silly mistakes. Regular practice with discipline will surely help the student to ace their exams.
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There are six basic types of quadrilaterals.
 Parallelogram
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NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals
NCERT solutions for class 8 maths chapter 3 understanding quadrilaterals define a polygon as a simple closed curve that is made up of straight lines. Thus, a quadrilateral can be defined as a polygon that has four sides, four angles, and four vertices. This chapter starts by introducing children to some very important concepts that they need to learn before moving on to studying quadrilaterals . These topics include the classification of polygons on the basis of sides, examining diagonals , concave, convex, regular, and irregular polygons as well as the angle sum property. The scope of NCERT solutions class 8 maths chapter 3 is very vast as there are several properties and types of quadrilaterals available. However, the explanation given in these solutions helps to simplify the learning process ensuring that students can build a strong geometrical foundation.
Class 8 maths NCERT solutions chapter 3 elaborates on special quadrilaterals such as squares , rectangles , parallelograms , kites , and rhombuses . They show kids how to solve problems based on these figures and intelligently utilize the associated properties to remove the complexities from such questions. In the NCERT solutions Chapter 3 Understanding Quadrilaterals we will take an indepth look at the basic elements and theories of these foursided polygons and also you can find some of these in the exercises given below.
 NCERT Solutions Class 8 Maths Chapter 3 Ex 3.1
 NCERT Solutions Class 8 Maths Chapter 3 Ex 3.2
 NCERT Solutions Class 8 Maths Chapter 3 Ex 3.3
 NCERT Solutions Class 8 Maths Chapter 3 Ex 3.4
NCERT Solutions for Class 8 Maths Chapter 3 PDF
Using the NCERT solutions class 8 maths children can solidify several concepts of quadrilaterals. They understand the conditions under which a special quadrilateral such as a parallelogram becomes a square, how to find the measure of an interior or exterior angle , and so on. The links to all these brief and precise solutions are given below and kids can use them to improve their mathematical acumen.
☛ Download Class 8 Maths NCERT Solutions Chapter 3 Understanding Quadrilaterals
NCERT Class 8 Maths Chapter 3 Download PDF
NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals
Quadrilaterals form a vital shape contributing to geometrical studies. Thus, children need to develop a robust conceptual foundation as they will require it in higher classes for solving more complicated problems and constructing this figure. They can do this by revising the solutions given above regularly. The following sections deal with an exercisewise detailed analysis of NCERT Solutions Class 8 Maths Chapter 3 understanding quadrilaterals.
 Class 8 Maths Chapter 3 Ex 3.1  7 Questions
 Class 8 Maths Chapter 3 Ex 3.2  6 Questions
 Class 8 Maths Chapter 3 Ex 3.3  12 Questions
 Class 8 Maths Chapter 3 Ex 3.4  6 Questions
☛ Download Class 8 Maths Chapter 3 NCERT Book
Topics Covered: Identifying the polygon, finding the measure of angles, and verifying the exterior angles of a polygon are topics under class 8 maths NCERT solutions chapter 3. Apart from this, there are many sections dealing with the various elements of trapeziums , parallelograms, rectangles, squares, etc.
Total Questions: There are a total of 31 fantastic sums in Class 8 maths chapter 3 Understanding Quadrilaterals. 7 are simple theorybased problems, 16 are inbetween and 8 are higherorder thinking sums.
List of Formulas in NCERT Solutions Class 8 Maths Chapter 3
The questions in the NCERT solutions class 8 maths chapter 3 are not only based on some formulas but also see the use of various vital properties. The sum of interior and exterior angles , along with theorems give the keys to attempting these sums. The angle sum property states that the sum of all the interior angles of a polygon is a multiple of the number of triangles that make up that polygon. Such pointers covered in NCERT solutions for class 8 maths chapter 3 make up the crux of this lesson and are given below.
 Angle Sum Property of a Quadrilateral: a + b + c + d = 360°. (a, b, c, d are the interior angles).
 The opposite sides and opposite angles of a parallelogram are equal in length.
 The adjacent angles in a parallelogram are supplementary.
 The diagonals of a parallelogram bisect each other.
 The diagonals of a rhombus are perpendicular bisectors of one another.
Important Questions for Class 8 Maths NCERT Solutions Chapter 3
CBSE Important Questions for Class 8 Maths Chapter 3 Exercise 3.1 

CBSE Important Questions for Class 8 Maths Chapter 3 Exercise 3.2 

CBSE Important Questions for Class 8 Maths Chapter 3 Exercise 3.3 

CBSE Important Questions for Class 8 Maths Chapter 3 Exercise 3.4 

NCERT Solutions for Class 8 Maths Video Chapter 3
NCERT Class 8 Maths Videos for Chapter 3  

Video Solutions for Class 8 Maths Exercise 3.1  
Video Solutions for Class 8 Maths Exercise 3.2  
Video Solutions for Class 8 Maths Exercise 3.3  
Video Solutions for Class 8 Maths Exercise 3.4  
FAQs on NCERT Solutions Class 8 Maths Chapter 3
Do i need to practice all questions provided in ncert solutions class 8 maths understanding quadrilaterals.
All the sums in the NCERT Solutions Class 8 Maths Understanding Quadrilaterals cover different subtopics of the lesson. These sums also pave a foundation for the geometrical topics in grades that are to follow. Thus, it is crucial for kids to practice all questions so as to get a clear idea of all the components in a quadrilateral.
What are the Important Topics Covered in Class 8 Maths NCERT Solutions Chapter 3?
Each exercise is based on a different topic such as angles of a polygon, rhombus, square, and rectangles; thus, each section that falls under the NCERT Solutions Class 8 Maths Chapter 3 must be given equal importance. Kids need to strategize their studies to focus more on learning properties and then applying them to questions.
How Many Questions are there in NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals?
There are a total of 31 questions in the NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals that are distributed among 4 exercises. There are different types of questions such as true and false sums, identifying the type of shape based on certain properties, and finding the measure of a particular angle using formulas.
What are the Important Formulas in Class 8 Maths NCERT Solutions Chapter 3?
Formulas such as the angle sum property of a quadrilateral, exterior angle property of a polygon, and other associated theories form the foundation of the NCERT Solutions Class 8 Maths Chapter 3. Students must spend a good amount of time practicing questions so as to get a good understanding of their application.
How CBSE Students can utilize NCERT Solutions Class 8 Maths Chapter 3 effectively?
To effectively utilize NCERT Solutions Class 8 Maths Chapter 3 it is advised that students go through the theory and solved examples associated with each exercise. They should then try to attempt the problem on their own. Finally, to get the best out of these solutions kids should crosscheck their answers and go through the steps so that they can organize their answers in a wellstructured manner.
Why Should I Practice NCERT Solutions Class 8 Maths Understanding Quadrilaterals Chapter 3?
The only way to ensure that a student has perfected his knowledge of a chapter is by practicing the questions periodically. The NCERT Solutions Class 8 Maths Understanding Quadrilaterals Chapter 3 has been given by experts with certain tips included to simplify the problems. By regular revision, kids will be confident with the topic and can get an amazing score in their examination.
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In this chapter, we will learn
 What are curves , open curves, closed curves, simple curves
 What are polygons , Different Types of Polygons
 Diagonal of a Polygon
 Convex and Concave Polygons
 Regular and Irregular Polygons
 Angle Sum Property of Polygons
 Sum of Exterior Angles of a Polygon
 Exterior Angles of a Regular Polygon
 What is a Quadrilateral
 Parallelogram
 Parallelogram propertie s  Opposite Angles are equal, Opposite sides are equal, Adjacent Angles are supplementary, Diagonals Bisect Each other
 Rhombus, Rectangle, Square are all parallelograms with additional properties
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NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals
NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals are provided below. Our solutions covered each questions of the chapter and explains every concept with a clarified explanation. To score good marks in Class 8 Mathematics examination, it is advised to solve questions provided at the end of each chapter in the NCERT book.
NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals are prepared based on Class 8 NCERT syllabus, taking the types of questions asked in the NCERT textbook into consideration. Further, all the CBSE Class 8 Solutions Maths Chapter 3 are in accordance with the latest CBSE guidelines and marking schemes.
Class 8 Maths Chapter 3 Exercise 3.1 Solutions
Class 8 Maths Chapter 3 Exercise 3.2 Solutions
Class 8 Maths Chapter 3 Exercise 3.3 Solutions
Class 8 Maths Chapter 3 Exercise 3.4 Solutions
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 CBSEUnderstanding Quadrilaterals
 Sample Questions
Understanding QuadrilateralsSample Questions
 STUDY MATERIAL FOR CBSE CLASS 8 MATH
 Chapter 1  Algebraic Expressions and Identities
 Chapter 2  Comparing Quantities
 Chapter 3  Cubes and Cube Roots
 Chapter 4  Data handling
 Chapter 5  Direct and Inverse Proportions
 Chapter 6  Exponents and Powers
 Chapter 7  Factorization
 Chapter 8  Introduction to Graphs
 Chapter 9  Mensuration
 Chapter 10  Playing with Numbers
 Chapter 11  Practical Geometry
 Chapter 12  Squares and Square Roots
 Chapter 13  Visualizing Solid Shapes
 Chapter 14  Linear Equations in One Variable
 Chapter 15  Rational Numbers
 Chapter 16  Understanding Quadrilaterals
Class 8 Maths Chapter 3 Important Question Answers  Understanding Quadrilaterals
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Q1: What is the maximum exterior angle possible for a regular polygon? Sol: To find: The maximum exterior angle possible for a regular polygon. A polygon with minimum sides is an equilateral triangle. So, the number of sides = 3 The sum of all exterior angles of a polygon is 360° Exterior angle = 360°/Number of sides Therefore, the maximum exterior angle possible will be = 360°/3 = 120°
Q2: Find the measure of angles P and S if SP and RQ are parallel. Sol: ∠P + ∠Q = 180° (angles on the same side of transversal) ∠P + 130° = 180° ∠P = 180° – 130° = 50° also, ∠R + ∠S = 180° (angles on the same side of transversal) ⇒ 90° + ∠S = 180° ⇒ ∠S = 180° – 90° = 90° Thus, ∠P = 50° and ∠S = 90° Yes, there is more than one method to find m∠P. PQRS is a quadrilateral. The sum of measures of all angles is 360°. Since we know the measurement of ∠Q, ∠R and ∠S. ∠Q = 130°, ∠R = 90° and ∠S = 90° ∠P + 130° + 90° + 90° = 360° ⇒ ∠P + 310° = 360° ⇒ ∠P = 360° – 310° = 50°
Sol: AD and DC are drawn in such a way that AD is parallel to BC and AB is parallel to DC AD = BC and AB = DC ABCD is a rectangle since the opposite sides are equal and parallel to each other, and the measure of all the interior angles is altogether 90°. In a rectangle, all the diagonals bisect each other and are of equal length. Therefore, AO = OC = BO = OD Hence, O is equidistant from A, B and C. Q4: State whether true or false. (a) All the rectangles are squares. (b) All the rhombuses are parallelograms. (c) All the squares are rhombuses and also rectangles. (d) All the squares are not parallelograms. (e) All the kites are rhombuses. (f) All the rhombuses are kites. (g) All the parallelograms are trapeziums. (h) All the squares are trapeziums. Sol: (a) This statement is false. A rectangle has opposite sides equal and all angles equal to 90 degrees, but for it to be a square, all four sides must be equal. Therefore, not all rectangles are squares.
(b) This statement is true.
A rhombus is a type of parallelogram where all four sides are of equal length. Since it has both pairs of opposite sides parallel, it is always a parallelogram.
(c) This statement is true.
A square has all properties of a rhombus (all sides equal) and a rectangle (all angles 90 degrees). Therefore, all squares are both rhombuses and rectangles.
(d) This statement is false.
A square is a specific type of parallelogram where all sides are equal and all angles are right angles. Therefore, all squares are parallelograms.
(e) This statement is false.
A kite has two pairs of adjacent sides equal, but not all four sides need to be equal. For it to be a rhombus, all four sides must be equal. Therefore, not all kites are rhombuses.
(f) This statement is true.
A rhombus has all four sides equal, which satisfies the condition of a kite having two pairs of adjacent sides equal. Therefore, all rhombuses are kites.
(g) This statement is false.
A parallelogram has both pairs of opposite sides parallel, and a trapezium is a quadrilateral with exactly one pair of parallel sides. Since parallelograms have two pairs of parallel sides, they do not meet this criterion and are not considered trapeziums.
(h) This statement is true.
A square has both pairs of opposite sides parallel, which means it satisfies the condition of a trapezium having at least one pair of parallel sides. Therefore, all squares are trapeziums.
Q5: The two adjacent angles of a parallelogram are the same. Find the measure of each and every angle of the parallelogram. Sol: A parallelogram with two equal adjacent angles. To find: the measure of each of the angles of the parallelogram. The sum of all the adjacent angles of a parallelogram is supplementary. ∠A + ∠B = 180° 2∠A = 180° ∠A = 90° ∠B = ∠A = 90° In a parallelogram, the opposite sides are the same. Therefore, ∠C = ∠A = 90° ∠D = ∠B = 90° Hence, each angle of the parallelogram measures 90°. Q6: ABCD is a parallelogram in which ∠A = 110 ° . Find the measure of the angles B, C and D, respectively. Sol: The measure of angle A = 110° the sum of all adjacent angles of a parallelogram is 180° ∠A + ∠B = 180 110°+ ∠B = 180° ∠B = 180° 110° = 70°. Also ∠B + ∠C = 180° [Since ∠B and ∠C are adjacent angles] 70° + ∠C = 180° ∠C = 180° 70° = 110°. Now ∠C + ∠D = 180° [Since ∠C and ∠D are adjacent angles] 110° + ∠D = 180° ∠D = 180° 110° = 70° Q7: The measure of the two adjacent angles of the given parallelogram is the ratio of 3:2. Then, find the measure of each angle of the parallelogram. Sol: A parallelogram with adjacent angles in the ratio of 3:2 To find: The measure of each of the angles of the parallelogram. Let the measure of angle A be 3x Let the measure of angle B be 2x Since the sum of the measures of adjacent angles is 180° for a parallelogram, ∠A+∠B=180° 3x+2x=180° 5x=180° x=36° ∠A=∠C =3x=108° ∠B=∠D =2x=72° (Opposite angles of a parallelogram are equal). Hence, the angles of a parallelogram are 108°, 72°,108°and 72°
Q7: Find x in the following figure. Sol: The two interior angles in the given figures are right angles = 90° 70° + m = 180° m = 180° – 70° = 110° (In a linear pair, the sum of two adjacent angles altogether measures up to 180°) 60° + n = 180° n = 180° – 60° = 120° (In a linear pair, the sum of two adjacent angles altogether measures up to 180° The given figure has five sides, and it is a pentagon. Thus, the sum of the angles of the pentagon = 540° 90° + 90° + 110° + 120° + y = 540° 410° + y = 540° y = 540° – 410° = 130° x + y = 180°….. (Linear pair) x + 130° = 180° x = 180° – 130° = 50°
Q8: Adjacent sides of a rectangle are in the ratio 5: 12; if the perimeter of the given rectangle is 34 cm, find the length of the diagonal. Sol: The ratio of the adjacent sides of the rectangle is 5: 12 Let 5x and 12x be adjacent sides. The perimeter is the sum of all the given sides of a rectangle. 5x + 12x + 5x + 12x = 34 cm ……(since the opposite sides of the rectangle are the same) 34x = 34 x = 34/34 x = 1 cm Therefore, the adjacent sides of the rectangle are 5 cm and 12 cm, respectively. That is, Length =12 cm Breadth = 5 cm Length of the diagonal = √( l2 + b2) = √( 122 + 52) = √(144 + 25) = √169 = 13 cm Hence, the length of the diagonal of a rectangle is 13 cm.
Q9: Find the measure of all the exterior angles of a regular polygon with (i) 9 sides and (ii) 15 sides. Sol: (i) Total measure of all exterior angles = 360° Each exterior angle =sum of exterior angle = 360° = 40° number of sides 9 Each exterior angle = 40° (ii) Total measure of all exterior angles = 360° Each exterior angle = sum of exterior angle =360° = 24° number of sides 15 Each exterior angle = 24°
Q10: A quadrilateral has three acute angles, each measuring 80°. What is the measure of the fourth angle of the quadrilateral? Sol: – Let x be the measure of the fourth angle of a quadrilateral. The sum of all the angles of a quadrilateral + 360° 80° + 80° + 80° + x = 360° …………(since the measure of all the three acute angles = 80°) 240° + x = 360° x = 360° – 240° x = 120° Hence, the fourth angle made by the quadrilateral is 120°. Q11: How many sides do regular polygons consist of if each interior angle is 165 ° ? Sol: A regular polygon with an interior angle of 165° We need to find the sides of the given regular polygon: The sum of all exterior angles of any given polygon is 360°. Formula Used: Number of sides = 360 ∘ /Exterior angle Exterior angle = 180 ∘ −Interior angle Thus, Each interior angle = 165° Hence, the measure of every exterior angle will be = 180° − 165° = 15° Therefore, the number of sides of the given polygon will be = 360°/15° = 24° Q12: ABCD is a parallelogram with ∠A = 80°. The internal bisectors of ∠B and ∠C meet each other at O. Find the measure of the three angles of ΔBCO. Sol: The measure of angle A = 80°. In a parallelogram, the opposite angles are the same. Hence, ∠A = ∠C = 80° And ∠OCB = (1/2) × ∠C = (1/2) × 80° = 40° ∠B = 180° – ∠A (the sum of interior angles situated on the same side of the transversal is supplementary) = 180° – 80° = 100° Also, ∠CBO = (1/2) × ∠B ∠CBO= (1/2) × 100° ∠CBO= 50°. By the property of the sum of the angle BCO, we get, ∠BOC + ∠OBC + ∠CBO = 180° ∠BOC = 180° – (∠OBC + CBO) = 180° – (40° + 50°) = 180° – 90° = 90° Hence, the measure of all the angles of triangle BCO is 40°, 50° and 90°. Q13: Is it ever possible to have a regular polygon, each of whose interior angles is 100? Sol: The sum of all the exterior angles of a regular polygon is 360° As we also know, the sum of interior and exterior angles are 180° Exterior angle + interior angle = 180  100 = 80° When we divide the exterior angle, we will get the number of exterior angles since it is a regular polygon means the number of exterior angles equals the number of sides. Therefore n = 360/ 80 = 4.5 And we know that 4.5 is not an integer, so having a regular polygon is impossible. Whose exterior angle is 100° Q14: A diagonal and a side of a rhombus are of equal length. Find the measure of the angles of the rhombus. Sol: Let ABCD be the rhombus. All the sides of a rhombus are the same. Thus, AB = BC = CD = DA. The side and diagonal of a rhombus are equal. AB = BD Therefore, AB = BC = CD = DA = BD Consider triangle ABD, Each side of a triangle ABD is congruent. Hence, ΔABD is an equilateral triangle. Similarly, ΔBCD is also an equilateral triangle. Thus, ∠BAD = ∠ABD = ∠ADB = ∠DBC = ∠BCD = ∠CDB = 60° ∠ABC = ∠ABD + ∠DBC = 60° + 60° = 120° And ∠ADC = ∠ADB + ∠CDB = 60° + 60° = 120° Hence, all angles of the given rhombus are 60°, 120°, 60° and 120°, respectively. Q15: The measures of the two adjacent angles of a parallelogram are in the given ratio 3: 2. Find the measure of every angle of the parallelogram. Sol: Let the measures of two adjacent angles ∠A and ∠B be 3x and 2x, respectively, in parallelogram ABCD. ∠A + ∠B = 180° ⇒ 3x + 2x = 180° ⇒ 5x = 180° ⇒ x = 36° The opposite sides of a parallelogram are the same. ∠A = ∠C = 3x = 3 × 36° = 108° ∠B = ∠D = 2x = 2 × 36° = 72°
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NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals
 NCERT Solutions
 Chapter 3 Understanding Quadrilaterals
NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals  FREE PDF Download
The NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals covers all the chapter's questions (All Exercises). These NCERT Solutions for Class 8 Maths have been carefully compiled and created in accordance with the most recent CBSE Syllabus 202425 updates. Students can use these NCERT Solutions for Class 8 to reinforce their foundations. Subject experts at Vedantu have created the continuity and differentiability class 8 NCERT solutions to ensure they match the current curriculum and help students while solving or practising problems.
Glance of NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals  Vedantu
In this article, we will learn about different quadrilaterals like squares, rectangles, parallelograms, rhombuses, and trapeziums, along with their properties.
This chapter dives into the world of quadrilaterals, which are foursided closed figures.
This chapter explains effective methods to solve problems concerning quadrilaterals.
Each type of quadrilateral is discussed in terms of its defining properties including side lengths, angle measurements, diagonals, and symmetry.
The chapter also highlights special properties of certain quadrilaterals, like the properties of diagonals in rectangles and squares, and the diagonals of rhombuses.
This article contains chapter notes, formulas, exercise links, and important questions for chapter 3  Understanding Quadrilaterals.
There are four exercises (26 fully solved questions) in Class 8th Maths Chapter 3 Understanding Quadrilaterals.
Access Exercise Wise NCERT Solutions for Chapter 3 Maths Class 8
Current Syllabus Exercises of Class 8 Maths Chapter 3 




Exercises Under NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals
Exercise 3.1 introduces polygons, covering their basic definition and classification based on the number of sides, such as triangles, quadrilaterals, pentagons, etc. It also distinguishes between convex and concave polygons, helping students understand the differences between these types of polygons.
Exercise 3.2 delves into the properties of quadrilaterals, exploring various types like trapeziums, kites, and parallelograms. This exercise helps students learn to identify different quadrilaterals and understand their specific properties.
Exercise 3.3 examines the properties of parallelograms, such as opposite sides being equal and parallel, opposite angles being equal, and diagonals bisecting each other. It includes problems for identifying parallelograms based on these properties and proving certain properties using theorems.
Exercise 3.4 looks at special parallelograms like rhombuses, rectangles, and squares. It highlights their unique properties, such as all sides being equal in a rhombus and all angles being 90 degrees in a rectangle. This exercise helps students understand and differentiate between these specific types of parallelograms.
List of Formulas
There are two major kinds of formulas related to quadrilaterals  Area and Perimeter. The following tables depict the formulas related to the areas and perimeters of different kinds of quadrilaterals.
Area of Quadrilaterals
Area of a Square  Side x Side 
Area of a Rectangle  Length x Width 
Area of a Parallelogram  Base x Height 
Area of a Rhombus  1/2 x 1st Diagonal x 2nd Diagonal 
Area of a Kite  1/2 x 1st Diagonal x 2nd Diagonal 
Perimeter of Quadrilaterals
Perimeter of any quadrilateral is equal to the sum of all its sides, that is, AB + BC + CD + AD.
Name of the Quadrilateral  Perimeter 
Perimeter of a Square  4 x Side 
Perimeter of a Rectangle  2 (Length + Breadth) 
Perimeter of a Parallelogram  2 (Base + Side) 
Perimeter of a Rhombus  4 x Side 
Perimeter of a Kite  2 (a + b), where a and b are the adjacent pairs 
Access NCERT Solutions for Class 8 Maths Chapter 3 – Understanding Quadrilaterals
Exercise 3.1.
1. Given here are some figures.
Classify each of them on the basis of following.
Simple Curve
Ans: Given: the figures $(1)$to $(8)$
We need to classify the given figures as simple curves.
We know that a curve that does not cross itself is referred to as a simple curve.
Therefore, simple curves are $1,2,5,6,7$.
Simple Closed Curve
We need to classify the given figures as simple closed curves.
We know that a simple closed curve is one that begins and ends at the same point without crossing itself.
Therefore, simple closed curves are $1,2,5,6,7$.
We need to classify the given figures as polygon.
We know that any closed curve consisting of a set of sides joined in such a way that no two segments
cross is known as a polygon.
Therefore, the polygons are $1,2$.
Convex Polygon
We need to classify the given figures as convex polygon.
We know that a closed shape with no vertices pointing inward is called a convex polygon.
Therefore, the convex polygon is $2$.
Concave Polygon
We need to classify the given figures as concave polygon.
We know that a polygon with at least one interior angle greater than 180 degrees is called a concave
Therefore, the concave polygon is $1$.
2. What is a regular polygon?
State the name of a regular polygon of
(i) 3 sides (ii) 4 sides (iii) 6 sides
Solution: A regular polygon is a flat shape with all sides of equal length and all interior 5angles equal in measure. In simpler terms, all the sides are the same size and all the corners look the same.
Here are the names of regular polygons based on the number of sides:
(i) 3 sides  Equilateral Triangle (all three angles are also 60 degrees each)
(ii) 4 sides  Square
(iii) 6 sides  Hexagon
Exercise3.2
1. Find ${\text{x}}$in the following figures.
We need to find the value of ${\text{x}}{\text{.}}$
We know that the sum of all exterior angles of a polygon is ${360^ \circ }.$
$ {\text{x}} + {125^ \circ } + {125^ \circ } = {360^ \circ } $
$ \Rightarrow {\text{x}} + {250^ \circ } = {360^ \circ } $
$ \Rightarrow {\text{x}} = {360^ \circ }  {250^ \circ } $
$ \Rightarrow {\text{x}} = {110^ \circ } $
$ {\text{x}} + {90^ \circ } + {60^ \circ } + {90^ \circ } + {70^ \circ } = {360^ \circ } $
$ \Rightarrow {\text{x}} + {310^ \circ } = {360^ \circ } $
$ \Rightarrow {\text{x}} = {360^ \circ }  {310^ \circ } $
$ \Rightarrow {\text{x}} = {50^ \circ } $
2. Find the measure of each exterior angle of a regular polygon of
Given: a regular polygon with $9$ sides
We need to find the measure of each exterior angle of the given polygon.
We know that all the exterior angles of a regular polygon are equal.
The sum of all exterior angle of a polygon is ${360^ \circ }$.
Formula Used: ${\text{Exterior}}\;{\text{angle}} = \dfrac{{{{360}^ \circ }}}{{{\text{Number}}\;{\text{of}}\;{\text{sides}}}}$
Sum of all angles of given regular polygon $ = {360^ \circ }$
Number of sides $ = 9$
Therefore, measure of each exterior angle will be
$ = \dfrac{{{{360}^ \circ }}}{9} $
$ = {40^ \circ } $
Given: a regular polygon with $15$ sides
Number of sides $ = 15$
$ = \dfrac{{{{360}^ \circ }}}{{15}} $
$ = {24^ \circ } $
3. How many sides does a regular polygon have if the measure of an exterior angle is ${24^ \circ }$?
Ans: Given: A regular polygon with each exterior angle ${24^ \circ }$
We need to find the number of sides of given polygon.
We know that sum of all exterior angle of a polygon is ${360^ \circ }$.
Formula Used: ${\text{Number}}\;{\text{of}}\;{\text{sides}} = \dfrac{{{{360}^ \circ }}}{{{\text{Exterior}}\;{\text{angle}}}}$
Each angle measure $ = {24^ \circ }$
Therefore, number of sides of given polygon will be
$ = \dfrac{{{{360}^ \circ }}}{{{{24}^ \circ }}} $
$ = 15 $
4. How many sides does a regular polygon have if each of its interior angles is ${165^ \circ }$?
Ans: Given: A regular polygon with each interior angle ${165^ \circ }$
We need to find the sides of the given regular polygon.
${\text{Exterior}}\;{\text{angle}} = {180^ \circ }  {\text{Interior}}\;{\text{angle}}$
Each interior angle $ = {165^ \circ }$
So, measure of each exterior angle will be
$ = {180^ \circ }  {165^ \circ } $
$ = {15^ \circ } $
Therefore, number of sides of polygon will be
$ = \dfrac{{{{360}^ \circ }}}{{{{15}^ \circ }}} $
$ = 24 $
Is it possible to have a regular polygon with measure of each exterior angle as ${22^ \circ }$?
Given: A regular polygon with each exterior angle ${22^ \circ }$
We need to find if it is possible to have a regular polygon with given angle measure.
We know that sum of all exterior angle of a polygon is ${360^ \circ }$. The polygon will be possible if ${360^ \circ }$ is a perfect multiple of exterior angle.
$\dfrac{{{{360}^ \circ }}}{{{{22}^ \circ }}}$ does not give a perfect quotient.
Thus, ${360^ \circ }$ is not a perfect multiple of exterior angle. So, the polygon will not be possible.
Can it be an interior angle of a regular polygon? Why?
Ans: Given: Interior angle of a regular polygon $ = {22^ \circ }$
We need to state if it can be the interior angle of a regular polygon.
And, ${\text{Exterior}}\;{\text{angle}} = {180^ \circ }  {\text{Interior}}\;{\text{angle}}$
Thus, Exterior angle will be
$ = {180^ \circ }  {22^ \circ } $
$ = {158^ \circ } $
$\dfrac{{{{158}^ \circ }}}{{{{22}^ \circ }}}$ does not give a perfect quotient.
Thus, ${158^ \circ }$ is not a perfect multiple of exterior angle. So, the polygon will not be possible.
What is the minimum interior angle possible for a regular polygon?
Ans: Given: A regular polygon
We need to find the minimum interior angle possible for a regular polygon.
A polygon with minimum number of sides is an equilateral triangle.
So, number of sides $ = 3$
${\text{Exterior}}\;{\text{angle}} = \dfrac{{{{360}^ \circ }}}{{{\text{Number}}\;{\text{of}}\;{\text{sides}}}}$
Thus, Maximum Exterior angle will be
$ = \dfrac{{{{360}^ \circ }}}{3} $
$ = {120^ \circ } $
We know, ${\text{Interior}}\;{\text{angle}} = {180^ \circ }  {\text{Exterior}}\;{\text{angle}}$
Therefore, minimum interior angle will be
$ = {180^ \circ }  {120^ \circ } $
$ = {60^ \circ } $
What is the maximum exterior angel possible for a regular polygon?
Ans: Given: A regular polygon
We need to find the maximum exterior angle possible for a regular polygon.
Therefore, Maximum Exterior angle possible will be
$ = {120^ \circ } $
Exercise 3.3
1. Given a parallelogram ABCD. Complete each statement along with the definition or property used.
$\;{\text{AD}}$ = $...$
Given: A parallelogram ${\text{ABCD}}$
We need to complete each statement along with the definition or property used.
We know that opposite sides of a parallelogram are equal.
Hence, ${\text{AD}}$ = ${\text{BC}}$
$\;\angle {\text{DCB }} = $ $...$
Given: A parallelogram ${\text{ABCD}}$.
${\text{ABCD}}$ is a parallelogram, and we know that opposite angles of a parallelogram are equal.
Hence, $\angle {\text{DCB = }}\angle {\text{DAB}}$
${\text{OC}} = ...$
${\text{ABCD}}$ is a parallelogram, and we know that diagonals of parallelogram bisect each other.
Hence, ${\text{OC = OA}}$
$m\angle DAB\; + \;m\angle CDA\; = \;...$
Given : A parallelogram ${\text{ABCD}}$.
${\text{ABCD}}$ is a parallelogram, and we know that adjacent angles of a parallelogram are supplementary to each other.
Hence, $m\angle DAB\; + \;m\angle CDA\; = \;180^\circ $
2. Consider the following parallelograms. Find the values of the unknowns x, y, z.
Given: A parallelogram ${\text{ABCD}}$
We need to find the unknowns ${\text{x,y,z}}$
The adjacent angles of a parallelogram are supplementary.
Therefore, ${\text{x} + 100^\circ = 180^\circ }$
${\text{x} = 80^\circ }$
Also, the opposite angles of a parallelogram are equal.
Hence, ${\text{z}} = {\text{x}} = 80^\circ $ and ${\text{y}} = 100^\circ $
Given: A parallelogram.
We need to find the values of ${\text{x,y,z}}$
The adjacent pairs of a parallelogram are supplementary.
Hence, $50^\circ + {\text{y}} = 180^\circ $
${\text{y}} = 130^\circ $
Also, ${\text{x}} = {\text{y}} = 130^\circ $(opposite angles of a parallelogram are equal)
And, ${\text{z}} = {\text{x}} = 130^\circ $ (corresponding angles)
(iii)
Given: A parallelogram
${\text{x}} = 90^\circ $(Vertically opposite angles)
Also, by angle sum property of triangles
${\text{x}} + {\text{y}} + 30^\circ = 180^\circ $
${\text{y}} = 60^\circ $
Also,${\text{z}} = {\text{y}} = 60^\circ $(alternate interior angles)
Given: A parallelogram
Corresponding angles between two parallel lines are equal.
Hence, ${\text{z}} = 80^\circ $ Also,${\text{y}} = 80^\circ $ (opposite angles of parallelogram are equal)
In a parallelogram, adjacent angles are supplementary
Hence,${\text{x}} + {\text{y}} = 180^\circ $
$ {\text{x}} = 180^\circ  80^\circ $
$ {\text{x}} = 100^\circ $
As the opposite angles of a parallelogram are equal, therefore,${\text{y}} = 112^\circ $
Also, by using angle sum property of triangles
$ {\text{x}} + {\text{y}} + 40^\circ = 180^\circ $
$ {\text{x}} + 152^\circ = 180^\circ $
$ {\text{x}} = 28^\circ $
And ${\text{z}} = {\text{x}} = 28^\circ $(alternate interior angles)
3. Can a quadrilateral ${\text{ABCD}}$be a parallelogram if
(i) $\angle {\text{D}}\;{\text{ + }}\angle {\text{B}} = 180^\circ ?$
Given: A quadrilateral ${\text{ABCD}}$
We need to find whether the given quadrilateral is a parallelogram.
For the given condition, quadrilateral ${\text{ABCD}}$ may or may not be a parallelogram.
For a quadrilateral to be parallelogram, the sum of measures of adjacent angles should be $180^\circ $ and the opposite angles should be of same measures.
(ii) ${\text{AB}} = {\text{DC}} = 8\;{\text{cm}},\;{\text{AD}} = 4\;{\text{cm}}\;$and ${\text{BC}} = 4.4\;{\text{cm}}$
As, the opposite sides ${\text{AD}}$and ${\text{BC}}$are of different lengths, hence the given quadrilateral is not a parallelogram.
(iii) $\angle {\text{A}} = 70^\circ $and $\angle {\text{C}} = 65^\circ $
As, the opposite angles have different measures, hence, the given quadrilateral is a parallelogram.
4. Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
Given: A quadrilateral.
We need to draw a rough figure of a quadrilateral that is not a paralleloghram but has exactly two opposite angles of equal measure.
A kite is a figure which has two of its interior angles, $\angle {\text{B}}$and $\angle {\text{D}}$of same measures. But the quadrilateral ${\text{ABCD}}$is not a parallelogram as the measures of the remaining pair of opposite angles are not equal.
5. The measures of two adjacent angles of a parallelogram are in the ratio 3:2. Find the measure of each of the angles of the parallelogram.
Ans: Given: A parallelogram with adjacent angles in the ratio $3:2$
We need to find the measure of each of the angles of the parallelogram.
Let the angles be $\angle {\text{A}} = 3{\text{x}}$and $\angle {\text{B}} = 2{\text{x}}$
As the sum of measures of adjacent angles is $180^\circ $ for a parallelogram.
$ \angle {\text{A}} + \angle {\text{B}} = 180^\circ $
$ 3{\text{x}} + 2{\text{x}} = 180^\circ $
$ 5{\text{x}} = 180^\circ $
$ {\text{x}} = 36^\circ $
$~\angle A=$ $\angle {\text{C}}$ $= 3{\text{x}} = 108^\circ$and $~\angle B=$ $\angle {\text{D}}$ $= 2{\text{x}} = 72^\circ$(Opposite angles of a parallelogram are equal).
Hence, the angles of a parallelogram are $108^\circ ,72^\circ ,108^\circ $and $72^\circ $.
6. Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Given: A parallelogram with two equal adjacent angles.
The sum of adjacent angles of a parallelogram are supplementary.
$ \angle {\text{A}} + \;\angle {\text{B}} = 180^\circ $
$ 2\angle {\text{A}}\;{\text{ = 180}}^\circ $
$ \angle {\text{A}}\;{\text{ = }}\;{\text{90}}^\circ $
$ \angle {\text{B}}\;{\text{ = }}\angle {\text{A}}\;{\text{ = }}\;{\text{90}}^\circ $
Also, opposite angles of a parallelogram are equal
$ \angle {\text{C}} = \angle {\text{A}} = 90^\circ $
$ \angle {\text{D}} = \angle {\text{B}} = 90^\circ $
Hence, each angle of the parallelogram measures $90^\circ $.
7. The adjacent figure ${\text{HOPE}}$is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.
Given: A parallelogram ${\text{HOPE}}$.
We need to find the measures of angles ${\text{x,y,z}}$and also state the properties used to find these angles.
$\angle {\text{y}} = 40^\circ $(Alternate interior angles)
And $\angle {\text{z}} + 40^\circ = 70^\circ $(corresponding angles are equal)
$\angle {\text{z}} = 30^\circ $
Also, ${\text{x}} + {\text{z}} + 40^\circ = 180^\circ $(adjacent pair of angles)
${\text{x}} = 110^\circ $
8. The following figures ${\text{GUNS}}$and ${\text{RUNS}}$are parallelograms. Find ${\text{x}}$and${\text{y}}$. (Lengths are in cm).
Given: Parallelogram ${\text{GUNS}}$.
We need to find the measures of ${\text{x}}$and ${\text{y}}$.
${\text{GU = SN}}$(Opposite sides of a parallelogram are equal).
$ 3{\text{y }}  {\text{ }}1{\text{ }} = {\text{ }}26{\text{ }} $
$ 3{\text{y }} = {\text{ }}27{\text{ }} $
$ {\text{y }} = {\text{ }}9{\text{ }} $
Also,${\text{SG = NU}}$
Therefore,
$ 3{\text{x}} = 18 $
$ {\text{x}} = 3 $
Given: Parallelogram ${\text{RUNS}}$
We need to find the value of ${\text{x}}$and ${\text{y}}{\text{.}}$
The diagonals of a parallelogram bisect each other, therefore,
$ {\text{y }} + {\text{ }}7{\text{ }} = {\text{ }}20{\text{ }} $
$ {\text{y }} = {\text{ }}13 $
$ {\text{x }} + {\text{ y }} = {\text{ }}16 $
$ {\text{x }} + {\text{ }}13{\text{ }} = {\text{ }}16 $
$ {\text{x }} = {\text{ }}3{\text{ }} $
9. In the above figure both ${\text{RISK}}$and ${\text{CLUE}}$are parallelograms. Find the value of ${\text{x}}{\text{.}}$
Given: Parallelograms ${\text{RISK}}$and ${\text{CLUE}}$
As we know that the adjacent angles of a parallelogram are supplementary, therefore,
In parallelogram ${\text{RISK}}$
$ \angle {\text{RKS + }}\angle {\text{ISK}} = 180^\circ $
$ 120^\circ + \angle {\text{ISK}} = 180^\circ $
As the opposite angles of a parallelogram are equal, therefore,
In parallelogram ${\text{CLUE}}$,
$\angle {\text{ULC}} = \angle {\text{CEU}} = 70^\circ $
Also, the sum of all the interior angles of a triangle is $180^\circ $
$ {\text{x }} + {\text{ }}60^\circ {\text{ }} + {\text{ }}70^\circ {\text{ }} = {\text{ }}180^\circ $
$ {\text{x }} = {\text{ }}50^\circ $
10. Explain how this figure is a trapezium. Which of its two sides are parallel?
We need to explain how the given figure is a trapezium and find its two sides that are parallel.
If a transversal line intersects two specified lines in such a way that the sum of the angles on the same side of the transversal equals $180^\circ $, the two lines will be parallel to each other.
Here, $\angle {\text{NML}} = \angle {\text{MLK}} = 180^\circ $
Hence, ${\text{NM}}{\text{LK}}$
Hence, the given figure is a trapezium.
11. Find ${\text{m}}\angle {\text{C}}$in the following figure if ${\text{AB}}\parallel {\text{CD}}$${\text{AB}}\parallel {\text{CD}}$.
Given: ${\text{AB}}\parallel {\text{CD}}$ and quadrilateral
We need to find the measure of $\angle {\text{C}}$
$\angle {\text{B}} + \angle {\text{C}} = 180^\circ $(Angles on the same side of transversal).
$ 120^\circ + \angle {\text{C}} = 180^\circ $
$ \angle {\text{C}} = 60^\circ $
12. Find the measure of $\angle {\text{P}}$and$\angle {\text{S}}$, if ${\text{SP}}\parallel {\text{RQ}}$in the following figure. (If you find${\text{m}}\angle {\text{R}}$, is there more than one method to find${\text{m}}\angle {\text{P}}$?)
Given: ${\text{SP}}\parallel {\text{RQ}}$and
We need to find the measure of $\angle {\text{P}}$and $\angle {\text{S}}$.
The sum of angles on the same side of transversal is $180^\circ .$
$\angle {\text{P}} + \angle {\text{Q}} = 180^\circ $
$ \angle {\text{P}} + 130^\circ = 180^\circ $
$ \angle {\text{P}} = 50^\circ
$\angle {\text{R }} + {\text{ }}\angle {\text{S }} = {\text{ }}180^\circ {\text{ }} $
$ {\text{ }}90^\circ {\text{ }} + {\text{ }}\angle {\text{S }} = {\text{ }}180^\circ $
${\text{ }}\angle {\text{S }} = {\text{ }}90^\circ {\text{ }} $
Yes, we can find the measure of ${\text{m}}\angle {\text{P}}$ by using one more method.
In the question,${\text{m}}\angle {\text{R}}$and ${\text{m}}\angle {\text{Q}}$are given. After finding ${\text{m}}\angle {\text{S}}$ we can find ${\text{m}}\angle {\text{P}}$ by using angle sum property.
Exercise 3.4
1. State whether True or False.
(a) All rectangles are squares.
(b) All rhombuses are parallelograms.
(c) All squares are rhombuses and also rectangles.
(d) All squares are not parallelograms.
(e) All kites are rhombuses.
(f) All rhombuses are kites.
(g) All parallelograms are trapeziums.
(h) All squares are trapeziums.
Every square is indeed a type of rectangle, not every rectangle can be called a square.
It's correct to say that all squares can be classified as parallelograms due to their shared characteristic of having opposite sides that are parallel and opposite angles that are equal.
Because, a kite shape is that its adjacent sides are not necessarily equal in length, unlike those of a square.
2. Identify all the quadrilaterals that have.
(a) four sides of equal length
(b) four right angles
(a) Rhombus and square have all four sides of equal length.
(b) Square and rectangles have four right angles.
3. Explain how a square is
(i) a quadrilateral
(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle
(i) Square is a quadrilateral because it has four sides.
(ii) A square is a parallelogram because its opposite sides are parallel and opposite angles are equal.
(iii) Square is a rhombus because all four sides are of equal length and diagonals bisect at right angles.
(iv)Square is a rectangle because each interior angle, of the square, is 90°
4. Name the quadrilaterals whose diagonals.
(i) bisect each other
(ii) are perpendicular bisectors of each other
(iii) are equal
(i) Parallelogram, Rhombus, Square and Rectangle
(ii) Rhombus and Square
(iii)Rectangle and Square
5. Explain why a rectangle is a convex quadrilateral.
A rectangle is a convex quadrilateral because both of its diagonals lie inside the rectangle.
6. ABC is a rightangled triangle and O is the midpoint of the side opposite to the right angle. Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you).
AD and DC are drawn so that AD  BC and AB  DC
AD = BC and AB = DC
ABCD is a rectangle as opposite sides are equal and parallel to each other and all the interior angles are of 90°.
In a rectangle, diagonals are of equal length and also bisect each other.
Hence, AO = OC = BO = OD
Thus, O is equidistant from A, B and C.
Overview of Deleted Syllabus for CBSE Class 8 Maths Understanding Quadrilaterals
Chapter  Dropped Topics 
Understanding Quadrilaterals  3.1 Introduction 
3.2 Polygons  
3.2.1 Classification of polygons  
3.2.2 Diagonals  
3.2.5 Angle sum property. 
Class 8 Maths Chapter 3: Exercises Breakdown
Exercise  Number of Questions 
Exercise 3.1  2 Questions & Solutions (1 Long Answer, 1 Short Answer) 
Exercise 3.2  6 Questions & Solutions (6 Short Answers) 
Exercise 3.3  12 Questions & Solutions (6 Long Answers, 6 Short Answers) 
Exercise 3.4  6 Questions & Solutions (1 Long Answer, 5 Short Answers) 
In conclusion, NCERT Solutions for Class 8 Maths Chapter 3  Understanding Quadrilaterals provides a comprehensive and detailed understanding of the properties and characteristics of various types of quadrilaterals. By studying this chapter and using the NCERT solutions, students can enhance their knowledge of quadrilaterals and develop their problemsolving abilities. The chapter starts by introducing quadrilaterals and their diverse types, including parallelograms, rectangles, squares, rhombuses, and trapeziums. It goes on to explain each type, detailing their characteristic properties like side lengths, angles, diagonals, and symmetry. Students that practice these kinds of questions will gain confidence and perform well on tests.
Other Study Material for CBSE Class 8 Maths Chapter 3
S.No.  Important links for Class 8 Maths Chapter 3 Understanding Quadrilaterals 
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ChapterSpecific NCERT Solutions for Class 8 Maths
Given below are the chapterwise NCERT Solutions for Class 8 Maths . Go through these chapterwise solutions to be thoroughly familiar with the concepts.
S.No.  NCERT Solutions Class 8 ChapterWise Maths PDF 
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FAQs on NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals
1. What is the Area of a Field in the Shape of a Rectangle with Dimensions of 20 Meters and 40 Meters?
We know that the field is rectangular. Hence, we can apply the area of a rectangle to find the field area.
Length of the field = 40 Metre
Width of the field = 20 Metre
Area of the rectangular field = Length × Width = 40 × 20 = 800 Sq. Meters.
We know if the length of the rectangle is L and breadth is B then,
Area of a rectangle = Length × Breadth or L × B
Perimeter = 2 × (L + B)
So, the properties and formulas of quadrilaterals that are used in this question:
Area of the Rectangle = Length × Width
So, we used only a specific property to find the answer.
2. Find the Rest of the Angles of a Parallelogram if one Angle is 80°?
For a parallelogram ABCD, as we know the properties:
Opposite angles are equal.
Opposite sides are equal and parallel.
Diagonals bisect each other.
The summation of any two adjacent angles = 180 degrees.
So, the angles opposite to the provided 80° angle will likewise be 80°.
Like we know, know that the Sum of angles of any quadrilateral = 360°.
So, if ∠A = ∠C = 80° then,
Sum of ∠A, ∠B, ∠C, ∠D = 360°
Also, ∠B = ∠D
Sum of 80°, ∠B, 80°, ∠D = 360°
Or, ∠B +∠ D = 200°
Hence, ∠B = ∠D = 100°
Now, we found all the angles of the quadrilateral, which are:
3. Why are the NCERT Solutions for Class 8 Maths Chapter 3 important?
The questions included in NCERT Solutions for Chapter 3 of Class 8 Maths are important not only for the exams but also for the overall understanding of quadrilaterals. These questions have been answered by expert teachers in the subject as per the NCERT (CBSE) guidelines. As the students answer the exercises, they will grasp the topic more comfortably and in a better manner.
4. What are the main topics covered in NCERT Solutions for Class 8 Maths Chapter 3?
All the topics of the syllabus of Class 8 Maths Chapter 3 have been dealt with in detail in the NCERT Solutions by Vedantu. The chapter is Understanding Quadrilaterals and has four exercises. All the important topics in Quadrilaterals have also been carefully covered. Students can also refer to the important questions section to get a good idea about the kind of questions usually asked in the exam.
5. Do I need to practice all the questions provided in the NCERT Solutions Class 8 Maths “Understanding Quadrilaterals”?
It helps to solve as many questions as possible because Mathematics is all about practice. If you solve all the practice questions and exercises given in NCERT Solutions for Class 8 Maths, you will be able to score very well in your exams comfortably. This will also help you understand the concepts clearly and allow you to apply them logically in the questions.
6. What are the most important concepts that I need to remember in Class 8 Maths Chapter 3?
For Class 8 Maths Chapter 3, you must remember the definition, characteristics and properties of all the quadrilaterals prescribed in the syllabus, namely, parallelogram, rhombus, rectangle, square, kite, and trapezium. Also know the properties of their angles and diagonals. Regular practise will help students learn the chapter easily.
7. Is Class 8 Maths Chapter 3 Easy?
Class 8 chapter 3 of Maths is a really interesting but critical topic. It's important not only for the Class 8 exams but also for understanding future concepts in higher classes. So, to stay focused and get a good grip of all concepts, it is advisable to download the NCERT Solutions for Class 8 Maths from the Vedantu website or from the Vedantu app at free of cost. This will help the students to clear out any doubts and allow them to excel in the exams.
8. In Maths Class 8 Chapter 3, how are quadrilaterals used in everyday life?
Quadrilaterals are everywhere! Here are some examples:
Shapes in your house: Doors, windows, tabletops, picture frames, book covers, even slices of bread are quadrilaterals (mostly rectangles).
Construction and design: Architects use rectangles and squares for walls, floors, and windows. Roads and bridges often involve trapezoids and other quadrilaterals for support.
Everyday objects: Stop signs, traffic signals, and many sports fields (like baseball diamonds) are quadrilaterals.
9. How many quadrilaterals are there in Class 8 Chapter 3 Maths?
There are many types of quadrilaterals mentioned in Class 8 Understanding Quadrilaterals, but some of the most common include:
Rectangle (all four angles are 90 degrees, opposite sides are equal and parallel)
Square (a special rectangle with all sides equal)
Parallelogram (opposite sides are parallel)
Rhombus (all four sides are equal)
Trapezoid (one pair of parallel sides)
10. What are real examples of quadrilaterals in Class 8 Chapter 3 Maths?
Some real examples of quadrilaterals are:
Rectangle: Doorway, window pane, sheet of paper, tabletop, chocolate bar, playing card (most common)
Square: Dice, coaster, napkin, wall tiles (when all sides are equal)
Parallelogram: Textbook cover, kite (when opposite sides are parallel), solar panel
Rhombus: Traffic warning sign (diamond shape with all sides equal)
Trapezoid: Slice of pizza, roof truss (one pair of parallel sides)
Irregular Quadrilateral: Flag (many flags like the US flag are not perfectly symmetrical quadrilaterals)
11. How do you identify a quadrilateral in Maths Class 8 Chapter 3?
A quadrilateral has the following properties:
Four straight sides
Four angles (interior angles add up to 360 degrees)
Four vertices (corners where two sides meet)
NCERT Solutions for Class 8 Maths
Ncert solutions for class 8.
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Case Study Questions for Class 8 Maths
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Table of Contents
Here in this article, we are providing case study questions for class 8 maths.
Maths Class 8 Chapter List
Latest chapter list (202324).
Chapter 1 Rational Numbers Chapter 2 Linear Equations in One Variable Chapter 3 Understanding Quadrilaterals Chapter 4 Data Handling Chapter 5 Squares and Square Roots Chapter 6 Cubes and Cube Roots Chapter 7 Comparing Quantities Chapter 8 Algebraic Expressions and Identities Chapter 9 Mensuration Chapter 10 Exponents and Powers Chapter 11 Direct and Indirect proportions Chapter 12 Factorisation Chapter 13 Introduction to Graphs
Old Chapter List
Chapter 1 Rational Numbers Chapter 2 Linear Equations in One Variable Chapter 3 Understanding Quadrilaterals Chapter 4 Practical Geometry Chapter 5 Data Handling Chapter 6 Squares and Square Roots Chapter 7 Cubes and Cube Roots Chapter 8 Comparing Quantities Chapter 9 Algebraic Expressions and Identities Chapter 10 Visualising Solid Shapes Chapter 11 Mensuration Chapter 12 Exponents and Powers Chapter 13 Direct and Indirect proportions Chapter 14 Factorisation Chapter 15 Introduction to Graphs Chapter 16 Playing with Numbers
Tips for Answering Case Study Questions for Class 8 Maths in Exam
1. Comprehensive Reading for Context: Prioritize a thorough understanding of the provided case study. Absorb the contextual details and data meticulously to establish a strong foundation for your solution.
2. Relevance Identification: Pinpoint pertinent mathematical concepts applicable to the case study. By doing so, you can streamline your thinking process and apply appropriate methods with precision.
3. Deconstruction of the Problem: Break down the complex problem into manageable components or steps. This approach enhances clarity and facilitates organized problemsolving.
4. Highlighting Key Data: Emphasize critical information and data supplied within the case study. This practice aids quick referencing during the problemsolving process.
5. Application of Formulas: Leverage pertinent mathematical formulas, theorems, and principles to solve the case study. Accuracy in formula selection and unit usage is paramount.
6. Transparent Workflow Display: Document your solution with transparency, showcasing intermediate calculations and steps taken. This not only helps track progress but also offers insight into your analytical process.
7. Variable Labeling and Definition: For introduced variables or unknowns, offer clear labels and definitions. This eliminates ambiguity and reinforces a structured solution approach.
8. Step Explanation: Accompany each step with an explanatory note. This reinforces your grasp of concepts and demonstrates effective application.
9. Realistic Application: When the case study pertains to realworld scenarios, infuse practical reasoning and logic into your solution. This ensures alignment with reallife implications.
10. Thorough Answer Review: Postsolving, meticulously review your answer for accuracy and coherence. Assess its compatibility with the case study’s context.
11. Solution Recap: Before submission, revisit your solution to guarantee comprehensive coverage of the problem and a wellorganized response.
12. Previous Case Study Practice: Boost your confidence by practicing with past case study questions from exams or textbooks. This familiarity enhances your readiness for the question format.
13. Efficient Time Management: Strategically allocate time for each case study question based on its complexity and the overall exam duration.
14. Maintain Composure and Confidence: Approach questions with poise and selfassurance. Your preparation equips you to conquer the challenges presented.
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NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12
NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals
October 4, 2019 by Sastry CBSE
 Class 8 Maths Understanding Quadrilaterals Exercise 3.1
 Class 8 Maths Understanding Quadrilaterals Exercise 3.2
 Class 8 Maths Understanding Quadrilaterals Exercise 3.3
 Class 8 Maths Understanding Quadrilaterals Exercise 3.4
 Understanding Quadrilaterals Class 8 Extra Questions
NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.1
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NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals
Class 8 Maths Chapter 3 Try These Class 8 Maths Exercise 3.1 Solutions Class 8 Maths Exercise 3.2 Solutions Class 8 Maths Exercise 3.3 Solutions Class 8 Maths Exercise 3.4 Solutions
The NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals and Class 8 Maths Chapter 3 Try These Solutions in English and Hindi Medium modified and updated for session 202425. As per the revised syllabus, the of exercises in chapter 3 of class 8th Maths are four only, which are given here with solutions.
8th Maths Chapter 3 Solutions in English Medium
 Class 8 Maths Chapter 3 Try These Solutions
 Class 8 Maths Exercise 3.1 in English
 Class 8 Maths Exercise 3.2 in English
 Class 8 Maths Exercise 3.3 in English
 Class 8 Maths Exercise 3.4 in English
Class: 8  Mathematics 
Chapter 3:  Understanding Quadrilaterals 
Number of Exercises:  4 (Four) 
Content:  Exercises Solution 
Mode of Content:  Online Images, Text and Videos 
Session:  202425 
Medium:  Hindi and English Medium 
8th Maths Chapter 3 Solutions in Hindi Medium
 Class 8 Maths Exercise 3.1 in Hindi
 Class 8 Maths Exercise 3.2 in Hindi
 Class 8 Maths Exercise 3.3 in Hindi
 Class 8 Maths Exercise 3.4 in Hindi
 Class 8 Maths Chapter 3 NCERT Book
 Class 8th Maths Solutions Page
 Class 8 all Subjects Solutions
Class VIII Mathematics chapter 3 is based on latest NCERT Books for current year, useful to all students. Download Prashnavali 3.1, Prashnavali 3.2, Prashnavali 3.3 and Prashnavali 3.4 in Hindi Medium and Exercise 3.1, Exercise 3.2, Exercise 3.3 and Exercise 3.4 in English Medium free to download in PDF format. These NCERT Solutions are applicable for all board using NCERT Books for their academic session. All the solutions are updated on the basis of latest CBSE Curriculum 202425. This chapter is based on closed figures like triangles, quadrilaterals and other polygons.
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More about Class 8 Maths Chapter 3
What is a regular polygon state the name of a regular polygon of: (a) 3 sides (b) 4 sides (c) 6 sides.
A regular polygon: A polygon having all sides of equal length and the interior angles of equal size is known as regular polygon. (i) 3 sides Polygon having three sides is called a triangle.
(ii) 4 sides Polygon having four sides is called a quadrilateral.
(iii) 6 sides Polygon having six sides is called a hexagon.
How many sides does a regular polygon have, if the measure of an exterior angle is 24 degree?
Let number of sides be n. Sum of exterior angles of a regular polygon = 360 Number of sides = n = 360/24 = 15 Hence, the regular polygon has 15 sides.
Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.1, Exercise 3.2, Exercise 3.3 and Exercise 3.4 in English Medium as well as Hindi Medium are given below to download in PDF form. Download Class 8 Maths App in English for offline use and Kaksha 8 Ganit App in Hindi Medium for the academic session 202425.
In Chapter 3 Understanding Quadrilaterals, we will learn about plane surface and plane figures, different types of polygons like Triangles, Quadrilaterals, Pentagon, Hexagon, Heptagon, etc. Number of diagonals in each polygon, a brief description about CONVEX and CONCAVE polygons. Regular and irregular polygons having 3, 4, 5 and 6. Angle sum property of triangle and the questions based on the same fact to find the missing term. Questions based on sum of measure of exterior angles and questions related to Trapezium, Parallelogram, Kite to find the missing side or angle.
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There are exercises based on the properties of different types of quadrilaterals in this lesson. Just as all the sides and all the angles of the square are equal, the opposite sides of the rectangle are equal, the opposite angle and the diagonal are also equal. Quadrilateral and rectangle have the difference of angles and diagonals. If the diagonals of a rhombus are equal, then it is square. Every angle of a square is a right angle. All the questions in chapter 3 of 8th Maths are mainly based on the properties of polygons.
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NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals
Ncert solutions for class 8 maths chapter 3 understanding quadrilaterals pdf download.
Study Materials for Class 8 Maths Chapter 3 Understanding Quadrilaterals 

 Exercise 3.1 Chapter 3 Class 8 Maths NCERT Solutions
 Exercise 3.2 Chapter 3 Class 8 Maths NCERT Solutions
 Exercise 3.3 Chapter 3 Class 8 Maths NCERT Solutions
 Exercise 3.4 Chapter 3 Class 8 Maths NCERT Solutions
NCERT Solutions for Class 8 Maths Chapters:
How many exercises in Chapter 3 Understanding Quadrilaterals
What is equilateral triangle, in a quadrilateral abcd, the angles a, b, c and d are in the ratio 1 : 2 : 3 : 4. find the measure of each angle of the quadrilateral., the interior angle of a regular is 108°. find the number of sides of the polygon., contact form.
Extra Questions – Class 8 Maths Chapter 3 Understanding Quadrilaterals
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In Class 8 Mathematics, Chapter 3 focuses on Understanding Quadrilaterals , which are shapes with four sides. To help students practice and understand this chapter better, extra questions have been created. These extra questions are like bonus exercises that give students more practice and help them explore quadrilaterals in more detail.
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The Class 8 Maths Chapter 3 Extra Questions cover various aspects of quadrilaterals, such as their properties, types, and how they are used. By solving these extra questions, students can improve their knowledge and skills in working with quadrilaterals. The questions also come with solutions, making it easier for students to check their answers and learn from their mistakes.
These extra questions provide a fun and engaging way for students to learn more about quadrilaterals. They can practice identifying different types of quadrilaterals, understanding their features, and solving problems related to them. By working through these extra questions , students can boost their confidence in math and be better prepared for tests and exams.
Class 8 Maths Chapter 3 Extra Questions with Solutions – Understanding Quadrilaterals
For Class 8 students learning about quadrilaterals, extra questions with solutions are a helpful tool. These extra questions from chapter 3 class 8 maths cover different aspects of quadrilaterals and come with answers to check your work. By practicing with these questions, students can improve their understanding of quadrilaterals and how to solve related problems.
The solutions provided not only give the correct answers but also explain how to solve each question step by step. This resource helps students identify areas where they need more practice and enhances their overall grasp of quadrilaterals in a clear and straightforward manner.
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Very Short Answer Type
3x + 5 = 5x – 1
⇒ 3x – 5x = 1 – 5
x + y + z = 360°
(x + 10)° + (3x + 5)° + (2x + 15)° = 180°
⇒ x + 10 + 3x + 5 + 2x + 15 = 180
⇒ 6x + 30 = 180
⇒ 6x = 180 – 30
Question 4. The angles of a quadrilateral are in the ratio of 2 : 3 : 5 : 8. Find the measure of each angle.
Solution: The sum of a quadrilateral’s internal angles equals 360°.
Let the quadrilateral’s angles be 2x°, 3x°, 5x°, and 8x°.
2x + 3x + 5x + 8x = 360°
⇒ 18x = 360°
Hence the angles are
2 × 20 = 40°,
3 × 20 = 60°,
5 × 20 = 100°
and 8 × 20 = 160°.
Question 5. Find the measure of an interior angle of a regular polygon of 9 sides.
Solution: Measure of an interior angle of a regular polygon
Question 6. Length and breadth of a rectangular wire are 9 cm and 7 cm respectively. If the wire is bent into a square, find the length of its side.
Side of the square = \(\frac { 32 }{ 4 }\) = 8 cm.
Hence, the length of the side of square = 8 cm.
Then m∠S = 110° (Opposite angles are equal)
Since ∠P and ∠Q are supplementary.
Then m∠P + m∠Q = 180°
⇒ m∠P + 110° = 180°
⇒ m∠P = 180° – 110° = 70°
⇒ m∠P = m∠R = 70° (Opposite angles)
Hence m∠P = 70, m∠R = 70°
and m∠S = 110°
Since the diagonals of a rhombus bisect each other
z = 5 and y = 12
Hence, x = 13 cm, y = 12 cm and z = 5 cm.
⇒ 125° + ∠D = 180°
⇒ ∠D = 180° – 125°
⇒ 125° = y + 56°
⇒ y = 125° – 56°
∠z + ∠y = 180° (Adjacent angles)
⇒ ∠z + 69° = 180°
⇒ ∠z = 180° – 69° = 111°
Hence the angles x = 55°, y = 69° and z = 111°
Now, sum of exterior angles of a polygon is 360°, therefore,
x + 60° + 90° + 90° + 40° = 360°
⇒ x + 280° = 360°
Short Answer Type
3y + 2y – 5 = 180°
⇒ 5y – 5 = 180°
⇒ 5y = 180 + 5°
⇒ 5y = 185°
3y = 3x + 3
⇒ 3 × 37 = 3x + 3
⇒ 111 = 3x + 3
⇒ 111 – 3 = 3x
Hence, x = 36° and y – 37°.
∠ABC = ∠ADC (Opposite angles of a rhombus)
∠ADC = 126°
∠ODC = \(\frac { 1 }{ 2 }\) × ∠ADC (Diagonal of rhombus bisects the respective angles)
⇒ ∠ODC = \(\frac { 1 }{ 2 }\) × 126° = 63°
⇒ ∠DOC = 90° (Diagonals of a rhombus bisect each other at 90°)
∠OCD + ∠ODC + ∠DOC = 180° (Angle sum property)
⇒ ∠OCD + 63° + 90° = 180°
⇒ ∠OCD + 153° = 180°
⇒ ∠OCD = 180° – 153° = 27°
Hence ∠OCD or ∠ACD = 27°
x + 8 = 16 – x
⇒ x + x = 16 – 8
Similarly, OB = OD
5y + 4 = 2y + 13
Hence, x = 4 and y = 3
The Class 8 HOTS Course enhances critical thinking and problemsolving skills through engaging activities and advanced learning techniques, ensuring academic excellence. Class 8 HOTS Course
Question 15. Write true and false against each of the given statements.
(a) Diagonals of a rhombus are equal.
(b) Diagonals of rectangles are equal.
(c) Kite is a parallelogram.
(d) Sum of the interior angles of a triangle is 180°.
(e) A trapezium is a parallelogram.
(f) Sum of all the exterior angles of a polygon is 360°.
(g) Diagonals of a rectangle are perpendicular to each other.
(h) Triangle is possible with angles 60°, 80° and 100°.
(i) In a parallelogram, the opposite sides are equal.
Question 16. The sides AB and CD of a quadrilateral ABCD are extended to points P and Q respectively. Is ∠ADQ + ∠CBP = ∠A + ∠C? Give reason.
Join AC, then
∠CBP + ∠ADQ = ∠BCA + ∠BAC + ∠ACD + ∠DAC
= (∠BCA + ∠ACD) + (∠BAC + ∠DAC)
Higher Order Thinking Skills (HOTS)
Let AD = x cm
diagonal BD = 3x cm
In rightangled triangle DAB,
AD 2 + AB 2 = BD 2 (Using Pythagoras Theorem)
x 2 + AB 2 = (3x) 2
⇒ x 2 + AB 2 = 9x 2
⇒ AB 2 = 9x 2 – x 2
⇒ AB 2 = 8x 2
⇒ AB = √8x = 2√2x
Required ratio of AB : AD = 2√2x : x = 2√2 : 1
Question 18. If AM and CN are perpendiculars on the diagonal BD of a parallelogram ABCD, Is ∆AMD = ∆CNB? Give reason. (NCERT Exemplar)
AD = BC (opposite sides of parallelogram)
∠AMB = ∠CNB = 90°
∠ADM = ∠NBC (AD  BC and BD is transversal.)
So, ∆AMD = ∆CNB (AAS)
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 Chapter 3: Understanding Quadrilaterals
NCERT Solutions for Class 8 Maths Chapter 3  Understanding Quadrilaterals
NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals are provided here, which can be downloaded for free in PDF format. The NCERT Solutions for the chapter Understanding Quadrilaterals are prepared by the mathematics experts at BYJU’S, from an examination point of view. These solutions explain the accurate method of solving problems. By understanding the concepts used in NCERT Solutions for Class 8, students will be able to clear all their doubts related to Understanding Quadrilaterals.
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Access Answers of Maths NCERT Class 8 Chapter 3 – Understanding Quadrilaterals
Exercise 3.1 page: 41.
1. Given here are some figures.
Classify each of them on the basis of the following.
Simple curve (b) Simple closed curve (c) Polygon
(d) Convex polygon (e) Concave polygon
a) Simple curve: 1, 2, 5, 6 and 7
b) Simple closed curve: 1, 2, 5, 6 and 7
c) Polygon: 1 and 2
d) Convex polygon: 2
e) Concave polygon: 1
2. How many diagonals does each of the following have?
a) A convex quadrilateral (b) A regular hexagon (c) A triangle
a) A convex quadrilateral: 2.
b) A regular hexagon: 9.
c) A triangle: 0
3. What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? (Make a nonconvex quadrilateral and try!)
Let ABCD be a convex quadrilateral.
From the figure, we infer that the quadrilateral ABCD is formed by two triangles,
i.e. ΔADC and ΔABC.
Since we know that sum of the interior angles of a triangle is 180°,
the sum of the measures of the angles is 180° + 180° = 360°
Let us take another quadrilateral ABCD which is not convex .
Join BC, such that it divides ABCD into two triangles ΔABC and ΔBCD. In ΔABC,
∠1 + ∠2 + ∠3 = 180° (angle sum property of triangle)
∠4 + ∠5 + ∠6 = 180° (angle sum property of triangle)
∴, ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 180° + 180°
⇒ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 360°
⇒ ∠A + ∠B + ∠C + ∠D = 360°
Thus, this property holds if the quadrilateral is not convex.
4. Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)
What can you say about the angle sum of a convex polygon with number of sides? (a) 7 (b) 8 (c) 10 (d) n
The angle sum of a polygon having side n = (n2)×180°
Here, n = 7
Thus, angle sum = (72)×180° = 5×180° = 900°
Here, n = 8
Thus, angle sum = (82)×180° = 6×180° = 1080°
Here, n = 10
Thus, angle sum = (102)×180° = 8×180° = 1440°
Here, n = n
Thus, angle sum = (n2)×180°
5. What is a regular polygon?
State the name of a regular polygon of
(i) 3 sides (ii) 4 sides (iii) 6 sides
Regular polygon: A polygon having sides of equal length and angles of equal measures is called a regular polygon. A regular polygon is both equilateral and equiangular.
(i) A regular polygon of 3 sides is called an equilateral triangle.
(ii) A regular polygon of 4 sides is called a square.
(iii) A regular polygon of 6 sides is called a regular hexagon.
6. Find the angle measure of x in the following figures.
a) The figure has 4 sides. Hence, it is a quadrilateral. Sum of angles of the quadrilateral = 360°
⇒ 50° + 130° + 120° + x = 360°
⇒ 300° + x = 360°
⇒ x = 360° – 300° = 60°
b) The figure has 4 sides. Hence, it is a quadrilateral. Also, one side is perpendicular forming a right angle.
Sum of angles of the quadrilateral = 360°
⇒ 90° + 70° + 60° + x = 360°
⇒ 220° + x = 360°
⇒ x = 360° – 220° = 140°
c) The figure has 5 sides. Hence, it is a pentagon.
Sum of angles of the pentagon = 540° Two angles at the bottom are a linear pair.
∴, 180° – 70° = 110°
180° – 60° = 120°
⇒ 30° + 110° + 120° + x + x = 540°
⇒ 260° + 2x = 540°
⇒ 2x = 540° – 260° = 280°
⇒ 2x = 280°
d) The figure has 5 equal sides. Hence, it is a regular pentagon. Thus, all its angles are equal.
⇒ x = 540°/5
a) Sum of all angles of triangle = 180°
One side of triangle = 180° (90° + 30°) = 60°
x + 90° = 180° ⇒ x = 180° – 90° = 90°
y + 60° = 180° ⇒ y = 180° – 60° = 120°
z + 30° = 180° ⇒ z = 180° – 30° = 150°
x + y + z = 90° + 120° + 150° = 360°
b) Sum of all angles of quadrilateral = 360°
One side of quadrilateral = 360° (60° + 80° + 120°) = 360° – 260° = 100°
x + 120° = 180° ⇒ x = 180° – 120° = 60°
y + 80° = 180° ⇒ y = 180° – 80° = 100°
z + 60° = 180° ⇒ z = 180° – 60° = 120°
w + 100° = 180° ⇒ w = 180° – 100° = 80°
x + y + z + w = 60° + 100° + 120° + 80° = 360°
Exercise 3.2 Page: 44
1. Find x in the following figures.
125° + m = 180° ⇒ m = 180° – 125° = 55° (Linear pair)
125° + n = 180° ⇒ n = 180° – 125° = 55° (Linear pair)
x = m + n (The exterior angle of a triangle is equal to the sum of the two opposite interior angles)
⇒ x = 55° + 55° = 110°
Two interior angles are right angles = 90°
70° + m = 180° ⇒ m = 180° – 70° = 110° (Linear pair)
60° + n = 180° ⇒ n = 180° – 60° = 120° (Linear pair) The figure is having five sides and is a pentagon.
Thus, sum of the angles of a pentagon = 540°
⇒ 90° + 90° + 110° + 120° + y = 540°
⇒ 410° + y = 540° ⇒ y = 540° – 410° = 130°
x + y = 180° (Linear pair)
⇒ x + 130° = 180°
⇒ x = 180° – 130° = 50°
2. Find the measure of each exterior angle of a regular polygon of
(i) 9 sides (ii) 15 sides
Sum of the angles of a regular polygon having side n = (n2)×180°
(i) Sum of the angles of a regular polygon having 9 sides = (92)×180°= 7×180° = 1260°
Each interior angle=1260/9 = 140°
Each exterior angle = 180° – 140° = 40°
Each exterior angle = Sum of exterior angles/Number of angles = 360/9 = 40°
(ii) Sum of angles of a regular polygon having side 15 = (152)×180°
= 13×180° = 2340°
Each interior angle = 2340/15 = 156°
Each exterior angle = 180° – 156° = 24°
Each exterior angle = sum of exterior angles/Number of angles = 360/15 = 24°
3. How many sides does a regular polygon have if the measure of an exterior angle is 24°?
Each exterior angle = sum of exterior angles/Number of angles
24°= 360/ Number of sides
⇒ Number of sides = 360/24 = 15
Thus, the regular polygon has 15 sides.
4. How many sides does a regular polygon have if each of its interior angles is 165°?
Interior angle = 165°
Exterior angle = 180° – 165° = 15°
Number of sides = sum of exterior angles/exterior angles
⇒ Number of sides = 360/15 = 24
Thus, the regular polygon has 24 sides.
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?
a) Exterior angle = 22°
Number of sides = sum of exterior angles/ exterior angle
⇒ Number of sides = 360/22 = 16.36
No, we can’t have a regular polygon with each exterior angle as 22° as it is not a divisor of 360.
b) Interior angle = 22°
Exterior angle = 180° – 22°= 158°
No, we can’t have a regular polygon with each exterior angle as 158° as it is not a divisor of 360.
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?
a) An equilateral triangle is the regular polygon (with 3 sides) having the least possible minimum interior angle because a regular polygon can be constructed with minimum 3 sides.
Since the sum of interior angles of a triangle = 180°
Each interior angle = 180/3 = 60°
b) An equilateral triangle is the regular polygon (with 3 sides) having the maximum exterior angle because the regular polygon with the least number of sides has the maximum exterior angle possible. Maximum exterior possible = 180 – 60° = 120°
Exercise 3.3 Page: 50
1. Given a parallelogram ABCD. Complete each statement along with the definition or property used.
(i) AD = …… (ii) ∠DCB = ……
(iii) OC = …… (iv) m ∠DAB + m ∠CDA = ……
(i) AD = BC (Opposite sides of a parallelogram are equal)
(ii) ∠DCB = ∠DAB (Opposite angles of a parallelogram are equal)
(iii) OC = OA (Diagonals of a parallelogram are equal)
(iv) m ∠DAB + m ∠CDA = 180°
2. Consider the following parallelograms. Find the values of the unknown x, y, z
y = 100° (opposite angles of a parallelogram)
x + 100° = 180° (adjacent angles of a parallelogram)
⇒ x = 180° – 100° = 80°
x = z = 80° (opposite angles of a parallelogram)
∴, x = 80°, y = 100° and z = 80°
50° + x = 180° ⇒ x = 180° – 50° = 130° (adjacent angles of a parallelogram) x = y = 130° (opposite angles of a parallelogram)
x = z = 130° (corresponding angle)
x = 90° (vertical opposite angles)
x + y + 30° = 180° (angle sum property of a triangle)
⇒ 90° + y + 30° = 180°
⇒ y = 180° – 120° = 60°
also, y = z = 60° (alternate angles)
z = 80° (corresponding angle) z = y = 80° (alternate angles) x + y = 180° (adjacent angles)
⇒ x + 80° = 180° ⇒ x = 180° – 80° = 100°
y = 112o z = 28o
3. Can a quadrilateral ABCD be a parallelogram if (i) ∠D + ∠B = 180°?
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
(iii)∠A = 70° and ∠C = 65°?
(i) Yes, a quadrilateral ABCD can be a parallelogram if ∠D + ∠B = 180° but it should also fulfil some conditions, which are:
(a) The sum of the adjacent angles should be 180°.
(b) Opposite angles must be equal.
(ii) No, opposite sides should be of the same length. Here, AD ≠ BC
(iii) No, opposite angles should be of the same measures. ∠A ≠ ∠C
4. Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
ABCD is a figure of quadrilateral that is not a parallelogram but has exactly two opposite angles, that is, ∠B = ∠D of equal measure. It is not a parallelogram because ∠A ≠ ∠C.
5. The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.
Let the measures of two adjacent angles ∠A and ∠B be 3x and 2x, respectively in
parallelogram ABCD.
∠A + ∠B = 180°
⇒ 3x + 2x = 180°
⇒ 5x = 180°
We know that opposite sides of a parallelogram are equal.
∠A = ∠C = 3x = 3 × 36° = 108°
∠B = ∠D = 2x = 2 × 36° = 72°
6. Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Let ABCD be a parallelogram.
Sum of adjacent angles of a parallelogram = 180°
⇒ 2∠A = 180°
also, 90° + ∠B = 180°
⇒ ∠B = 180° – 90° = 90°
∠A = ∠C = 90°
∠B = ∠D = 90
7. The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.
y = 40° (alternate interior angle)
∠P = 70° (alternate interior angle)
∠P = ∠H = 70° (opposite angles of a parallelogram)
z = ∠H – 40°= 70° – 40° = 30°
∠H + x = 180°
⇒ 70° + x = 180°
⇒ x = 180° – 70° = 110°
8. The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)
(i) SG = NU and SN = GU (opposite sides of a parallelogram are equal) 3x = 18
3y – 1 = 26
⇒ 3y = 26 + 1
⇒ y = 27/3=9
x = 6 and y = 9
(ii) 20 = y + 7 and 16 = x + y (diagonals of a parallelogram bisect each other) y + 7 = 20
⇒ y = 20 – 7 = 13 and,
⇒ x + 13 = 16
⇒ x = 16 – 13 = 3
x = 3 and y = 13
9. In the above figure both RISK and CLUE are parallelograms. Find the value of x.
∠K + ∠R = 180° (adjacent angles of a parallelogram are supplementary)
⇒ 120° + ∠R = 180°
⇒ ∠R = 180° – 120° = 60°
also, ∠R = ∠SIL (corresponding angles)
⇒ ∠SIL = 60°
also, ∠ECR = ∠L = 70° (corresponding angles) x + 60° + 70° = 180° (angle sum of a triangle)
10. Explain how this figure is a trapezium. Which of its two sides are parallel? (Fig 3.32)
When a transversal line intersects two lines in such a way that the sum of the adjacent angles on the same side of transversal is 180°, then the lines are parallel to each other. Here, ∠M + ∠L = 100° + 80° = 180°
Thus, MN  LK
As the quadrilateral KLMN has one pair of parallel lines, it is a trapezium. MN and LK are parallel lines.
11. Find m∠C in Fig 3.33 if AB  DC.
m∠C + m∠B = 180° (angles on the same side of transversal)
⇒ m∠C + 120° = 180°
⇒ m∠C = 180° 120° = 60°
12. Find the measure of ∠P and ∠S if SP  RQ ? in Fig 3.34. (If you find m∠R, is there more than one method to find m∠P?)
∠P + ∠Q = 180° (angles on the same side of transversal)
⇒ ∠P + 130° = 180°
⇒ ∠P = 180° – 130° = 50°
also, ∠R + ∠S = 180° (angles on the same side of transversal)
⇒ 90° + ∠S = 180°
⇒ ∠S = 180° – 90° = 90°
Thus, ∠P = 50° and ∠S = 90°
Yes, there are more than one method to find m∠P.
PQRS is a quadrilateral. Sum of measures of all angles is 360°.
Since, we know the measurement of ∠Q, ∠R and ∠S.
∠Q = 130°, ∠R = 90° and ∠S = 90°
∠P + 130° + 90° + 90° = 360°
⇒ ∠P + 310° = 360°
⇒ ∠P = 360° – 310° = 50°
Exercise 3.4 Page: 55
1. State whether True or False.
(a) All rectangles are squares.
(b) All rhombuses are parallelograms.
(c) All squares are rhombuses and also rectangles.
(d) All squares are not parallelograms.
(e) All kites are rhombuses.
(f) All rhombuses are kites.
(g) All parallelograms are trapeziums.
(h) All squares are trapeziums.
Because all squares are rectangles but all rectangles are not squares.
Because all squares are parallelograms as opposite sides are parallel and opposite angles are equal.
Because, for example, the length of the sides of a kite are not of the same length.
2. Identify all the quadrilaterals that have.
(a) four sides of equal length (b) four right angles
(a) Rhombus and square have all four sides of equal length.
(b) Square and rectangle have four right angles.
3. Explain how a square is
(i) a quadrilateral (ii) a parallelogram (iii) a rhombus (iv) a rectangle
(i) Square is a quadrilateral because it has four sides.
(ii) Square is a parallelogram because it’s opposite sides are parallel and opposite angles are equal.
(iii) Square is a rhombus because all the four sides are of equal length and diagonals bisect at right angles.
(iv)Square is a rectangle because each interior angle, of the square, is 90°
4. Name the quadrilaterals whose diagonals.
(i) bisect each other (ii) are perpendicular bisectors of each other (iii) are equal
(i) Parallelogram, Rhombus, Square and Rectangle
(ii) Rhombus and Square
(iii)Rectangle and Square
5. Explain why a rectangle is a convex quadrilateral.
A rectangle is a convex quadrilateral because both of its diagonals lie inside the rectangle.
6. ABC is a rightangled triangle and O is the midpoint of the side opposite to the right angle. Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you).
AD and DC are drawn so that AD  BC and AB  DC
AD = BC and AB = DC
ABCD is a rectangle as opposite sides are equal and parallel to each other and all the interior angles are of 90°.
In a rectangle, diagonals are of equal length and also bisect each other.
Hence, AO = OC = BO = OD
Thus, O is equidistant from A, B and C.
NCERT Solutions for Class 8 Maths Chapter 3 – Understanding Quadrilaterals
The major concepts covered in this chapter include: 3.1 Introduction 3.2 Polygons 3.2.1 Classification of Polygons 3.2.2 Diagonals 3.2.3 Convex and Concave Polygons 3.2.4 Regular and Irregular Polygons 3.2.5 Angle sum property 3.3 Sum of the Measures of the Exterior Angles of a Polygon 3.4 Kinds of Quadrilaterals 3.4.1 Trapezium 3.4.2 Kite 3.4.3 Parallelogram 3.4.4 Elements of a parallelogram 3.4.5 Angles of parallelogram 3.4.6 Diagonals of a parallelogram 3.5 Some special parallelograms 3.5.1 Rhombus 3.5.2 A rectangle 3.5.3 A square Exercise 3.1 Solutions 7 Questions (1 Long Answer Question, 6 Short Answer Questions) Exercise 3.2 Solutions 6 Questions (6 Short Answer Questions) Exercise 3.3 Solutions 12 Questions (6 Long Answer Questions, 6 Short Answer Questions) Exercise 3.4 Solutions 6 Questions (1 Long Answer Question, 5 Short Answer Questions)
Chapter 3 of NCERT Solutions for Class 8 Maths helps you understand the fundamental concepts related to quadrilaterals. From explanations of quadrilaterals to different types of quadrilaterals, the chapter mainly discusses the following concepts: 1. Parallelogram: A quadrilateral with each pair of opposite sides parallel. Properties: Opposite sides are equal. Opposite angles are equal. Diagonals bisect one another 2. Rhombus: A parallelogram with sides of equal length. Properties: All the properties of a parallelogram. Diagonals are perpendicular to each other 3. Rectangle: A parallelogram with a right angle. Properties: All the properties of a parallelogram. Each of the angles is a right angle. Diagonals are equal. 4. Square: A rectangle with sides of equal length. Properties: All the properties of a parallelogram, rhombus and a rectangle. 5. Kite: A quadrilateral with exactly two pairs of equal consecutive sides Properties: The diagonals are perpendicular to one another. One of the diagonals bisects the other.
Learning the chapter Understanding Quadrilaterals enables the students to understand the following:
 Properties of quadrilaterals: Sum of angles of a quadrilateral is equal to 360 o
 Properties of parallelogram: Opposite sides of a parallelogram are equal. Opposite angles of a parallelogram are equal.
 Diagonals of a parallelogram bisect each other.
 Diagonals of a rectangle are equal and bisect each other.
 Diagonals of a rhombus bisect each other at right angles.
 Diagonals of a square are equal and bisect each other at right angles.
Disclaimer:
Dropped Topics – 3.1 Introduction, 3.2 Polygons, 3.2.1 Classification of polygons, 3.2.2 Diagonals and 3.2.5 Angle sum property.
Frequently Asked Questions on NCERT Solutions for Class 8 Maths Chapter 3
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Home » Understanding Quadrilaterals Class 8 MCQs with Answers – Mathematics Chapter 3 – Free PDF
Understanding Quadrilaterals Class 8 MCQs with Answers – Mathematics Chapter 3 – Free PDF
Update on: 05 Apr 2024, 04:05 PM
NCERT class 8 mathematics chapter 3 – “Understanding Quadrilaterals” teaches us about quadrilaterals, which are a closed figure with four sides and four angles. They are of various kinds and have different properties. Practice multiplechoice questions (MCQs) for the CBSE exams to prepare for the objective questions. We have provided Class 8 MCQ Questions on “Understanding Quadrilaterals” paired with comprehensive explanations. CBSE is emphasizing the role of MCQs as they assist in understanding the concepts completely.
As compared to subjective questions, MCQs are very different so practicing and understanding how to get appropriate answers in MCQs is very essential. To revise the main concepts, students should practice all the MCQs with the answers given. This will also help them familiarize themselves with the kinds of questions that might appear in the board exams.
Recent – Linear Equations in one Variable Class 8 MCQs with Answers – Mathematics Chapter 2 – Free PDF
Also Check – Class 8 Mathematics MCQs with Answers
Topics covered in Understanding Quadrilaterals
 Convex and Concave Polygons
 Regular and Irregular Polygons
 Angle sum Property
 Sum of the Measures of the Exterior Angles of a Polygon
 Kinds of Quadrilaterals
 Parallelogram
 Elements of a Parallelogram
 Angles of a Parallelogram
 Diagonals of a Parallelogram
 A Rectangle
CBSE Class 8 Mathematics Understanding Quadrilaterals MCQs – PDF Download
Answers –
Summary for NCERT class 8 mathematics chapter 3 – Understanding Quadrilaterals
 Opposite sides are equal.
 Opposite angles are equal.
 Diagonals bisect one another.
 All the properties of a parallelogram.
 Diagonals are perpendicular to each other.
 All the properties of parallelogram.
 Each of the angles is a right angle.
 Diagonals are equal.
 All the properties of a parallelogram, rhombus and a rectangle.
 The diagonals are perpendicular to one another.
 One of the diagonals bisect the other.
Best Reference Books for Class 8 Mathematics
 NCERT Textbook + Exemplar Problems Solutions Mathematics
 NCERT at your Fingertips Mathematics
 Foundation Course Mathematics
 PracticecumWorkbook Mathematics
 Integrated Learning Mathematics
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 Heron’s Formula Class 9 Case Study Questions Maths Chapter 10
Last Updated on September 8, 2024 by XAM CONTENT
Hello students, we are providing case study questions for class 9 maths. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 9 maths. In this article, you will find case study questions for CBSE Class 9 Maths Chapter 10 Heron’s Formula. It is a part of Case Study Questions for CBSE Class 9 Maths Series.
Heron’s Formula  
Case Study Questions  
Competency Based Questions  
CBSE  
9  
Maths  
Class 9 Studying Students  
Yes  
Mentioned  
Table of Contents
Case Study Questions on Heron’s Formula
Mayank bought a triangle shape field and wants to grow potato and wheat on his field. He divided his field by joining opposite sides. On the largest park he grew wheat and on the rest part he grew potato. The dimensions of a park are shown in the park.
On the basis of the above information, solve the following questions:
Q 1. Find the length of AC in ΔABC.
Q 2. Find the area of ΔABC.
Q 3. If the cost of ploughing park is ₹5 per cm 2 , then find the total cost of ploughing the park.
1. In right angled $\triangle A B C$, use Pythagoras theorem,
$$ \begin{aligned} A C & =\sqrt{(A B)^2+(B C)^2}=\sqrt{(12)^2+(5)^2} \\ \vdots & =\sqrt{144+25}=\sqrt{169}=13 \mathrm{~m} \end{aligned} $$
Hence, length of $A C$ is 13 m .
2. Area of $\triangle \mathrm{ABC}$
$$ \begin{aligned} & =\frac{1}{2} \times A B \times B C \\ & =\frac{1}{2} \times 12 \times 5=30 \mathrm{~m}^2 \end{aligned} $$
3. Since, the total area of the park $=30 \mathrm{~m}^2$ $\because$ The cost of ploughing the park in $1 \mathrm{~m}^2=5$ $\therefore$ The cost of ploughing the park in $30 \mathrm{~m}^2$
$$ \begin{aligned} & = 5\times 30\\ & =150 \end{aligned} $$
Understanding Heron’s Formula
Area of Triangle: The total space occupied inside the boundary of the triangle is said to be an area of triangle.
Perimeter of Triangle: Sum of lengths of all three sides of a triangle.
$\begin{aligned} 2 s & =a+b+c \\ s & =\frac{a+b+c}{2}\end{aligned}$
Rightangled Triangle: It is a triangle with one right angle.
1. Area $=\frac{1}{2} \times a \times b$ 2. Altitude $=a$ 3. Perimeter$=a+b+\sqrt{a^2+b^2}$
where ‘a’ and ‘b’ are the sides that includes to the right angle.
Isosceles Triangle: Triangle that has two equal sides and corresponding two equal angles.
1. Area $=\frac{b}{4} \sqrt{4 a^2b^2}$ 2. Perimeter $=2 a+b$ 3. Altitude $=\frac{1}{2} \sqrt{4 a^2b^2}$
where ‘a’ is length of two equal sides and ‘b’ is base.
Equilateral Triangle: Triangle with all sides and all angles equal (each being 60°)
1. Area $=\frac{\sqrt{3}}{4} a^2$ 2. Perimeter $=3 a$ 3. Altitude $=\frac{\sqrt{3}}{2} a$
where ‘a’ is side.
Heron‘s Formula: The formula given by Heron about the area of a triangle.
Area of triangle $=\sqrt{s(sa)(sb)(sc)}$
where a, b and c are the sides of triangle and s is its semiperimeter.
Boost your knowledge
(i) The length of longest altitude is the perpendicular distance from the opposite vertex to the smallest side of a triangle. (ii) The length of smallest altitude is the perpendicular distance from the opposite vertex to the largest side of a triangle. (iii) Heron‘s formula is helpful when it is not possible to find the height of the triangle easily. (iv) Heron‘s formula is applicable to all types of triangles whether it is a right triangle or an isosceles or an equilateral triangle
 Circles Class 9 Case Study Questions Maths Chapter 9
 Quadrilaterals Class 9 Case Study Questions Maths Chapter 8
 Triangles Class 9 Case Study Questions Maths Chapter 7
 Lines and Angles Class 9 Case Study Questions Maths Chapter 6
 Introduction to Euclid’s Geometry Class 9 Case Study Questions Maths Chapter 5
 Linear Equations in Two Variables Class 9 Case Study Questions Maths Chapter 4
 Coordinate Geometry Class 9 Case Study Questions Maths Chapter 3
Polynomials Class 9 Case Study Questions Maths Chapter 2
Number systems class 9 case study questions maths chapter 1, topics from which case study questions may be asked.
 Quadrilaterals
 Parallelograms
 Properties of a parallelogram
 Midpoint theorem
 Converse of Midpoint theorem
A trapezium is not a parallelogram (as only one pair of opposite sides is parallel in a trapezium and we require both pairs to be parallel in a parallelogram)
Case study questions from the above given topic may be asked.
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Frequently Asked Questions (FAQs) on Heron’s Formula Case Study
Q1: what is heron’s formula.
A1: Heron’s Formula is used to calculate the area of a triangle when the lengths of all three sides are known. The formula is: Area of triangle $=\sqrt{s(sa)(sb)(sc)}$
Q2: How is Heron’s Formula derived?
A2: Heron’s formula is derived from the general area formula of a triangle. It simplifies the calculation of the area without needing the height. This is particularly useful for triangles where the height is difficult to determine directly. The formula is based on the triangle’s semiperimeter and each of its sides.
Q3: When should Heron’s Formula be used?
A3: Heron’s Formula should be used when the lengths of all three sides of a triangle are given, and the height is unknown. It allows you to find the area of any triangle (scalene, isosceles, or equilateral) as long as you know the sides.
Q4: Can Heron’s Formula be used for rightangled triangles?
A4: Yes, Heron’s Formula can be used for rightangled triangles, although for rightangled triangles, a simpler method involving the base and height (half the product of the base and height) can also be used to find the area.
Q5: What is the significance of the semiperimeter in Heron’s Formula?
A5: The semiperimeter (s) is half the perimeter of the triangle. It acts as a key variable in the formula, simplifying the calculation of the area by incorporating all three sides of the triangle in a single expression.
Q6: Can Heron’s Formula be applied to a quadrilateral?
A6: No, Heron’s Formula specifically applies to triangles. However, if a quadrilateral can be divided into two triangles, the area of each triangle can be found using Heron’s Formula, and the sum of the areas will give the total area of the quadrilateral.
Q7: How is Heron’s Formula applied in reallife problems?
A7: Heron’s Formula is used in fields like civil engineering, architecture, and land surveying where determining the area of irregular land plots or structures is necessary, especially when only the lengths of the boundaries are known.
Q8: Are there any online resources or tools available for practicing Circles case study questions?
A9: We provide case study questions for CBSE Class 9 Maths on our website. Students can visit the website and practice sufficient case study questions and prepare for their exams. If you need more case study questions, then you can visit Physics Gurukul website. they are having a large collection of case study questions for all classes.
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Understanding Quadrilaterals Class 8 Case Study Questions Maths Chapter 3. By XAM CONTENT / April 30, 2024. Reading Time: 8 minutes. Last Updated on September 8, 2024 by XAM CONTENT. Hello students, we are providing case study questions for class 8 maths. Case study questions are the new question format that is introduced in CBSE board.
Here we are providing Case Study questions for Class 8 Maths Chapter 3 Understanding Quadrilaterals. Maths Class 8 Chapter 3 Understanding Quadrilaterals. Maths: CBSE Class 8: Chapter Covered: Class 8 Maths Chapter 3: Topics: Sum of the measures of exterior angles of a Polygon
Students can also reach Important Questions for Class 8 Maths to get important questions for all the chapters here. Class 8 Chapter 3 Important Questions. Questions and answers are given here based on important topics of class 8 Maths Chapter 3. Q.1: A quadrilateral has three acute angles, each measure 80°. What is the measure of the fourth angle?
03_CBSE_CLASS_VIII_CASESTUDY_QUESTIONS (1)  Free download as PDF File (.pdf), Text File (.txt) or read online for free. Sanmesh earns Rs. 150000 per month. He spends 10% on food, 25% on shopping with family, and 20% on education for his two children. The remaining amount he saves. Priya wants to make a square box with area 2916 sqm. Each side of the square box will be the square root of 2916 ...
CBSE Class 8 Maths Important Questions for Understanding Quadrilaterals  Free PDF Download. Important Questions for Class 8 Chapter 3  Understanding Quadrilaterals is based upon the basic concepts of Quadrilaterals and the questions given in the segment by Vedantu will help students prepare for final exams.
Chapter 3 of Class 8 Mathematics is called 'Understanding Quadrilaterals'. A quadrilateral is a closed shape and also a type of polygon that has four sides, four vertices and four angles. It is formed by joining four noncollinear points. The sum of all the interior angles of a quadrilateral is always equal to 360 degrees.
Question 3. In the given figure, find x. Question 4. The angles of a quadrilateral are in the ratio of 2 : 3 : 5 : 8. Find the measure of each angle. Let the angles of the quadrilateral be 2x°, 3x°, 5x° and 8x°. and 8 × 20 = 160°. Question 5. Find the measure of an interior angle of a regular polygon of 9 sides.
Class 8 Maths Chapter 3  Understanding Quadrilaterals  Case Study QuestionIn this video, I have solved case study question of class 8 maths chapter 3 Under...
NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals. Quadrilaterals form a vital shape contributing to geometrical studies. Thus, children need to develop a robust conceptual foundation as they will require it in higher classes for solving more complicated problems and constructing this figure.
Updated for new NCERT Book. Get NCERT Solutions of Chapter 3 Class 8 Understanding Quadrilaterals free at teachoo. Answers to all exercise questions and examples have been solved, with concepts of the chapter explained. In this chapter, we will learn. To learn from the NCERT, click on an exercise or example link below to get started.
The classification of quadrilaterals are dependent on the nature of sides or angles of a quadrilateral and they are as follows: Trapezium. Kite. Parallelogram. Square. Rectangle. Rhombus. The figure given below represents the properties of different quadrilaterals.
NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals are provided below. Our solutions covered each questions of the chapter and explains every concept with a clarified explanation. To score good marks in Class 8 Mathematics examination, it is advised to solve questions provided at the end of each chapter in the NCERT book.
Why? 3. The perimeter of a parallelogram is 150 cm. One of its side is greater than the other by 25 cm. Find length of all sides of the parallelogram. 4. Lengths of adjacent sides of a parallelogram is 3 cm and 4 cm. Find its perimeter. 5. In a parallelogram, the ratio of the adjacent sides is 4 : 5 and its perimeter is 72 cm then, find the ...
The Important Questions: Understanding 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.
The NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals covers all the chapter's questions (All Exercises). These NCERT Solutions for Class 8 Maths have been carefully compiled and created in accordance with the most recent CBSE Syllabus 202425 updates. Students can use these NCERT Solutions for Class 8 to reinforce their ...
Tips for Answering Case Study Questions for Class 8 Maths in Exam. 1. Comprehensive Reading for Context: Prioritize a thorough understanding of the provided case study. Absorb the contextual details and data meticulously to establish a strong foundation for your solution. 2.
Ex 3.1 Class 8 Maths Question 5. What is a regular polygon? State the name of a regular polygon of (i) 3 sides (ii) 4 sides (iii) 6 sides Solution: A polygon with equal sides and equal angles is called a regular polygon. (i) Equilateral triangle (ii) Square (iii) Regular Hexagon. Ex 3.1 Class 8 Maths Question 6.
Class 8 Maths Chapter 3 Solutions. Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.1, Exercise 3.2, Exercise 3.3 and Exercise 3.4 in English Medium as well as Hindi Medium are given below to download in PDF form. Download Class 8 Maths App in English for offline use and Kaksha 8 Ganit App in Hindi Medium for the academic session ...
20. The kite has exactly two distinct consecutive pairs of sides of equal length. Answer: A. True. Explanation: A kite is a quadrilateral that has exactly two distinct consecutive pairs of sides of equal length. Class 8 Maths Chapter 3 Understanding Quadrilaterals MCQs Questions are provided online at BYJU'S, with answers. Also, the PDF of ...
MCQs Questions for Class 8 Maths Chapter 3 Understanding Quadrilaterals. Page No: 41. Exercise 3.1. 1. Given here are some figures. Classify each of them on the basis of the following. (a) Simple curve (b) Simple closed curve (c) Polygon. (d) Convex polygon (e) Concave polygon. Answer.
Question 3. In the given figure, find x. Question 4. The angles of a quadrilateral are in the ratio of 2 : 3 : 5 : 8. Find the measure of each angle. Solution: The sum of a quadrilateral's internal angles equals 360°. Let the quadrilateral's angles be 2x°, 3x°, 5x°, and 8x°. and 8 × 20 = 160°. Question 5.
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According to NCERT Solutions for Class 8 Maths Chapter 3, a quadrilateral is a plane figure that has four sides or edges and also has four corners or vertices. Quadrilaterals will typically be of standard shapes with four sides like rectangle, square, trapezoid, and kite or irregular and uncharacterized shapes. Q3.
NCERT class 8 mathematics chapter 3  "Understanding Quadrilaterals" teaches us about quadrilaterals, which are a closed figure with four sides and four angles. They are of various kinds and have different properties. Practice multiplechoice questions (MCQs) for the CBSE exams to prepare for the objective questions. We have provided Class 8 MCQ Questions on…
Reading Time: 8 minutes Last Updated on September 8, 2024 by XAM CONTENT. Hello students, we are providing case study questions for class 9 maths. Case study questions are the new question format that is introduced in CBSE board.