# Important Questions for CBSE Class 8 Maths Chapter 3 – Understanding Quadrilaterals

**Important Questions Class 8 Maths Chapter 3 – Understanding Quadrilaterals**

Maths deals with numbers of various forms, shapes, logic, quantity and arrangements. Maths also teaches us to solve problems based on numerical calculations and find solutions.

Chapter 3 of Class 8 Maths 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 non-collinear 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 two-dimensional. 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 Maths teachers have prepared a variety of NCERT solutions to help students in their studies and exam preparation. Students can refer to our **understanding quadrilaterals class 8 extra questions** to practise exam-oriented 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 Maths Chapter 3.

To ace their exams, students can register on the Extramarks website to access Class 8 Maths Chapter 3 important questions, **CBSE extra questions**, Maths formulas, and much more.

**Get Access to CBSE Class 8 Maths Important Questions 2022-23 with Chapter-Wise Solutions**

You can also find CBSE Class 8 Maths Chapter-by-Chapter Important Questions here:

CBSE Class 8 Maths Important Questions |
||

Sr No. |
Chapters |
Chapters Name |

1 | Chapter 1 | Rational Numbers |

2 | Chapter 2 | Linear Equations in One Variable |

3 | Chapter 3 | Understanding Quadrilaterals |

4 | Chapter 4 | Practical Geometry |

5 | Chapter 5 | Data Handling |

6 | Chapter 6 | Squares and Square Roots |

7 | Chapter 7 | Cubes and Cube Roots |

8 | Chapter 8 | Comparing Quantities |

9 | Chapter 9 | Algebraic Expressions and Identities |

10 | Chapter 10 | Visualising Solid Shapes |

11 | Chapter 11 | Mensuration |

12 | Chapter 12 | Exponents and Powers |

13 | Chapter 13 | Direct and Inverse Proportions |

14 | Chapter 14 | Factorisation |

15 | Chapter 15 | Introduction to Graphs |

16 | Chapter 16 | Playing with Numbers |

**Understanding Quadrilaterals Class 8 Important Questions with Answers**

Mentioned below are some sets of questions and their answers from our Chapter 3 Class 8 Maths 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°

x = 120°

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°

**Question 3: **

**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°

= 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°

**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,

∠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°

**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 = 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°

**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.

∠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°.

**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°

⇒ x = 36°

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

∠A + ∠B = 180°

But given that ∠A = ∠B

Now substituting, we get

∠A + ∠A = 180°

2∠A = 180°

∠A=180/2 = 90°

**Question 16:Triangle ABC is a right-angled 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

=360°/3

=120°

**Benefits of Solving Class 8 Maths Chapter 3 Important Questions**

Math 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 Maths 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 Maths subject teachers have carefully assembled the most important questions Class 8 Maths Chapter 3 by analysing many past exam questions.
- Maths 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 Maths Chapter 3.
- The questions covered in our set of Important Questions Class 8 Maths 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.
- By solving our important questions Class 8 Maths Chapter 3, students will get a gist of how the paper will be prepared. Practising questions similar to the exam questions would help the students gain confidence, perform better in their exams, and eventually ace them.

Extramarks provides comprehensive learning solutions for students from Class 1 to Class 12. We have other study resources on our website, along with important questions and solutions. Students can also click on the below-mentioned links and access some of these resources:

- NCERT books
**CBSE Revision Notes**- CBSE syllabus
- CBSE sample papers
- CBSE previous year’s question papers
- Important formulas
- CBSE extra questions

**Q.1 ****Two adjacent angles of a parallelogram are (3x 4) and (3x + 10). Find the angles of the parallelogram.**

**Marks:**3

**Ans**

Sum of the adjacent angles of a parallelogram = 180^{°}

(3x 4) + (3x + 10) = 180

6x + 6 = 180

6x = 174

x = 29

(3 × 29) 4 = 87 4 = 83

and, 3 ×29 + 10 = 87 + 10 = 97

Hence, four angles of a parallelogram are 83^{°}, 97^{°}, 83^{°}, 97^{°}.

**Q.2 **

$\begin{array}{l}\mathrm{ABCD}\text{}\mathrm{is}\text{}\mathrm{a}\text{}\mathrm{trapezium}\text{}\mathrm{in}\text{}\mathrm{which}\text{}\mathrm{AB}\text{}\mathrm{DC}.\text{}\mathrm{If}\text{}\mathrm{A}\text{}=\mathrm{B}\text{}=\text{}40\xb0,\\ \mathrm{find}\text{}\mathrm{the}\text{}\mathrm{measures}\text{}\mathrm{of}\text{}\mathrm{other}\text{}\mathrm{two}\text{}\mathrm{angles}.\end{array}$

**Marks:**1

**Ans**

$\begin{array}{l}\mathrm{A}\text{}+\mathrm{D}\text{}=\text{}180\xb0\left(\begin{array}{l}\mathrm{Sum}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{angleson}\text{}\mathrm{the}\text{}\mathrm{same}\text{}\mathrm{side}\text{}\\ \mathrm{of}\text{}\mathrm{transversal}\text{}\mathrm{is}\text{}\mathrm{supplementary}\end{array}\right)\\ \text{}\mathrm{D}\text{}=\text{}180\xb0\text{}40\xb0=\text{}140\xb0\\ \mathrm{Similarly},\text{}\mathrm{C}\text{}=\text{}140\xb0\end{array}$

**Q.3 ****Find the measure of each exterior angle of a regular polygon of 9 sides.**

**Marks:**1

**Ans**

$\begin{array}{l}\mathrm{Total}\text{}\mathrm{measure}\text{}\mathrm{of}\text{}\mathrm{all}\text{}\mathrm{exterior}\text{}\mathrm{angles}\text{}=\text{}360\xb0\\ \begin{array}{l}\mathrm{Number}\text{}\mathrm{of}\text{}\mathrm{sides}\text{}=\text{}9.\\ \mathrm{Measure}\text{}\mathrm{of}\text{}\mathrm{each}\text{}\mathrm{exterior}\text{}\mathrm{angle}\text{}=\frac{360}{9}=40\xb0\end{array}\text{}\end{array}$

**Q.4 ****The diagonals AC and BD of rectangle ABCD intersect each other at point O. If OA = 5 cm, find AC and BD.**

**Marks:**2

**Ans**

OA = 5 cm

AC = 2 x OA = 2 x 5 = 10 cm

AC = BD (diagonals of rectangle are of equal length)

Thus, AC = 10 cm

and, BD = 10 cm

**Q.5 ****For which of the folIowing polygon, the sum of the measures of its interior angles is 1080°**

A. Hexagon

B. Octagon

C. Nonagon

D. Pentagon

**Marks:**1

**Ans**

Pentagon

##### CBSE Class 8 Maths Important Questions

##### FAQs (Frequently Asked Questions)

## 1. What can I get from the Extramarks website?

Extramarks is one of the best educational platforms as it has its own archive of educational resources, which assists students in acing their exams. You can get all the NCERT-related material like NCERT solutions, solved exemplar solutions, NCERT-based mock tests, CBSE revision notes, and important questions Class 8 Maths Chapter 3 on the Extramarks website. Apart from this, you can get comprehensive guidance from our subject experts and doubt-clearing sessions once you sign up on our official website for any study resources.

## 2. How can I score well in Class 8 Maths exams?

Maths 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 Maths 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.

## 3. How many types of quadrilaterals are there?

There are six basic types of quadrilaterals.

They are:

- Trapezium
- Parallelogram
- Rectangle
- Rhombus
- Square
- Kite