NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.1

NCERT Solutions for Class 8 Maths Chapter 3 Exercise 3.1 teaches the fundamentals of polygons. The exercise introduces polygons, counting their sides, angles, and diagonals, and how they are classified into categories according to their number of sides. Exercise 3.1 is a foundation in determining properties of quadrilaterals, as it helps solve more involved geometry problems in the material forthcoming. These questions are simple as they are based on a direct concept of polygons, which means they will be useful in the context of CBSE exam and possible revision activities.

The NCERT Solutions for Class 8 Maths Chapter 3 Exercise 3.1 is based on the most recent NCERT syllabus and is prepared in a step by step manner for easy understanding. Practicing these questions will help you out on not only the school tests but to also help build foundation for related concepts in higher classes and other competitive exams like the olympiads.

1.0Download NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.1: Free PDF

Exercise 3.1 is about recognizing and classifying polygons with respect to the number of their sides and angles. The pdf of NCERT Solutions for Class 8 Maths Chapter 3 offers a straightforward explanation for each answer. Download the free PDF from below:

2.0Key Concepts in Exercise 3.1 of Class 8 Maths Chapter 3

This exercise introduces basic ideas related to polygons. Below are the key concepts covered in this exercise.

• Definition of polygons and their types
• Classifying polygons based on the number of sides
• Understanding convex and concave polygons
• Counting sides, vertices, and diagonals of a polygon
• Identifying regular and irregular polygons

3.0NCERT Class 8 Maths Chapter 3: Other Exercises

4.0NCERT Class 8 Maths Chapter 3 Exercise 3.1: Detailed Solutions

• 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 Sol. The classification of the given figures is as under : (a) Simple curve : (1), (2), (5), (6) and (7) (b) Simple closed curve : (1), (2), (5), (6) and (7) (d) Convex polygon : (2) (e) Concave polygon: (1)
• How many diagonals does each of the following have? (a) A convex quadrilateral (b) A regular hexagon (c) A triangle Sol. (a) A convex quadrilateral has two diagonals. (b) A regular hexagon has nine diagonals. (c) A triangle has no diagonal
• What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? Sol. The sum of measures of the angles of a convex quadrilateral is $360^{\circ }$ Yes, this property holds in case if the quadrilateral is not convex.
• Examine the table What can you say about the angle sum of a convex polygon with number of sides? (a) 7 (b) 8 (c) 10 (d) $n$ Sol. From the given table, clearly we observe that the sum of angles (interior angles) of a polygon with $n$ sides $=(n-2)\times 180^{\circ }$. (a) Here, $\mathrm{n}=7$ $\therefore$ The sum of the angles of a polygon of 7 sides $=(7-2)\times 180^{\circ }=5\times 180^{\circ }=900^{\circ }$ (b) Here $\mathrm{n}=8$ $\therefore$ The sum of the angles of a polygon of 8 sides $=(8-2)\times 180^{\circ }=6\times 180^{\circ }=1080^{\circ }$ (c) Here $\mathrm{n}=10$ $\therefore$ The sum of the angles of a polygon of 10 sides $=(10-2)\times 180^{\circ }=8\times 180=1440^{\circ }$ (d) The sum of the angles of a polygon of ' $n$ ' sides $=(n-2)\times 180^{\circ }$
• What is a regular polygon? State the name of a regular polygon of (i) 3 sides (ii) 4 sides (iii) 6 sides Sol. A polygon is said to be a regular polygon, if all its (i) interior angles are equal; (ii) sides are equal; and (iii) exterior angles are equal. The name of a regular polygon of (i) 3 sides is equilaterial triangle. (ii) 4 sides is square. (iii) 6 sides is regular hexagon.
• Find the angle measure x in the following figures:
  
  Sol. (a) Since the sum of angles (interior angles) of a quadrilateral is $360^{\circ }$. $\therefore \mathrm{x}+120^{\circ }+130^{\circ }+50^{\circ }=360^{\circ }$ $\Rightarrow \mathrm{x}+300^{\circ }=360^{\circ }$ $\Rightarrow \mathrm{x}=360^{\circ }-300^{\circ }=60^{\circ }$ (b) Since the sum of angles (interior angles) of a quadrilateral is $360^{\circ }$ $\therefore \mathrm{x}+70^{\circ }+60^{\circ }+90^{\circ }=360^{\circ }$ $\Rightarrow \mathrm{x}+220^{\circ }=360^{\circ }$ $\Rightarrow \mathrm{x}=360^{\circ }-220^{\circ }=140^{\circ }$ (c) Since the sum of angles (interior angles) of a pentagon is $540^{\circ }$. Let its fourth and fifth angle be y & z. $y=180^{\circ }-70^{\circ }=110^{\circ }(\because$ Linear pair $)$ $\mathrm{z}=180^{\circ }-60^{\circ }=120^{\circ }(\because$ Linear pair $)$ Now, $30^{\circ }+\mathrm{x}+\mathrm{x}+110^{\circ }+120^{\circ }=540^{\circ }$ $\Rightarrow 260+2\mathrm{x}=540^{\circ }$ $\Rightarrow 2\mathrm{x}=540^{\circ }-260^{\circ }$ $\Rightarrow 2\mathrm{x}=280^{\circ }\Rightarrow \mathrm{x}=140^{\circ }$ (d) Since the sum of angles (interior angles) of a pentagon is $540^{\circ }$ Here, its a regular pentagon. So all 5 angles are equal. $\therefore$ each angle $=\frac{540^{\circ }}{5}=108^{\circ }$
• (a) Find $x+y+z$
  
  (b) Find $\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{w}$
  
  Sol. (a) Since the sum of angles of a triangle is $180^{\circ }$.
  
  $\therefore \mathrm{m}\angle 1+30^{\circ }+90^{\circ }=180^{\circ }$ $\Rightarrow \mathrm{m}\angle 1+120^{\circ }=180^{\circ }$ $\Rightarrow \mathrm{m}\angle 1=180^{\circ }-120^{\circ }=60^{\circ }$ Also, $\mathrm{x}+90^{\circ }=180^{\circ }$ [Linear pair] $\Rightarrow \mathrm{x}=180^{\circ }-90^{\circ }=90^{\circ }$ and, $\mathrm{y}+\mathrm{m}\angle 1=180^{\circ }$ $\left[\because \mathrm{m}\angle 1=60^{\circ }\right]$ $\Rightarrow y+60^{\circ }=180^{\circ }$ $\Rightarrow y=180^{\circ }-60^{\circ }=120^{\circ }$ and $\mathrm{z}+30^{\circ }=180^{\circ }$ [Linear pair] $\Rightarrow \mathrm{z}=180^{\circ }-30^{\circ }=150^{\circ }$ Adding (1), (2) and (3), we get $x+y+z=90^{\circ }+120^{\circ }+150^{\circ }=360^{\circ }$ (b) Since the sum of angles of a quadrilateral is $360^{\circ }$.
  
  $\because \mathrm{m}\angle 1+120^{\circ }+80^{\circ }+60^{\circ }=360^{\circ }$ $\Rightarrow \mathrm{m}\angle 1+260^{\circ }=360^{\circ }$ $\Rightarrow \mathrm{m}\angle 1=360^{\circ }-260^{\circ }$ $\Rightarrow \mathrm{m}\angle 1=100^{\circ }$ Also, $x+120^{\circ }=180^{\circ }$ (Linear pair) $\mathrm{x}=180^{\circ }-120^{\circ }$ $\mathrm{x}=60^{\circ }$ and $y+80^{\circ }=180^{\circ }$ (Linear pair) $\Rightarrow \mathrm{y}=180^{\circ }-80^{\circ }$ $\Rightarrow \mathrm{y}=100^{\circ }$ and $\mathrm{z}+60^{\circ }=180^{\circ }$ (Linear pair) $\Rightarrow \mathrm{z}=180^{\circ }-60^{\circ }$ $\Rightarrow \mathrm{z}=120^{\circ }$ and $\mathrm{w}+\mathrm{m}\angle 1=180^{\circ }$ (Linear pair) $\Rightarrow \mathrm{w}+100^{\circ }=180^{\circ }$ $\Rightarrow \mathrm{w}=180^{\circ }-100^{\circ }$ $\Rightarrow \mathrm{w}=80^{\circ }$ Adding (1), (2), (3) and (4), we get $\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{w}=60^{\circ }+100^{\circ }+120^{\circ }+80^{\circ }=360^{\circ }$

5.0Key Features and Benefits of Class 8 Maths Chapter 3 Exercise 3.1

• The exercise explains the topic in a simple manner covering concepts of polygon, side, and angles.
• They follow the current NCERT syllabus of Class 8 Maths.
• Helps in developing good foundations in geometry that's applicable in Maths Olympiad and other competitions.
• Improves visualizing and thinking skills and contributes to problem solving with shapes and figures.
• Helps students gain accuracy, and confidence while solving geometry problems.