Get the most accurate GSEB Solutions for Class 6 Mathematics Chapter 05 પાયાના આકારોની સમજૂતી here. Updated for the 2026-27 academic session, these solutions are based on the latest GSEB textbooks for Class 6 Mathematics. Our expert-created answers for Class 6 Mathematics are available for free download in PDF format.
Detailed Chapter 05 પાયાના આકારોની સમજૂતી GSEB Solutions for Class 6 Mathematics
For Class 6 students, solving GSEB textbook questions is the most effective way to build a strong conceptual foundation. Our Class 6 Mathematics solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 05 પાયાના આકારોની સમજૂતી solutions will improve your exam performance.
Class 6 Mathematics Chapter 05 પાયાના આકારોની સમજૂતી GSEB Solutions PDF
Question 1. તપાસો કે નીચેનામાંથી કયા બહુકોણ છે? તેમાંનો કોઈ ન હોય, તો કહો કે તે શા માટે નથી?
Answer:(a) અહીં, આપેલી આકૃતિ એક ખુલ્લી આકૃતિ છે, તેથી તે બહુકોણ નથી. (Here, the given figure is an open shape, so it is not a polygon.)
(b) અહીં, આપેલી આકૃતિ એક બંધ આકૃતિ છે, તેથી તે બહુકોણ છે. આ બહુકોણને છ બાજુઓ છે, તેને ષટ્કોણ પણ કહેવાય. (Here, the given figure is a closed shape, so it is a polygon. This polygon has six sides, so it is also called a hexagon.)
(c) અહીં, આપેલી આકૃતિ એક બંધ આકૃતિ છે, પરંતુ તે રેખાખંડો વડે બનેલી આકૃતિ નથી, તેથી તેને બહુકોણ ન કહેવાય. (Here, the given figure is a closed shape, but it is not made of line segments, so it cannot be called a polygon.)
(d) અહીં, આપેલી આકૃતિ એક બંધ આકૃતિ છે પરંતુ આખી આકૃતિ રેખાખંડો વડે બનેલી નથી. આકૃતિમાં બે રેખાખંડો છે જ્યારે ત્રીજો વક્ર છે, તેથી તે બહુકોણ નથી. (Here, the given figure is a closed shape, but the entire figure is not made of line segments. The figure has two line segments, while the third is a curve, so it is not a polygon.)
In simple words: To be a polygon, a shape must be closed and made only of straight line segments. We check each figure against these rules.
Exam Tip: Always remember the two key characteristics of a polygon: it must be a closed figure, and all its sides must be straight line segments. If any part is open or curved, it is not a polygon.
Question 2. દરેક બહુકોણનું નામ લખો:
આ દરેકનાં વધુ બે ઉદાહરણો આપો.
Answer:
(a) ચતુષ્કોણ (Quadrilateral)
આ બહુકોણને ચાર બાજુઓ છે, તેથી તેને ચતુષ્કોણ કહેવાય. (This polygon has four sides, so it is called a quadrilateral.)
ઉદાહરણ: (Examples:)
(b) ત્રિકોણ (Triangle)
આ બહુકોણને ત્રણ બાજુઓ છે, તેથી તેને ત્રિકોણ કહેવાય.
(This polygon has three sides, so it is called a triangle.)
ઉદાહરણ: (Examples:)
In simple words: We name polygons by counting their sides. Four sides mean a quadrilateral, and three sides mean a triangle. We give more examples of each.
Exam Tip: The number of sides directly determines the name of a polygon (e.g., tri-angle for 3 sides, quadri-lateral for 4 sides, hexa-gon for 6 sides). Always check the side count.
Question 3. નિયમિત ષટ્કોણની કાચી આકૃતિ દોરો. તેમાં કોઈ પણ ત્રણ શિરોબિંદુઓને જોડી ત્રિકોણ રચો. તમે દોરેલો ત્રિકોણ કયા પ્રકારનો છે તે કહો.
Answer: અહીં, નિયમિત ષટ્કોણ બતાવ્યો છે. આ ષટ્કોણનાં વારાફરતી બિંદુઓ A, C અને E જોડતાં આપણને \( \triangle \) ACE મળે છે. આ બાજુઓ \( \overline{\mathrm{AC}} \), \( \overline{\mathrm{CE}} \) અને \( \overline{\mathrm{EA}} \) સમબાજુ ત્રિકોણ છે.
(Here, a regular hexagon is shown. By joining the non-consecutive vertices A, C, and E of this hexagon, we get triangle ACE. These sides \( \overline{\mathrm{AC}} \), \( \overline{\mathrm{CE}} \), and \( \overline{\mathrm{EA}} \) form an equilateral triangle.)
In simple words: We draw a regular hexagon. Then, we connect every other vertex, specifically A, C, and E. The triangle that forms, triangle ACE, turns out to be an equilateral triangle because all its sides are equal in length.
Exam Tip: In a regular polygon, connecting alternate vertices can often create another regular polygon or a special type of triangle, like an equilateral triangle, due to its inherent symmetry.
Question 4. નિયમિત અષ્ટકોણની કાચી આકૃતિ દોરો. (તમે ઇચ્છો તો ચોરસ પેપરનો ઉપયોગ કરી શકો.) અષ્ટકોણનાં બરાબર ચાર શિરોબિંદુઓને જોડીને લંબચોરસ બનાવો.
Answer: અહીં, ABCDEFGH એક નિયમિત અષ્ટકોણ છે. તેનાં ચોથાં-ચોથાં બિંદુઓ સામે જોડતાં, એટલે કે શિરોબિંદુઓ A અને D જોડતાં \( \overline{\mathrm{AD}} \) તથા H અને E જોડતાં \( \overline{\mathrm{HE}} \) મળે છે. આ રીતે લંબચોરસ ADEH મળે છે. વળી, શિરોબિંદુઓ B અને C જોડતાં \( \overline{\mathrm{BG}} \) તથા C અને F જોડતાં \( \overline{\mathrm{CF}} \) મળે છે. આ રીતે લંબચોરસ BCFG મળે છે.
(Here, ABCDEFGH is a regular octagon. By joining every fourth vertex, that is, by joining vertices A and D, we get \( \overline{\mathrm{AD}} \), and by joining H and E, we get \( \overline{\mathrm{HE}} \). In this way, we form the rectangle ADEH. Also, by joining vertices B and G, we get \( \overline{\mathrm{BG}} \), and by joining C and F, we get \( \overline{\mathrm{CF}} \). This creates the rectangle BCFG.)
In simple words: We start with a regular octagon and mark its corners. By joining vertices A, D, E, and H, we create a rectangle. Similarly, joining vertices B, C, F, and G also makes another rectangle. These connections use specific non-adjacent vertices.
Exam Tip: When forming shapes inside a regular polygon, remember that symmetry allows for predictable connections. For a rectangle, opposite vertices are joined to form parallel sides of equal length.
Question 5. વિકર્ણ એ એવો રેખાખંડ છે કે જે બહુકોણનાં કોઈ પણ બે શિરોબિંદુને જોડે છે અને તે બહુકોણની કોઈ જ બાજુ નથી. પંચકોણની કાચી આકૃતિ દોરી તેના વિકર્ણો દોરો.
Answer: અહીં, ABCDE એક પંચકોણ છે. બબ્બે શિરોબિંદુઓ છે જોડતાં આપણને વિકર્ણો \( \overline{\mathrm{AC}} \), \( \overline{\mathrm{AD}} \), \( \overline{\mathrm{BD}} \), \( \overline{\mathrm{BE}} \) અને \( \overline{\mathrm{CE}} \) મળે છે.
(Here, ABCDE is a pentagon. By joining pairs of non-consecutive vertices, we get the diagonals \( \overline{\mathrm{AC}} \), \( \overline{\mathrm{AD}} \), \( \overline{\mathrm{BD}} \), \( \overline{\mathrm{BE}} \), and \( \overline{\mathrm{CE}} \).)
In simple words: We draw a five-sided shape called a pentagon. A diagonal is a line that links two corners that are not next to each other. For a pentagon, we can draw five such lines: AC, AD, BD, BE, and CE.
Exam Tip: To draw all diagonals of a polygon, pick each vertex one by one and connect it to all other non-adjacent vertices. Be careful not to draw the same diagonal twice.
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GSEB Solutions Class 6 Mathematics Chapter 05 પાયાના આકારોની સમજૂતી
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FAQs
The complete and updated GSEB Class 6 Maths Solutions Chapter 5 પાયાના આકારોની સમજૂતી Exercise 5.8 is available for free on StudiesToday.com. These solutions for Class 6 Mathematics are as per latest GSEB curriculum.
Yes, our experts have revised the GSEB Class 6 Maths Solutions Chapter 5 પાયાના આકારોની સમજૂતી Exercise 5.8 as per 2026 exam pattern. All textbook exercises have been solved and have added explanation about how the Mathematics concepts are applied in case-study and assertion-reasoning questions.
Toppers recommend using GSEB language because GSEB marking schemes are strictly based on textbook definitions. Our GSEB Class 6 Maths Solutions Chapter 5 પાયાના આકારોની સમજૂતી Exercise 5.8 will help students to get full marks in the theory paper.
Yes, we provide bilingual support for Class 6 Mathematics. You can access GSEB Class 6 Maths Solutions Chapter 5 પાયાના આકારોની સમજૂતી Exercise 5.8 in both English and Hindi medium.
Yes, you can download the entire GSEB Class 6 Maths Solutions Chapter 5 પાયાના આકારોની સમજૂતી Exercise 5.8 in printable PDF format for offline study on any device.