Warning: Undefined array key "https://nctbsolution.com/mathsight-class-6-solutions/" in /home/862143.cloudwaysapps.com/hpawmczmfj/public_html/wp-content/plugins/wpa-seo-auto-linker/wpa-seo-auto-linker.php on line 192
Mathsight Class 6 Solutions Chapter 11 Basic Geometrical Concepts
Welcome to NCTB Solutions. Here with this post we are going to help 6th class students to solve MathSight Class 6 Mathematics Book, Chapter 11, Basic Geometrical Concepts. Here students can easily get all the exercise questions solution for Chapter 11, Basic Geometrical Concepts Exercise 11.1, 11.2, 11.3 and 11.4
Basic Geometrical Concepts Exercise 11.1 Solution :
Question no – (1)
Solution :
Infinity points
Question no – (2)
Solution :
Infinity line
Question no – (3)
Solution :
Two lines.
Question no – (4)
Solution :
Q lies between L and T
Question no – (5)
Solution :
Question no – (6)
Solution :
(a) BC, CB, BD, DB, DC, CD, AD, DA,
AB, BA, AC, CA .
Total number of line = 12
(b) AB, BC, AC, CA, AD, DA, BC, CB, DC, CD .
Total number of line = 10
(c) AB, BC, CD, DC, AC, CA, BD, DB, CE, EC, ED, DE,
CF, FC, DF, FD
Total number of line = 16
Question no – (7)
Solution :
Line → A line is a collection of pints that can be entered endless on both the sides.
Line segment → A line segment is a part of a line that is bonded by two distinct end points.
Question no – (8)
Solution :
(a) X, W, Y
(b) P, Q, R
(c) M, N, Z, O
Question no – (9)
Solution :
(a) PR
(b) QP, QR, QS, QT
(c) PQ, QR
(d) P, Q, R
Question no – (10)
Solution :
Question no – (11)
Solution :
(a) Line segment
(b) Ray
(c) Line
(d) Line segment
(e) Ray
Question no – (12)
Solution :
(a) Polygon
(b) Curved line
(c) open figure .
(d) pentagon .
(e) curvilinear boundary
(f) polygon
Basic Geometrical Concepts Exercise 11.2 Solution :
Question no – (1)
Solution :
(a) ∠EFG, ∠HFG
(b) KLM, MLN
(c) WXY, ZXY
Question no – (2)
Solution :
(a) Obtuse
(b) Obtuse
(c) Obtuse
Question no – (3)
Solution :
(a) A, H, P
(b) L, C, M, D
(c) Q
Question no – (4)
Solution :
(a) PR
(b) R
(c) QR
(d) ∠R
Question no – (5)
Solution :
Question no – (6)
Solution :
Let, common factor = x
We know fat .
2x + 3x + 70° = 180°
Or, 5x = 180 – 70° = 110°
Or, x = 110°/5 = 22°
∴ 2x = 2 × 22 = 44°
∴ 3x = 3 × 22 = 66°
Basic Geometrical Concepts Exercise 11.3 Solution :
Question no – (1)
Solution :
(a) Side = AB, AD, DC, CD
Angles = ∠A. ∠B, ∠C, ∠D
Vertices = A, B, C, D
(b) Side = PQ, QR, RS, SP
Angles = ∠P, ∠Q, ∠R, ∠S
Vertices = P, Q, R, S
Question no – (2)
Solution :
(a) OD, OC, OB, OA
(b) ∠DOC, ∠COB, ∠BOA, ∠DOA
Question no – (3)
Solution :
(a) Convex
(b) Concave
Question no – (4)
Solution :
| Side | Angle |
| (a) AB opposite to DC
AD opposite to BC |
∠A opposite to ∠C
∠B opposite to ∠D |
| (b) QR opposite to PS
PQ opposite to RS |
∠Q opposite to ∠S
∠P opposite to ∠R |
| (C) LM opposite to ON
LO opposite to MN |
∠L opposite to ∠N
∠M opposite to ∠O |
Question no – (5)
Solution :
(a) D
(b) B, E, F
(c) A, C
Question no – (6)
Solution :
In the given figure,
(a) There are total 8 triangles.
(b) 4 angles are formed at “T”
Question no – (7)
Solution :
(a) A, B, C
(b) D, E, F
(c) AB, BC, AC
(d) AC
Question no – (8)
Solution :
X = 360° – (100+95+60)
= 360° – 255°
= 105°
Question no – (9)
Solution :
(a) – 7 + 4
= – 3
(b) – 7 – (1)
= – 7 + 1
= 6
(c) 0 – (-9)
= – + 9
= 9
Question no – (10)
Solution :
The length of each side of the parallelogram should be less than Diagonals.
Basic Geometrical Concepts Exercise 11.4 Solution :
Question no – (1)
Solution :
(a) OB, OC, OA
(b) AC
(c) AB, BC
(d) AHC, ABC
(e) PQ
(f) QHP
Question no – (2)
Solution :
Question no – (3)
Solution :
Question no – (4)
Solution :
Lia = 7.8 × 2
= 15.6 cm
Question no – (5)
Solution :
It should be to practically
Question no – (6)
Solution :
(a) CD, DG
(b) HF
(c) AE
(d) C, D, H, F, G
(e) B, J, G
Previous Chapter Solution :
