# Number Line Prime Class 7 Solutions Chapter 16

## Number Line Prime Class 7 Solutions Chapter 16 Properties of Triangles

Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Number Line Prime Class 7 Math Book, Chapter 16, Properties of Triangles. Here students can easily find step by step solutions of all the problems for Properties of Triangles, Exercise 16A, 16B, 16C, 16D and 16E Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 16 solutions. Here in this post all the solutions are based on latest Syllabus.

Properties of Triangles Exercise 16A Solution :

Question no – (1)

Solution :

(a) ST

(b) ∠R

(C) S

Question no – (2)

Solution :

(a) m, n, o points

(b) S, T Points

(c) u, v, w points.

Question no – (3)

Solution :

(a) All lengths are non equal

∴ It is a scalene triangle.

(b) All lengths are equal.

∴ It is a equilateral triangle.

Question no – (4)

Solution :

(a) All angle is 90°

∴ It is a right triangle.

(b)Here are angle is greater than 90°

∴ It is a obtuse – Angled triangle.

(c) All angle are less than 90°

∴ It is a acute – angled triangle

Question no – (5)

Solution :

∴ In △ABC, AB = BC, BL = LC, AL ⊥ BC

∴ It is a altitude = median = AL

Properties of Triangles Exercise 16C Solution :

Question no – (1)

Solution :

(a)

∴ <ACB = 180° – 110°

∴ <ACB = 70°

∴ <ABC + <BAC + <ACB = 180°

∴ x + 60° + 70° = 180°

∴ x = 180° + 150°

∴ x = 30°

(b

∴ y = 180° – 45°

∴ y = 135°

Question no – (2)

Solution :

∴ <R = 180° – (2P + LQ)

∴ <R = 180° (80° + 65°)

∴ <R = 180° – 145°

∴ <R = 35°

∴ Vertex R° = 180° – 35°

∴ R° = 145°

Question no – (3)

Solution :

∴ LC = 180° – <BCD = 180° – 3x

∴ x + 80° + 180° – 3x = 180°

∴ – 2x = – 80°

∴ x = 40°

Question no – (4)

Solution :

∴ Let, interior angles – x°

∴ 2x° = 120°

∴ x = 60°

∴ Angles of the triangle – 60°, 60°, 60°

Question no – (5)

Solution :

∴ Let the interior opposite angles – x°

∴ Are angle should be,

= 180° – 100°

= 80°

∴ 80° + x + x = 180°

∴ 2x = 100

∴ x = 50°

∴ Opposite angle = 50°, 50°

Properties of Triangles Exercise 16D Solution :

Question no – (1)

Solution :

(a) Given, AB = 6cm, BC = 6 cm, AC = 6 cm

∴ In △ABC,

∴ AB – BC < CA

= 6 – 6 < 6

= 0 < 6,

∴ BC – CA < AB

= 6 – 6 < 6

= 0 < 6,

∴ CA – AB < BC

= 6 – 6 < 6

= 0 < 6

∴ Lengths are sides of triangle.

(b) Given, AB = 4 cm, BC = 7 cm and CA = 13 cm

∴ 4 – 7 < 13

= -3 < 13,

∴ 7 – 13 < 4

= – 6 < 4,

∴ 13 – 4 < 7

= 9 < 7

∴ Lengths are not side of a triangles.

(c) Given, AB = 9 cm, BC = 10 cm and CA = 20 cm

∴ 9 – 10 < 20

= – 1 < 20,

∴ 10 – 20 < 9

= – 10 < 9,

∴ 20 – 9 < 10

= 11 < 10

∴ Lengths are not side of a triangle.

Question no – (2)

Solution :

NO, Sum of any two lengths is always greeter than third angle.

Question no – (3)

Solution :

No, because they will from a straight line.

Properties of Triangles Exercise 16E Solution :

Question no – (1)

Solution :

(a) 9= 81

= 62 + 82

= 36 + 64

∴ 100 ≠ 81

∴ It is not a right angled triangle.

(b) ∴ (13)2

= 169,

∴ 122 + 52

= 144 + 25

= 169

∴ It is a right angled triangle.

Question no – (2)

Solution :

(a) x2 = 102 + 242

∴ x2 = 100 + 576

∴ x2 = 676

∴ x = √676

∴ x = 26

(b) (25)2 = x+ 72

∴ x2 = 252 – 72

∴ x2 = 625 – 49

∴ x = √676

∴ x = 24

Question no – (3)

Solution :

(a) ∴ 152

= 225

∴ 122 + 132

= 144 + 169

= 313

∴ Not satisfied.

(b) ∴ 52

= 25

∴ 32 + 42

= 9 + 16

= 25

∴ Pythagorean satisfied

Question no – (4)

Solution :

Let, the perpendicular – x

∴  (15)2 = (12)2 + x2

∴ x2 = 225 – 144

∴ x= 81

∴ x = √81

∴ x = 9

∴ Perpendicular – 9cm

Question no – (5)

Solution :

∴ PR = √(PQ)2 + (QR)2

= √64 + 225

= √289

= 17 cm

Therefore, PR will be 17 cm.

Question no – (6)

Solution :

∴ (26)2

= 676

∴ (10)2 + (24)2

= 100 + 576

= 676

∴ It is a right- angles triangle.

Question no – (7)

Solution :

∴ Let the distance x

∴ x2 = √144 + 25

∴ x2 = √169

∴ x2 = 13 cm

Therefore, the distance between initial and final position will be 13 cm.

Question no – (8)

Solution :

∴ BD = √(13)2 – (12)2

= √169 – 144

= √25

= 5

∴ DC = √(15)2 – (12)2

= √225 – 144

DC = √81

= 9

∴ BC = BD + DC

= 5 +9

= 14 cm

Therefore, the length of BC will be 14 cm.

Question no – (9)

Solution :

∴ Let the distance – x

∴ x = √ (10)2 + (24)2

∴ x = √100 + 576

∴ x = √676

∴ x = 26 m

Therefore, the distance between starting point and end point will be 26 m.

Next Chapter Solution :

Updated: July 25, 2023 — 6:11 am