RBSE Solutions for Class 7 Maths Chapter 8 Triangle and its Properties Ex 8.1

RBSE Solutions for Class 7 Maths Chapter 8 Triangle and its Properties Ex 8.1 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 8 Triangle and its Properties Exercise 8.1.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 7
Subject	Maths
Chapter	Chapter 8
Chapter Name	Triangle and its Properties
Exercise	Ex 8.1
Number of Questions	8
Category	RBSE Solutions

Rajasthan Board RBSE Class 7 Maths Chapter 8 Triangle and its Properties Ex 8.1

Question 1
Find the value of unknown (RBSESolutions.com) angle x in the following figures :

Solution:
Sum of all angles of (RBSESolutions.com) a triangle is 180°.
(i) x + 60° + 85° = 180°
⇒ x + 145° = 180°
⇒ x = 180° – 145° = 35°

(ii) x + 40° + 20° = 180°
⇒ x + 60° = 180°
⇒ x = 180° – 60° = 120°

(ii) x + 40° + 90° = 180°
⇒ x + 130° = 180°
⇒ x = 180° – 130° = 50°

(iv) x + x + x = 180°
⇒ 3x = 180°
⇒ x = \(\frac { { 180 }^{ 0 } }{ 3 }\) = 180°

Question 2
Find the unknown angle x for (RBSESolutions.com) the following figures

Solution:
In any triangle exterior angle is (RBSESolutions.com) always equal to the sum of two interior angles.
(i) x = 80° + 50° = 130°
(ii) x + 35° = 110°
⇒ x = 110° – 35° = 75°
(iii) x = 30° + 35° = 65°
(iv) x + 70° = 120°
⇒ x = 120° – 70° = 50°
(v) x + 35° = 65°
⇒ 65° – 35° = 30°
(vi) x = 55° + 40° = 95°

Question 3
Find the values of the unknown (RBSESolutions.com) angles x and y in the following figures :

Solution:
(i) In a triangle, exterior (RBSESolutions.com) angle and adjecent internal angle make a linear pair.
x + 110° = 180°
⇒ x = 180° – 110° = 70°
In a triangle, sum of all angles is 180°.
∴ x + y + 60° = 180°
⇒ 70° + y + 60° = 180°
⇒ y + 130° = 180°
⇒ y = 180° – 130° = 50°
∴ x = 70°, y = 50°

(ii) In a triangle, exterior angle and (RBSESolutions.com) adjecent internal angle make a linear pair.
x = 45° + 30° = 75°
∵ In a triangle, sum of all angles is 180°.
∴ y + 45° + 30° = 180°
⇒ y + 75° = 180°
∴ y = 180° – 75° = 105°
∴ x = 75°,y= 105°

(iii) In a triangle, exterior angle and adjecent internal (RBSESolutions.com) angle make a linear pair.
∴ y + 80° = 180°
⇒ y = 180° – 80° = 100°
In a triangle, sum of all angles is 180°
∴ x + x + y = 180°
⇒ 2x + 100° = 180°
⇒ 2x = 180° – 100°
⇒ 2x = 80°
∴ x = \(\frac { { 80 }^{ 0 } }{ 2 }\) = 40°
∴ x = 40°, y = 100°

(iv) Here x = 70° (vertically opposite angle)
∵ Sum of (RBSESolutions.com) three angles of a triangle = 180°
∴ x + y + 40° = 180°
⇒ 70° + y + 40° = 180°
⇒ 110° + y = 180°
∴ y = 180° – 110° = 70°
∴ x = 70°,y = 70°

Question 4
If one acute angle of a right angled triangle is 45° then find (RBSESolutions.com) another acute angle.
Solution:
Sum of all angles of a triangle is 180°.
∴ 90° + one acute angle + other acute angle = 180°
⇒ 90° + 45° + other acute angle = 180°
⇒ other acute angle = 180 – 135°
⇒ = 45°

Question 5
If two angles of (RBSESolutions.com) a triangle are 50° each then find the third angle.
Solution:
∵ Sum of three angles of a triangle = 180°
∴ 50° + 50° + third angle = 180°
⇒ 100° + third angle = 180°
⇒ third angle = 180° – 100°
⇒ = 80°

Question 6
If the angles of a triangle are in ratio 1 : 2 : 3, then find each angle (RBSESolutions.com) of the triangle.
Solution:
Let the angles of the triangle are x°, 2x° and 3x°
∴ x° + 2x° + 3x° = 180°
⇒ 6x° = 180°
∴ x° = \(\frac { { 180 }^{ 0 } }{ 6 }\) = 30°
∴ I angle = x° = 30°
II angle = 2x° = 2 x 30° = 60°
III angle = 3x° = 3 x 30° = 90°

Question 7
Is it possible to construct a right angled triangle (RBSESolutions.com) whose other two angles are 70° and 21° ? If not, then why ? Justify.
Solution:
No, ∴ Sum of three angles = 180
⇒ 70° + 21 + x° = 180°
⇒ x = 180 – 90°
⇒ x = 89°
≠ 9
∵ Third angle is not a right angle.
Hence. Such right angle triangle is not possible.

Question 8
Given below are the triads. Which of the (RBSESolutions.com) following triads represents the angles of a triangle ?
(i) 100°, 30°, 40°
(ii) 30°, 59°,91°
(iii) 45°, 45°, 90°
(iv) 120°, 30°, 50°
Solution:
(i) (100°, 30°, 40°)
∵ Sum of all angles of a triangle = 180°
(i) 100° + 30 + 40° ≠ 180° ⇒ 170 ≠ 180
∴ Triangle is not possible.
(ii) 30° + 59° + 91°= 180° ⇒ 180° = 180°
∴ Triangle is possible.
(iii) 45° + 45° + 90° = 180° ⇒ 180° = 180°
∴ Triangle is (RBSESolutions.com) possible.
(iv) 120° + 30° + 50° = 180° ⇒  200°= 180°
∴ Triangle is not possible.

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