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

RBSE Solutions

Question 2
Find the unknown angle x for (RBSESolutions.com) the following figures
RBSE Solutions for Class 7 Maths Chapter 8 Triangle and its Properties Ex 8.1
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 :
RBSE Solutions for Class 7 Maths Chapter 8 Triangle and its Properties Ex 8.1
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°

RBSE Solutions

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.

RBSE Solutions

We hope the RBSE Solutions for Class 7 Maths Chapter 8 Triangle and its Properties Ex 8.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 8 Triangle and its Properties Exercise 8.1, drop a comment below and we will get back to you at the earliest.

Leave a Comment

Step into high-class excitement at hell spin casino, where glittering reels, lavish bonuses, and thrilling jackpots create nonstop luxury. Each spin delivers pulse-raising suspense, elegance, and the electrifying chance of big Australian online casino wins.

Indulge in elite thrills at joefortune-casino.net, offering dazzling gameplay, sparkling rewards, and adrenaline-pumping jackpots. Every moment immerses players in glamour, high-stakes excitement, and the intoxicating pursuit of substantial casino victories.

Discover top-tier sophistication at neospin casino, with vibrant reels, generous bonuses, and luxurious jackpots. Each spin captivates with elegance, thrill, and the electrifying potential for extraordinary wins in the premium Australian casino environment.

Enter a world of luxury at rickycasino-aus.com, where high-class slots, sparkling bonuses, and pulse-racing jackpots create unforgettable moments. Every wager delivers excitement, sophistication, and the premium thrill of chasing massive casino wins.