NCERT Solutions for Class 9 Maths Chapter 7
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1, Exercise 7.2, Exercise 7.3, Exercise 7.4 and Exercise 7.5 in English Medium as well as Hindi Medium updated for new academic session 2020-2021. According to Utter Pradesh Board (Prayagraj) students of class 9 will use NCERT Books for 9th Maths as course books. So, UP Board Solutions for Class 9 Maths Tribhuj ki Prashnavali 7.1, Prashnavali 7.2, Prashnavali 7.3, Prashnavali 7.4 and Prashnavali 7.5 in Hindi Medium PDF format are given to free download for academic session 2020-21. NCERT Solutions in Hindi Medium and English Medium or View in Video Format, for the students studying NCERT Books 2020-21, are based on the CBSE Curriculum 2020-2021.
Questions and solutions of all the five exercise of 9th Maths Chapter 7 are given below. Steps for solving questions are described properly by subject experts. Easy and simplified solutions are given, so that student can understand easily.NCERT Solutions for Class 9 Maths Chapter 7
Class: 9 | Maths (English and Hindi Medium) |
Chapter 7: | Triangles |
9th Maths Exercise 7.1 Solutions
9th Maths Exercise 7.2 Solutions
9th Maths Exercise 7.3 Solutions
9th Maths Exercise 7.4 Solutions
9th Maths Exercise 7.5 Solutions
Study Material for Class 9 Maths
9th Maths Chapter 7 Solutions in English & Hindi Medium
NCERT Solutions for class 9 Maths chapter 7 all exercises in English & Hindi medium free download for session 2020-21. Download UP Board App Apk and NCERT Solutions Offline Apps 2020-21 updated on the basis of NCERT Books 2020-21 free to use offline.
Class 9 Maths Chapter 7 Exercise 7.1 Solution in Videos
Class 9 Maths Chapter 7 Exercise 7.2 Solution in Videos
Class 9 Maths Chapter 7 Exercise 7.3 Solution in Videos
Class 9 Maths Chapter 7 Exercise 7.4 Solution in Videos
Class 9 Maths Chapter 7 Exercise 7.5 Solution in Videos
Important Notes on 9th Maths Chapter 7
Rules for Congruence of Triangle: Two triangles are congruent, if sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle.
- SSS (Side – Side – Side) Congruence Rule: Two triangles are congruent, if three sides of one triangle are equal to three sides of the other triangle.
- RHS (Right angle – Hypotenuse – Side) Congruence Rule: Two right triangles are congruent, if the hypotenuse and one side of one triangle are equal to hypotenuse and one side of the other triangle.
- ASA (Angle – Side – Angle) Congruence Rule: Two triangles are congruent, if two angles and the included side of one triangle are equal to two angles and the included side of the other triangle.
- SAS (Side – Angle – Side) Congruence Rule: Two triangles are congruent, if two sides and the included angle of one triangle are equal to two sides and the included angle of the other triangle.
What are the properties of a Triangle?
Properties of a Triangle
1. Angle opposite to equal sides of an isosceles triangle are equal.
2. The sides opposite to equal angles of a triangle are equal.
3. If two sides of a triangle are unequal then the angle opposite to the longer side is larger.
4. In any triangle, the side opposite to the larger angle is larger.
5. The sum of any two sides of a triangle is greater than the third side.
6. The difference between any two sides of a triangle is less than its third side.
7. If the bisector of the vertical angle of a triangle bisects the base, then triangle is an isosceles triangle.
8. The sum of any two sides of a triangle is greater than twice the median drawn to the third side.
Define a Triangle.
Triangle: A plane figure closed by three intersecting lines is called a triangle. Here, ‘Tri’ means THREE. A triangle has three sides, three angles and three vertices.
Important Questions on 9th Maths Chapter 7
इसलिए, ∠A < 90° और ∠C < 90° [∵∠A + ∠C = 90°] अतः, ΔABC में, ∠B > ∠C [∵∠B = 90° और ∠C < 90°] AC > AB … (1)
[∵ किसी त्रिभुज में, बड़े कोण की सम्मुख भुजा बड़ी होती है]
इसीप्रकार, ΔABC में, ∠B > ∠A
[∵∠B = 90° और ∠A < 90°] AC > BC … (2)
[∵ किसी त्रिभुज में, बड़े कोण की सम्मुख भुजा बड़ी होती है]
समीकरण (1) और (2) से
AC > AB और AC > BC
अतः, कर्ण AC सबसे लंबी भुजा है।
OP = OQ = OR.
∠ADB = ∠ADC [∵ Each 90°]
AB = AC [∵ Given]
AD = AD [∵ Common]
Hence, ΔABD ≅ ΔACD [∵RHS Congruency Rule]
BD = DC [∵ CPCT]
Hence, AD bisects BC
AB = AC [∵ दिया है]
∠B = ∠C [∵ समद्विभाहु त्रिभुज की बराबर भुजाओं के सम्मुख कोण बराबर होते हैं]
ΔABC में,
∠A + ∠B + ∠C = 180°
⇒ 90° + ∠B + ∠C = 180° [∵∠A=90°]
⇒ 90°+∠B + ∠B = 180° [∵∠C = ∠B]
⇒ 2∠B = 180°-90° = 90°
⇒ ∠B = (90°)/2 = 45°
अतः, ∠B = ∠C = 90°
AB = AC [∵ दिया है]
∠C = ∠B … (1)
[∵ समद्विभाहु त्रिभुज की बराबर भुजाओं के सम्मुख कोण बराबर होते हैं]
इसी प्रकार
ΔABC में,
AB = BC [∵ दिया है]
∠C = ∠A … (2)
[∵ समद्विभाहु त्रिभुज की बराबर भुजाओं के सम्मुख कोण बराबर होते हैं]
समीकरण (1) और (2) से
∠A = ∠B = ∠C … (3)
ΔABC में,
∠A + ∠B + ∠C = 180°
⇒∠A + ∠A + ∠A = 180° [∵ समीकरण (3) से]
⇒ 3∠A = 180°
⇒ ∠A = (180°)/3 = 60°
अतः, ∠A = ∠B = ∠C = 60°
∠APB = ∠APC [∵ Each 90°]
AB = AC [∵ Given]
AP = AP [∵ Common]
Hence, ΔABP ≅ ΔACP [∵RHS Congruency Rule]
∠B = ∠C [∵ CPCT]