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.

Download त्रिभुज की प्रश्नावली 7.1, प्रश्नावली 7.2, प्रश्नावली 7.3, प्रश्नावली 7.4 और प्रश्नावली 7.5 in Hindi Medium PDF form 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. Download Class 9 Maths App for Offline use or Download कक्षा 9 गणित App for offline use.

NCERT Solutions for Class 9 Maths Chapter 7

Class:	9
Subject:	Maths – गणित
Chapter 7:	Triangles

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 NCERT Solutions Offline Apps 2020-21 updated on the basis of NCERT Books 2020-21 free to use offline.

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
- Study Material for 2020-21 – English Medium
- Study Material for 2020-21 – Hindi Medium

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 Notes on 9th Maths Chapter 7

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.
1. 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.
2. 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.
3. 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.
4. 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.

Important Questions on 9th Maths Chapter 7

ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.

Draw perpendicular bisectors of AB, BC and AC, which intersects each other at G. Point G is equidistant from the three vertices of ΔABC.

दर्शाइए कि समकोण त्रिभुज में कर्ण सबसे लंबी भुजा होती है।

ΔABC में, ∠B = 90° [∵ दिया है]
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।

In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Draw the bisectors of ∠A, ∠B and ∠C, which intersect each other at O. Point O is equidistant from the three sides of ΔABC i.e.

OP = OQ = OR.

AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that AD bisects BC.

In ΔABD and ΔACD,
∠ADB = ∠ADC [∵ Each 90°]
AB = AC [∵ Given]
AD = AD [∵ Common]
Hence, ΔABD ≅ ΔACD [∵RHS Congruency Rule]
BD = DC [∵ CPCT]
Hence, AD bisects BC

ABC एक समकोण त्रिभुज है, जिसमें ∠A = 90° और AB = AC है। ∠B और ∠C ज्ञात कीजिए।

ΔABC में,
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°

दर्शाइए कि किसी समबाहु त्रिभुज का प्रत्येक कोण 60° होता है।

ΔABC में,
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°

ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

In ΔABP and ΔACP,
∠APB = ∠APC [∵ Each 90°]
AB = AC [∵ Given]
AP = AP [∵ Common]
Hence, ΔABP ≅ ΔACP [∵RHS Congruency Rule]
∠B = ∠C [∵ CPCT]