Class 9 maths chapter 7 exercise 7.1 in hindi

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1 Exercise 7.1 of Chapter 7 of NCERT Class 9 Maths deals with the “Congruence of Triangles”. You must be wondering why this topic has been included in the syllabus, although we rarely see its applications. Well, in real life, two triangles are rarely congruent. However, studying this topic becomes crucial because it is used in the construction of large buildings or other architectural designs. Mathematicians have found the Triangle to be a very stable shape, and congruence is needed to create even surfaces. Also, you must have noticed congruent triangles in geometric art, carpeting designs, stepping stone designs, architectural designs, etc. Thus, this chapter plays a major role in daily life, and students have to focus on it using The NCERT Class 9 Maths Exercise is full of proof. Therefore, it’s important that you first know theorems related to it and then go for the exercise questions. Here, we have provided step-by-step NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1. Students can refer to it whenever they find difficulty in solving the questions. The Download PDF carouselExampleControls112 Previous Next Access other Exercise Solutions of Chapter 7 – Triangles You should practise all the questions of Exercise 7.1. To get the detailed solutions of all the exercises of NCERT Class 9 Chapter 7, visit the links provided below: Access Answers to NCERT Solutions for Class 9 Maths Chap...

Free download NCERT Solutions for Class 9 Maths Chapter 7 triangles exercise 7.1, 7.2, 7.3, 7.4 and 7.5 in NCERT Solutions for Class 9 Maths Chapter 7 If you need solutions in Hindi, Click for 9 Maths Chapter 7 All Exercises in Hindi To get the solutions in English, Click for Class 9 Maths Chapter 7 – Important Questions Triangles – Questions with answers • If ∆ABC is an isosceles triangle such that AB = AC, then prove that altitude AD from A on BC bisects it. • In a ∆PQR, angle P = 110°, PQ = PR. Find angle Q and angle R. [Answer: angle Q = 35 and angle R = 35] • In ∆ABC, if angle A = 55°, angle B = 75° then find out the smallest and longest side of the triangle. [Answer: Smallest = AB, Longest = AC] • The vertex angle of an isosceles triangle is 80°. Find out the measure of base angles. [Answer: 50, 50] • ABC is a triangle and D is the • Prove that angles opposite to the equal sides of an isosceles triangle are equal. • S is any point in the interior of a ∆PQR. Prove that SQ + SR 1/2 (AB + BC + CA). • Prove that the perimeter of a triangle is greater than the sum of its three altitudes. • Two sides AB, BC and median AM of ∆ABC are respectively equal to sides PQ, QR and median PN of ∆PQR. Show that: ∆ABM and ∆PQN are congruent. • Ram has a land in the shape of a square PQRS. Its diagonals PR and QS intersect at O. Show that ∆POQ, ∆QOR, ∆ROS and ∆SOP are congruent. Ram donates two triangular parts of land for opening a hospital and a school. Which values are exhibited by ...

RBSE Class 9 Maths Solutions Chapter 7 Triangles Exercise 7.1 Question 1. In quadrilateral ACBD, AC = AD and AB bisects ∠A (see figure). Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD? Answer: Now, in ∆s ABC and ABD, we have : AC = AD (Given) ∠CAB = ∠BAD (∵ AB bisects ∠∆) and, AB = AB (Common) ∴ By SAS congruence criterion, we have: ∆ABC ≅ ∆ABD ⇒ BC = BD, i.e. they are equal. (∵ Corresponding parts of congruent triangles are equal.) Question 2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see figure). Prove that: (i) ∆ABD ≅ ∆BAC (ii) BD = AC (iii) ∠ABD = ∠BAC. Answer: In ∆s ABD and BAC, we have : AD = BC (Given) ∠DAB = ∠CBA (Given) AB = BA (Common) ∴ By SAS criterion of congruence, we have ∆ABD ≅ ∆BAC, which proves (i) From (i), (ii) BD = AC (C.P.C.T.) and, (iii) ∠ABD = ∠BAC (∵ Corresponding parts of congruent triangles (CPCT) are equal.) Question 3. AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB. Answer: Since AB and CD intersect at O, therefore ∠AOD = ∠BOC ….(1) (Vertically opp. angles) In ∆s AOD and BOC, we have: ∠AOD = ∠BOC [From (1)] ∠DAO = ∠CBO (Each = 90°) and, AD = BC (Given) ∴By AAS congruence criterion, we have : ∆AOD ≅ ∆BOC ⇒ OA = OB (∵ Corresponding parts of congruent triangles (CPCT) are equal.) i.e. O is the mid-point AB. Hence, CD bisects AB. Question 4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ∆ABC ≅ ∆CDA. Answer: Since l ...