Class 9 maths ch 7 ex 7.2

NCERT Solutions Class 9 Maths exercise 7.2 of the chapter 7-Triangles NCERT Solutions Class 9 Maths exercise 7.2 of the chapter 7-Triangles is based on the solutions of the questions related to congruency of triangles.All questions of the exercise 7.2 -Triangles are solved by an expert of CBSE Maths.All students of class 9 are required to go through each solutions for clearing their concept on Triangles.You can study here science and maths NCERT solutions and notes of each chapters from class 9-12, sample papers, solutions of previous year’s question papers ,solutions of important questions of science and maths ,articles on science and maths,government entrance exams and other competitive entrance exams and online jobs. Click for online shopping You can also study NCERT Solutions Q1.In an isosceles triangle ABC, with AB = AC, the bisector of angle ∠B and ∠C, intersect each other at O. Join A to O. Show that: (i) OB = OC (ii) AO bisects ∠A Ans. GIVEN: In ΔABC AB = AC BO is bisector of ∠B and CO is bisector of ∠C TO PROVE:(i) OB = OC (ii) AO bisects ∠A PROOF:In ΔABC AB = AC (given) ∠ABC = ∠ACB (angles opposite to equal sides) It is given to us that BO is bisector of ∠B and CO is bisector of ∠C 1/2(∠ABC) = 1/2(∠ACB) ∠OBC = ∠OCB OB = OC (sides opposite to equal angles) Hence proved (ii) OB = OC (proved above) AO = AO (common) AB = AC (given) ΔABO ≅ΔACO (SSS rule) ∠OAB = ∠OCA (by CPCT) Therefore AO is the bisector of ∠A. Hence proved Q2. In ΔABC , AD is perpendicular bisector o...

Transcript Ex 7.2, 15 Integrate the function: 𝑥﷮9 − 4𝑥2﷯ Step 1: Let 9 – 4𝑥2 = 𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 0−4.2 𝑥﷮2−1﷯= 𝑑𝑡﷮𝑑𝑥﷯ −8𝑥.𝑑𝑥=𝑑𝑡 𝑑𝑥 = 𝑑𝑡﷮−8𝑥﷯ Step 2: Integrating the function ﷮﷮ 𝑥﷮9 − 4𝑥2﷯﷯ . 𝑑𝑥 Putting 9 − 4𝑥2=𝑡 & 𝑑𝑥= 𝑑𝑡﷮−8𝑥﷯ = ﷮﷮ 𝑥﷮𝑡﷯﷯ . 𝑑𝑡﷮−8𝑥﷯ = − 1﷮8﷯ ﷮﷮ 1﷮𝑡﷯﷯ . 𝑑𝑡 = − 1﷮8﷯ log ﷮ 𝑡﷯﷯+𝐶 = − 𝟏﷮𝟖﷯ 𝐥𝐨𝐠 ﷮ 𝟗 − 𝟒𝒙𝟐﷯﷯+𝑪 Show More

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 • • NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Exercise 7.2 Ex 7.2 Class 7 Maths Question 1. Which congruence criterion do you use in the following? (a) Given: AC = DF AB = DE BC = EF So, ∆ABC = ∆DEF (b) Given: ZX = RP RQ = ZY ∠PRQ = ∠XZY So, ∆PQR ≅ ∆XYZ (c) Given: ∠MLN = ∠FGH ∠NML = ∠GFH ML = FG So, ∆LMN = ∆GFH (d) Given: EB = DB AE = BC ∠A = ∠C = 90° ∆ABE = ∆CDB Solution: (a) ∆ABC ≅ ∆DEF (BY SSS rule) (b) ∆PQR ≅ ∆XYZ (BY SAS rule) (c) ∆LMN ≅ ∆GFH (BY ASA rule) (d) ∆ABE ≅ ∆CDB (BY RHS rule) Ex 7.2 Class 7 Maths Question 2. You want to show that ∆ART = ∆PEN, (a) If you have to use SSS criterion, then you need to show (i) AR = (ii) RT = (iii) AT = (b) If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have (i) RT = and (ii) PN = (c) If it is given that AT = PN and you are to use ASA criterion, you need to have (i) ZA (ii) ZT Solution: (a) For SSS criterion, we need (i) AR = PE (ii) RT = EN (iii) AT = PN (b) For SAS criterion, we need (i) RT = EN and (ii) PN = AT (c) For ASA criterion, we need (i) ∠A = ∠P (ii) ∠T = ∠N Ex 7.2 Class 7 Maths Question 3. You have to show that ∆AMP ≅ ∆AMQ. In the fallowing proof, supply the missing reasons. Steps Reasons (i) PM = QM (i) (ii) ∠PMA – ∠QMA (ii) (iii) AM = AM (iii) (iv) ∆AMP = ∆AMQ (iv) Solution: Steps Reasons (i) PM = QM (i) Given (ii) ∠PMA = ∠QMA (ii) Given (iii) AM = AM (iii) Common (iv) ∆AMP = ∆AMQ (iv) SAS r...

Frank ICSE Solutions for Class 9 Maths Linear Equations Ex 7.2 Frank ICSE Solutions for Class 9 Maths Chapter 7 Linear Equations Ex 7.2 Answer 1. Answer 2. Answer 3. Answer 4. Answer 5. Answer 6. Answer 7. Answer 8. Answer 9. Answer 10. Answer 11. Answer 12. Answer 13. Answer 14. Answer 15. Answer 16. Answer 17. Answer 18. Answer 19. Answer 20. Answer 21. Answer 22. Answer 23. Answer 24. Answer 25. Answer 26. Answer 27. Answer 28. Filed Under: Tagged With:

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 are part of • • • • • Board CBSE Textbook NCERT Class Class 9 Subject Maths Chapter Chapter 7 Chapter Name Triangles Exercise Ex 7.2 Number of Questions Solved 8 Category NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 Ex 7.2 Class 9 Maths Question 1. In an Isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that (i) OB = OC (ii) AO bisects ∠A Solution: (i) In ΔABC, we have AB = AC (Given) => ∠B = ∠C \(\frac \) ∠C) => OB – OC (∵ Sides opposite to equal angles are equal) (ii) In Δ ABO and ΔACD, we have AB = AC (Given) ∠OBA = ∠OCA [From Eq. (i)] => OB = OC [From Eq. (ii)] ΔABO ≅ ΔACO (By SAS congruence axiom) => ∠BAO = ∠CAO (By CPCT) => AO is the bisector of ∠BAC. Ex 7.2 Class 9 Maths Question 2. In Δ ABC, AD is the perpendicular bisector of BC (see figure). Show that AABC is an isosceles triangle in which AB = AC. Solution: Given : AD is the perpendicular bisector of BC; To prove : In Δ ABC is an isosceles triangle. i.e., AB = AC Proof : In Δ ADB and Δ ADC, AD = AD [common] BD = DC (Given) and ∠ADB = ∠ADC [each= 90°] Δ ADB ≅ Δ ADC (By SAS congruence axiom) => AB = AC (By CPCT) So, Δ ABC is an isosceles triangle. Ex 7.2 Class 9 Maths Question 3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB, respectively (see figure). Show that these altitudes are equal. Solution: Given: ΔABC is an isosceles triang...