Last updated at May 29, 2018 by Teachoo

Transcript

Example 8 In Fig. 6.33, CM and RN are respectively the medians of Δ ABC and Δ PQR. If Δ ABC ~ Δ PQR, prove that : (i) Δ AMC ~ Δ PNR Given: Δ ABC and Δ PQR CM is the median of Δ ABC and RN is the median of Δ PQR Also , Δ ABC ~ Δ PQR To Prove: Δ AMC ~ Δ PNR Proof: CM is median of Δ ABC So, AM = MB = 1/2 AB Similarly, RN is the median of Δ PQR So, PN = QN = 1/2 PQ Given ∆ 𝐴𝐵𝐶 ~ ∆ 𝑃𝑄𝑅 𝐴𝐵/𝑃𝑄=𝐵𝐶/𝑄𝑅=𝐶𝐴/𝑅𝑃 𝐴𝐵/𝑃𝑄=𝐶𝐴/𝑅𝑃 (2 𝐴𝑀)/(2 𝑃𝑁)=𝐶𝐴/𝑅𝑃 𝐴𝑀/𝑃𝑁=𝐶𝐴/𝑅𝑃 Also, since ∆ 𝐴𝐵𝐶 ~ ∆ 𝑃𝑄𝑅 ∠ A = ∠ P In Δ AMC & ΔPNR ∠ A = ∠ P 𝐴𝑀/𝑃𝑁 " = " 𝐶𝐴/𝑅𝑃 Hence by SAS similarly ΔAMC ∼ ΔPNR Hence proved Example 8 In Fig. 6.33, CM and RN are respectively the medians of Δ ABC and Δ PQR. If Δ ABC ~ Δ PQR, prove that : (ii) 𝐶𝑀/𝑅𝑁=𝐴𝐵/𝑃𝑄 In part (i) we proved that Δ AMC ~ Δ PNR So, 𝐶𝑀/𝑅𝑁=𝐴𝐶/𝑃𝑅= 𝐴𝑀/𝑃𝑁 Therefore, 𝐶𝑀/𝑅𝑁= 𝐴𝑀/𝑃𝑁 𝐶𝑀/𝑅𝑁= 2𝐴𝑀/2𝑃𝑁 𝐶𝑀/𝑅𝑁=𝐴𝐵/𝑃𝑄 Hence Proved Example 8 In Fig. 6.33, CM and RN are respectively the medians of Δ ABC and Δ PQR. If Δ ABC ~ Δ PQR, prove that : (iii) Δ CMB ~ Δ RNQ Given ∆ 𝐴𝐵𝐶 ~ ∆ 𝑃𝑄𝑅 𝐴𝐵/𝑃𝑄=𝐵𝐶/𝑄𝑅=𝐶𝐴/𝑅𝑃 𝐴𝐵/𝑃𝑄=𝐵𝐶/𝑄𝑅 (2 𝐵𝑀)/(2 𝑄𝑁)=𝐵𝐶/𝑄𝑅 𝐵𝑀/𝑄𝑁=𝐵𝐶/𝑄𝑅 Also, since ∆ 𝐴𝐵𝐶 ~ ∆ 𝑃𝑄𝑅 ∠ B = ∠ Q Now in Δ CMB & ΔRNQ ∠𝐵=∠𝑄 𝐵𝑀/𝑄𝑁=𝐵𝐶/𝑄𝑅 Hence by SAS similarly ΔCMB ∼ ΔRNQ Hence proved

Example 5
Important

Example 8 Important You are here

Example 10 Important

Example 14 Important

Theorem 6.1 - Basic Proportionality Theorem (BPT) Important

Theorem 6.7 Important

Ex 6.2, 4 Important

Ex 6.2, 5 Important

Ex 6.2, 6 Important

Ex 6.2, 9 Important

Ex 6.3, 11 Important

Ex 6.3, 12 Important

Ex 6.3, 13 Important

Ex 6.3, 14 Important

Ex 6.3, 15 Important

Ex 6.4, 1 Important

Ex 6.4, 3 Important

Ex 6.4, 5 Important

Ex 6.5, 2 Important

Ex 6.5, 3 Important

Ex 6.5, 8 Important

Ex 6.5, 11 Important

Ex 6.5, 12 Important

Ex 6.5, 15 Important

Class 10

Important Questions for Exam - Class 10

- Chapter 1 Class 10 Real Numbers
- Chapter 2 Class 10 Polynomials
- Chapter 3 Class 10 Pair of Linear Equations in Two Variables
- Chapter 4 Class 10 Quadratic Equations
- Chapter 5 Class 10 Arithmetic Progressions
- Chapter 6 Class 10 Triangles
- Chapter 7 Class 10 Coordinate Geometry
- Chapter 8 Class 10 Introduction to Trignometry
- Chapter 9 Class 10 Some Applications of Trignometry
- Chapter 10 Class 10 Circles
- Chapter 11 Class 10 Constructions
- Chapter 12 Class 10 Areas related to Circles
- Chapter 13 Class 10 Surface Areas and Volumes
- Chapter 14 Class 10 Statistics
- Chapter 15 Class 10 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.