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Question 17 - CBSE Class 10 Sample Paper for 2018 Boards

Last updated at Nov. 27, 2018 by

In given figure ∠1 = ∠2 and ∆NSQ ≅ ∆MTR , then prove that ∆PTS ~ ∆ PRQ .

OR

In an equilateral triangle ABC, D is a point on the side BC such that BD = 1/3 BC. Prove that 9AD 2 = 7AB 2

This is a question of CBSE Sample Paper - Class 10 - 2017/18.

You can download the question paper here  https://www.teachoo.com/cbse/sample-papers/

In fig, angle 1 = angle 2 and NSQ congruent MTR, prove that PTS ~ PRQ

Question 17 - CBSE Class 10 Sample Paper for 2018 Boards - Part 2
Question 17 - CBSE Class 10 Sample Paper for 2018 Boards - Part 3

Question 17

In an equilateral triangle ABC, D is a point on the side BC such that BD = 1/3 BC. Prove that 9AD 2 = 7AB 2

 

This is exactly same as Ex 6.5, 15.

Check answer here

 

  1. Class 10
  2. Solutions of Sample Papers for Class 10 Boards

    3. CBSE Class 10 Sample Paper for 2018 Boards

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Transcript

Question 17 In given figure ∠1 = ∠2 and ∆NSQ ≅ ∆MTR , then prove that ∆PTS ~ ∆ PRQ . Given: ∠ 1 = ∠ 2 & ΔNSQ ≅ ΔMTR To Prove: ΔPTS ∼ PRQ Proof: Given ΔNSQ ≅ ΔMTR ∴ ∠ NQS = ∠ MRT i.e. ∠ PQR = ∠ PRQ Now, In Δ PST By angle sum property ∠ P + ∠ 1 + ∠ 2 = 180° Since ∠ 1 = ∠ 2 given ∠ P + ∠ 1 + ∠ 1 = 180° ∠ P + 2 ∠ 1 = 180° In Δ PQR By angle sum property ∠ P + ∠ PQR + ∠ PRQ = 180° From (1), ∠ PQR = ∠ PRQ ∠ P + ∠ PQR + ∠ PQR = 180° ∠ P + 2 ∠ PQR = 180° Right side of (2) and (3) are same So, from (2) and (3) ∠ P + 2 ∠ 1 = ∠ P + 2 ∠ PQR 2 ∠ 1 = 2 ∠ PQR ∠ 1 = ∠ PQR In ΔPTS & Δ PRQ ∠P = ∠P (Common angle) ∠ PST = ∠ PQR (From (4)) ∴ ΔPTS ∼ PRQ (AA similarity) Hence proved

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.