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Last updated at May 29, 2018 by Teachoo

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Ex 6.5,2 (Method 1) PQR is a triangle right angled at P and M is a point on QR such that PM โฅQR. Show that PM2 = QM . MR Given: โ ๐๐๐ where โ ๐ ๐๐=90ยฐ & PM โฅQR To prove: PM2 = QM .MR Proof: In ฮ PQR, โ ๐ ๐๐ = 90ยฐ So, ฮ PQR is a right triangle Using Pythagoras theorem in ฮ PQR Hypotenuse2 = (Height)2 + (Base)2 RQ2 = PQ2 + PR2 Now, in ฮ PMR, PM โฅ QR So, โ PMR = 90ยฐ โด ฮ PMR is a right triangle Using Pythagoras theorem in ฮ PMR Hypotenuse2 = (Height)2 + (Base)2 PR2 = PM2 + MR2 Similarly, In ฮ PMQ, โ PMQ = 90ยฐ โด ฮ PMR is a right triangle Using Pythagoras theorem in ฮ PMQ Hypotenuse2 = (Height)2 + (Base)2 PQ2 = PM2 + MQ2 So, our equations are RQ2 = PQ2 + PR2 โฆ(1) PR2 = PM2 + MR2 โฆ(2) PQ2 = PM2 + MQ2 โฆ(3) Putting (2) & (3) in (1) RQ2 = PQ2 + PR2 RQ2 = (PM2 + MQ2 ) + (PM2 + MR2 ) RQ2 = (PM2 + PM2 ) + (MQ2 + MR2 ) RQ2 = 2PM2 + (MQ2 + MR2 ) (MQ + MR)2= 2PM2 + (MQ2 + MR2 ) MQ2 + MR2 + 2 MQ ร MR = 2PM2 + (MQ2 + MR2 ) (MQ2 + MR2 ) โ (MQ2 + MR2 ) + 2 MQ ร MR = 2PM2 0 + 2 MQ ร MR = 2PM2 2 MQ ร MR = 2PM2 MQ ร MR = PM2 โ PM2 = MQ ร MR Hence proved Ex 6.5,2 (Method 2) PQR is a triangle right angled at P and M is a point on QR such that PM โฅQR. Show that PM2 = QM . MR Given: โ ๐๐๐ where โ ๐ ๐๐=90ยฐ & PM โฅQR To prove: PM2 = QM .MR i.e. ๐๐/๐๐ = ๐๐ /๐๐ Proof: From theorem 6.7, If a perpendicular is drawn from the vertex of the right angle to the hypotenuse then triangles on both sides of the Perpendicular are similar to the whole triangle and to each other So, โ ๐๐๐ ~ โ ๐๐๐ So, โ ๐๐๐ ~ โ ๐๐๐ If two triangles are similar , then the ratio of their corresponding sides are equal ๐๐/๐๐=๐๐ /๐๐ PM ร MP = MR ร QM PM2 = MR ร QM Hence proved

Ex 6.5

Ex 6.5, 1

Ex 6.5, 2 Important Not in Syllabus - CBSE Exams 2021 You are here

Ex 6.5, 3 Important Not in Syllabus - CBSE Exams 2021

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Ex 6.5, 8 Important

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Chapter 6 Class 10 Triangles

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.