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Ex 6.5, 2 - PQR is a triangle right angled at P and M is - Ex 6.5

  1. Chapter 6 Class 10 Triangles
  2. Serial order wise
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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 โŠฅQR. Show that PM2 = QM . MR Given: โˆ† ๐‘ƒ๐‘„๐‘… where โˆ  ๐‘…๐‘ƒ๐‘„=90ยฐ & PM โŠฅQR To prove: PM2 = QM .MR Proof: In ฮ” PQR, โˆ  ๐‘…๐‘ƒ๐‘„ = 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 โŠฅ QR So, โˆ  PMR = 90ยฐ โˆด ฮ” PMR is a right triangle Using Pythagoras theorem in ฮ” PMR Hypotenuse2 = (Height)2 + (Base)2 PR2 = PM2 + MR2 Similarly, In ฮ” PMQ, โˆ  PMQ = 90ยฐ โˆด 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 โ€ฆ(1) PR2 = PM2 + MR2 โ€ฆ(2) PQ2 = PM2 + MQ2 โ€ฆ(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 โŠฅQR. Show that PM2 = QM . MR Given: โˆ† ๐‘ƒ๐‘„๐‘… where โˆ  ๐‘…๐‘ƒ๐‘„=90ยฐ & PM โŠฅQR To prove: PM2 = QM .MR i.e. ๐‘ƒ๐‘€/๐‘„๐‘€ = ๐‘€๐‘…/๐‘ƒ๐‘€ 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, โˆ† ๐‘ƒ๐‘€๐‘… ~ โˆ† ๐‘„๐‘€๐‘ƒ So, โˆ† ๐‘ƒ๐‘€๐‘… ~ โˆ† ๐‘„๐‘€๐‘ƒ If two triangles are similar , then the ratio of their corresponding sides are equal ๐‘ƒ๐‘€/๐‘„๐‘€=๐‘€๐‘…/๐‘€๐‘ƒ PM ร— MP = MR ร— QM PM2 = MR ร— QM Hence proved

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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