Last updated at June 23, 2017 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

Example 5
Important

Example 8 Important

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 You are here

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

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .