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Ex 6.5

Ex 6.5, 1
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Chapter 6 Class 10 Triangles

Serial order wise

Last updated at March 22, 2023 by Teachoo

Ex 6.5,2 (Method 1) PQR is a triangle right angled at P and M is a point on QR such that PM β₯QR. Show that PM2 = QM . MR Given: β πππ where β π ππ=90Β° & PM β₯QR To prove: PM2 = QM .MR Proof: In Ξ PQR, β π ππ = 90Β° So, Ξ PQR is a right triangle Using Pythagoras theorem in Ξ PQR Hypotenuse2 = (Height)2 + (Base)2 RQ2 = PQ2 + PR2 Now, in Ξ PMR, PM β₯ QR So, β PMR = 90Β° β΄ Ξ PMR is a right triangle Using Pythagoras theorem in Ξ PMR Hypotenuse2 = (Height)2 + (Base)2 PR2 = PM2 + MR2 Similarly, In Ξ PMQ, β PMQ = 90Β° β΄ Ξ PMR is a right triangle Using Pythagoras theorem in Ξ PMQ Hypotenuse2 = (Height)2 + (Base)2 PQ2 = PM2 + MQ2 So, our equations are RQ2 = PQ2 + PR2 β¦(1) PR2 = PM2 + MR2 β¦(2) PQ2 = PM2 + MQ2 β¦(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 Γ MR = 2PM2 + (MQ2 + MR2 ) (MQ2 + MR2 ) β (MQ2 + MR2 ) + 2 MQ Γ MR = 2PM2 0 + 2 MQ Γ MR = 2PM2 2 MQ Γ MR = 2PM2 MQ Γ MR = PM2 β PM2 = MQ Γ 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 β₯QR. Show that PM2 = QM . MR Given: β πππ where β π ππ=90Β° & PM β₯QR To prove: PM2 = QM .MR i.e. ππ/ππ = ππ /ππ 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, β πππ ~ β πππ So, β πππ ~ β πππ If two triangles are similar , then the ratio of their corresponding sides are equal ππ/ππ=ππ /ππ PM Γ MP = MR Γ QM PM2 = MR Γ QM Hence proved