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

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.