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PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM.MR
Let's construct a diagram according to the given question.
We know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to each other and to the whole triangle.
In ΔPQR we have,
∠QPR = 90° and PM ⊥ QR
In ΔPQR and ΔMQP
∠QPR = ∠QMP = 90°
∠PQR = ∠MQP (commom angle)
⇒ ΔPQR ~ ΔMQP (AA Similarity) --------(1)
In ΔPQR and ΔMPR
∠QPR = ∠PMR = 90°
∠PRQ = ∠PRM (commom angle)
⇒ ΔPQR ~ ΔMPR (AA Similarity) --------(2)
From equation (1) and (2)
ΔMQP ~ ΔMPR
⇒ PM2 = QM.MR
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM.MR
NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.5 Question 2
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Hence proved that PM2 = QM.MR.
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