PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM.MR

Solution:

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

PM / MR = QM / PM (corresponding sides of similar triangles are proportional)

⇒ PM² = QM.MR

Hence proved.

☛ Check: NCERT Solutions for Class 10 Maths Chapter 6

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Video Solution:

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

Summary:

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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