NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 6: Triangles - Exercise 6.3

Question 1

Step 1: We are given that \(PR^2 - PQ^2 = QR^2\).

Rearranging: \(PR^2 = PQ^2 + QR^2\).

Step 2: This is the Pythagoras theorem form. So, \(\triangle PQR\) is a right-angled triangle with the right angle at \(Q\).

Step 3: Draw altitude \(QM\) from vertex \(Q\) perpendicular to hypotenuse \(PR\). Here, \(M\) lies on side \(PR\).

Step 4: When an altitude is drawn from the right angle of a right triangle to the hypotenuse, the following property holds: \[ QM^2 = PM \times MR. \]

Step 5 (Proof of property):

• Consider triangles \(\triangle PQM\), \(\triangle QRM\), and \(\triangle PQR\).
• All three are right-angled and share common angles.
• Therefore, they are similar triangles (AA similarity rule).

Step 6: From similarity, we get two relations:
(i) \( \dfrac{QM}{PM} = \dfrac{QR}{PR} \)
(ii) \( \dfrac{QM}{MR} = \dfrac{PQ}{PR} \)

Step 7: Multiply the two equations: \[ \left( \dfrac{QM}{PM} \right) \times \left( \dfrac{QM}{MR} \right) = \dfrac{QR}{PR} \times \dfrac{PQ}{PR} \]

Step 8: The right-hand side becomes: \( \dfrac{QR \times PQ}{PR^2} \). But from Step 2, \(PR^2 = PQ^2 + QR^2\), so this fits the relation.

Step 9: Simplifying the left-hand side gives: \( \dfrac{QM^2}{PM \times MR} \).

Step 10: Cross multiplying, we finally get: \[ QM^2 = PM \times MR. \]

Therefore proved.