If in a triangle the sum of the squares of two sides is equal to the square of the third side then the triangle is a right-angled triangle, the angle between the first two sides being a right angle.

Given In the ∆XYZ, XY\(^{2}\) + YZ\(^{2}\) = XZ\(^{2}\)

To prove ∠XYZ = 90°

**Construction:** Draw a ∆PQR in which ∠PQR
= 90° and PQ = XY, QR = YZ

**Proof:**

In the right-angled ∆PQR, PR\(^{2}\) = PQ\(^{2}\) + QR\(^{2}\)

Therefore, PR\(^{2}\) = XY\(^{2}\) + YZ\(^{2}\) = XZ\(^{2}\)

Therefore, PR = XZ

Now, in ∆XYZ and ∆PQR, XY = PQ, YZ = QR and XZ = PR

Therefore, ∆XYZ ≅ ∆PQR (by SSS criterion of congruency)

Therefore, ∠XYZ = ∠PQR = 90° (CPCTC)

Problems on Converse of Pythagoras’ Theorem

**1.** If the sides of a triangle are in the ratio 13:12:5, prove that the triangle is a right-angled triangle. Also state which angle is the right angle.

**Solution:**

Let the triangle be PQR.

Here the sides are PQ = 13k, QR = 12k and RP = 5k

Now, QR\(^{2}\) + RP\(^{2}\) = (12k)\(^{2}\) + (5k)\(^{2}\)

= 144k\(^{2}\) + 25k\(^{2}\)

= 169k\(^{2}\)

= (13k)\(^{2}\)

= PQ\(^{2}\)

Therefore, by converse of Pythagoras theorem, PQR is a right-angled triangle in which ∠R = 90°.

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