# Converse of Pythagoras’ Theorem

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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