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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