Question

in triangle pqr angle Q is an acute angle show that PR^2< PQ^2 + Q R^2

Answer 1

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Answer 2

Step-by-step explanation: We are a triangle PQR, where ∠Q is acute, i.e., ∠Q<90°. And we are to prove that  PR² < PQ² + QR².

For that, we take the help of a right angled triangle PQR (see the attached figure), where ∠P = 90°, ∠Q  is an acute angle.

Now, since ΔPQR ia a right angled one, so using Pythagoras theorem, we can write

QR² = PQ² + PR²

i.e., PR² = QR² -  PQ²

i.e., PR² + 2PQ² = QR² +  PQ².

Since the square of a number cannot be negative, so PQ²>0, implies 2PQ²>0.

Thus, PR² <  PQ² + QR².  Hence showed.

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