Question

Obtain all In ∆ PQR, ∠Q = 90°, Show that PR2 = 9PS2 – 8PQ2

Answer 1

Answer: Showed

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.