Question

in triangle PQR, angle Q is an acute angle. Show that PR^2<PQ^2+QR^2

Answer 1

Answer:   Showed.

Step-by-step explanation: Since we can consider any triangle with Q as acute angle, so let us consider right-angled triangle PQR with ∠Q acute<90°.

Also, ∠p = 90° = right angle and QR is the hypotenuse of ΔPQR.

Therefore, using Pythagoras theorem, we can write

$PR^2+PQ^2=QR^2\\\\\Rightarrow PR^2+PQ^2+PQ^2=QR^2+PQ^2\\\\\Rightarrow PR^2+2PQ^2=QR^2+PQ^2.$

Since PQ²>0, so PR²<PQ²+QR².

Hence showed.