Question

In the given figure, triangle PQR is right-angled at Q. S is the
mid-point of side QR. Prove that QR2 = 4 (PS2 - PQ2).​

Answer 1

Key points :

1) Read the question properly

2) Use pythagoras theoremin triangle PQS

3)Use QS=QR/2

4) Substitute the required value and keepon solving and your sum is DONE!!

Answer:

Given: In triangle PQR, ∠ PQR = 90° and S is the mid-point of QR.

To prove: QR2 = 4 (PS2 – PQ2)

Proof: In right-angled ∆ PQS, by Pythagoras theorem,

PQ2 + QS2 = PS2

⇒ QS^2 = PS^2 – PQ^2 … (i)

Since S is the mid-point of side QR,

∴QS=QR/2

Substituting the value of QS in equation (i),

=(QR/2)^2=PS^2−PQ^2

=QR^2/4=PS^2−PQ^2

Shifting the the 4 that is in the LHS to RHS

therefore we have,

QR^2=4(PS^2−PQ^2)