Question

PROVE
Pqsquar=4MPsquar - 3PRsquar​

Answer 1

EXPLANATION.

⇒ In ΔPQR.

M is the midpoint of QR.

⇒ ∠PRQ = 90°.

To prove : (PQ)² = 4PM² - 3PR².

As we know that,

Formula of Pythagoras Theorem.

⇒ [Hypotenuse]² = [Perpendicular]² + [Base]².

⇒ H² = P² + B².

In ΔPRM.

⇒ (PM)² = (PR)² + (RM)². - - - - - (1).

⇒ In ΔPRQ.

⇒ (PQ)² = (PR)² + (RQ)².

⇒ RQ = RM + MQ.

⇒ RQ = RM + RM.

⇒ RQ = 2RM.

⇒ (PQ)² = (PR)² + (RM + MQ)².

⇒ RM + MQ = 2RM.

⇒ (PQ)² = (PR)² + (2RM)².

⇒ (PQ)² = (PR)² + 4(RM)².

Put the value of (RM)² = (PM)² - (PR)².

⇒ (PQ)² = (PR)² + 4[PM² - PR²].

⇒ PQ² = PR² + 4PM² - 4PR².

⇒ PQ² = 4PM² - 3PR².

Hence Proved.

Answer 2

Step-by-step explanation:

given :

PROVE

Pqsquar=4MPsquar - 3PRsquar

to find :

Pqsquar=4MPsquar - 3PRsquar

solution :

• pq = 4pm - 3 pR