In the given figure, M is the midpoint of QR. &ang;PRQ = 90°. Prove that, PQ2 = 4PM2 – 3PR2.

Given: M is the midpoint of QR. &ang;PRQ = 90°. To prove: PQ2 = 4PM2 – 3PR2 Proof: In ∆PRM,&ang;PRM = 90° ...(Given)By Pythagoras theorem,&there4; PM2 = PR2 + RM2&there4; RM2 = PM2 − PR2 ...(1) In ∆PRQ,&ang;PRQ = 90° ...(Given)By Pythagoras theorem,&there4; PQ2 = PR2 + RQ2&there4; PQ2 = PR2 + (RM + MQ)2 ...[M is the midpoint of QR]&there4; PQ2 = PR2 + (RM + RM)2&there4; PQ2 = PR2 + (2RM)2&there4; PQ2 = PR2 + 4RM2&there4; PQ2 = PR2 + 4(PM2 − PR2) ...(from 1)&there4; PQ2 = PR2 + 4PM2 − 4PR2&there4; PQ2 = 4PM2 − 3PR2 Hence, PQ2 = 4PM2 – 3PR2.

In the given figure, M is the midpoint of QR. ∠PRQ = 90°. Prove that, PQ2 = 4PM2 – 3PR2

प्रश्न

In the given figure, M is the midpoint of QR. ∠PRQ = 90°. Prove that, PQ²= 4PM²– 3PR².

योग

उत्तर

Given: M is the midpoint of QR. ∠PRQ = 90°.

To prove: PQ²= 4PM²– 3PR²

Proof:

In ∆PRM,
∠PRM = 90° ...(Given)
By Pythagoras theorem,
∴ PM² = PR² + RM²∴ RM² = PM² − PR² ...(1)

In ∆PRQ,
∠PRQ = 90° ...(Given)
By Pythagoras theorem,
∴ PQ² = PR² + RQ²
∴ PQ² = PR² + (RM + MQ)² ...[M is the midpoint of QR]
∴ PQ² = PR² + (RM + RM)²
∴ PQ² = PR² + (2RM)²
∴ PQ² = PR² + 4RM²
∴ PQ² = PR² + 4(PM² − PR²) ...(from 1)
∴ PQ² = PR² + 4PM² − 4PR²
∴ PQ² = 4PM² − 3PR²