Question

the given figure, angle PQR 90 and QS perpedicular PR. Prove that RQ^2/PQ^2=RS/PS​

Answer 1

Answer:

Given data:

∠PQR = 90° ∴ ∆PQR is a right-angled triangle

QS ⊥ PR i.e., ∠QSP = ∠QSR = 90°

To prove: RQ²/PQ² = RS/PS

Case 1:  

Let’s consider, in right-angled ∆PQR and ∆QPS, we have

∠QPS = ∠QPR ….. [commom angle]

∠PQR = ∠PSQ = 90°   [given]

PQ is common side to both the triangles

∴ By AAS congruence,

∆PQR ~ ∆QPS  

Now, by using the property of area of similar triangles, we have

ar(∆QPS) / ar(∆PQR) = PS² / PQ²

⇒ [1/2 * PS * QS] / [1/2 * PQ * RQ] = PS² / PQ²

⇒ PQ² * QS = PS * PQ * RQ …… (i)

Case 2:

Let’s consider, in right-angled ∆PQR and ∆QSR, we have

∠QRS= ∠QRP ….. [commom angle]

∠PQR = ∠RSQ = 90°   [given]

RQ is common side to both the triangles  

∴ By AAS congruence,

∆PQR ~ ∆QSR  

Similarly, by using the property of area of similar triangles, we have

ar(∆QSR) / ar(∆PQR) = RS² / RQ²

or, [1/2 * RS * QS] / [1/2 * PQ * RQ] = RS² / RQ²

or, RQ² * QS = RS * PQ * RQ …… (ii)

Case 3:

On dividing eq.(ii) by (i), we get

[RQ² * QS] / [PQ² * QS] = [RS * PQ * RQ] / [PS * PQ * RQ]  

cancelling all the similar terms  

RQ² / PQ² = RS / PS

Hence proved