Question

Plz solve the 17th question it's urgent plzz solve it

Answer 1

Aditi your hands are so after and cute.

Answer 2

Answer:

Hello.  Hope this helps you.

(i)

∠PAQ = 180° - ∠APB - ∠ABP    [ sum of angles in triangle ]

  = 180° - 90° - ∠ABP                [ given BP ⊥ AC ]

  = 90° - ∠ABP

  = ∠PBR                                    [ given right angle at B ]

Since also ∠AQP = ∠BRP,        [ given right angles ]

the triangles AQP and BRP are similar.

Therefore

AQ / BR = PQ / PR   =>  AQ × PR = PQ × BR.

But BRPQ is a rectangle [ right angles as B, R, Q and therefore also at P ],

so BR = PQ.  Therefore

AQ × PR = PQ²

(ii)

Similarly to the above,

∠PCR = 90° - ∠RPC = ∠BPR

and ∠PRC = ∠BRP

so triangles PCR and BPR are similar.

Therefore

RB / RP = RP / RC  =>  RB × RC = RP².

But RB = PQ since BRPQ is a rectangle, so

PQ × RC = RP²