Question

In the given figure PQRS is a cyclic quadrilateral PQ and SR produced meet at T

1) Prove ΔTPS ~ ΔTRQ.

2) Find SP if TP = 18 cm, RQ = 4 cm and TR = 6 cm

3) Find the area of quadrilateral PQRS if the area of ΔPTS = 27 cm².

Answer 1

1) Since `square PQRS` is a cyclic quadrilateral

`angle QRT = angle SPT`   ....(1)(exterior angle is equal to interior opposite angle)

In ΔTPS and ΔTRQ

`angle PTS = angle RTQ` . . . .(common angle)

`angleQRT =  angle SPT` . . . .(from 1)

⇒ ΔTPS ~ ΔTRQ . . . .(AA similarity criterion)

2) Since ΔTPS ~ ΔTRQ, implies that corresponding sides are proportional

`i.e (SP)/(QR) = (TP)/(TR)`

`=> (SP)/4 = 18/6`

`=> (SP)= (18xx4)/6`

`=> SP =  12cm`

3) Since ΔTPS ~ ΔTRQ

`(Ar(ΔTPS))/(Ar(ΔTRQ)) = "SP"^2/"RQ"^2`

`=> 27/(Ar(ΔTRQ)) = (27xx4xx4)/(12xx12)`

`=> Ar(ΔTRQ) = 3cm^2`

Now, Ar(`square PQRS`) = Ar(ΔTPS) Ar(ΔTRQ)= 27 - 3 = 24 cm²