In the given figure, OQ : PQ = 3.4 and perimeter of Δ POQ = 60 cm. Determine PQ, QR and OP.

#### Solution

In the figure,

∠ PQD. Therefore we can use Pythagoras theorem to find the side *PO*.

`PO^2=PQ^2+OQ^2` …… (1)

In the problem it is given that,

`(OQ)/(PQ)=3/4`

`OQ= 3/4PQ`.....(2)

Substituting this in equation (1), we have,

`PQ^2=(9PQ^2)/16+PQ^2`

`PQ^2=(25PQ^2)/16`

`PQ^2=sqrt((25PQ^2)/16)`

`PQ=5/4PQ`........(3)

It is given that the perimeter of Δ POQis 60 cm. Therefore,

*PQ + OQ + PO* = 60

Substituting (2) and (3) in the above equation, we have,

`PQ+3/4PQ+5/4PQ=60`

`12/4PQ=60`

`3PQ=60`

`PQ=20`

Substituting for *PQ* in equation (2), we have,

`PD=5/4xx15`

`OQ=3/4xx20`

`OQ=15`

*OQ* is the radius of the circle and *QR* is the diameter. Therefore,

*QR* = *2OQ*

*QR* = 30

Substituting for *PQ* in equation (3), we have,

`PD=5/4xx20`

`PO=25`

Thus we have found that *PQ* = 20 cm, *QR* = 30 cm and *PO* = 25 cm.