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Question
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.
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