Question

construct APOR, such that Qp = 65 cm, LPQR=60
and PO-PR = 2.5​

Answer 1

Answer:

=> Here, PQ - PR = 2.5 cm

:- PQ > PR

=> As shown in the rough figure draw seg

=> QR = 6.5 CM

=> Draw a ray QT making on angles of 60⁰ with QR

=> Take a point S on ray QT, such that QS = 2.5 cm

=> Now, PQ - PS = QS [ Q - P - T ]

:- PQ - PS = 2.5 cm....1 (Given)

=> Also, PQ - PR = 2.5 cm (ii) [From (i) and (ii)

:- PQ - PS = PQ - PR

:- PS = PR

:- Point P is on the perpendicular bisector of seg RS

:- Points P is the intersection of ray QT and the perpendicular bisector of seg RS

Step of construction:-

1. Draw seg QR of length 6.5 cm.
2. Draw ray QT, such that angles RQT = 600.
3. Mark's point S on ray QT such that (QS) = 2.5 cm
4. Join point S and R.
5. Draw perpendicular bisector of seg SR interesting ray QT. Name the point as P.
6. Join the point P and R. Hence, angles PQR is the required triangle.

I hope this will help you dear..

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