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Answer variants:
Draw a line segment \(QR = 6 \ cm\).
Draw a perpendicular bisector \(AB\) of line segment \(RS\).
Take \(13 \ cm\) as the radius, with \(Q\) as centre, draw an arc that cuts \(QY\) at \(S\), then draw a line from \(R\) to \(S\) to connect the two points.
Mark a point \(P\), where the perpendicular bisector intersects \(QS\), then join \(PR\).
Make \(\angle YQR = 45^\circ\).
Construct a triangle \(PQR\) in which \(QR = 6 \ cm\), \(\angle Q = 45^\circ\) and \(PQ + PR = 13 \ cm\).
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Thus, \(PQR\) is the required triangle.
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