Construct a triangle similar to a given triangle \(PQR\) with its sides equal to \(\frac{5}{2}\) of the corresponding sides of the triangle \(PQR\).
Construction:
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
Answer variants:
Draw a ray \(QX\), making an acute angle with \(QR\) on the side opposite to vertex \(P\).
Draw a line through \(R'\) parallel to the line \(RP\) intersecting the extended line segment \(QP\) at \(P'\). Then, \(P'QR'\) is the required triangle, each of whose side is five - twos of the corresponding sides of \(\triangle PQR\).
Construct a triangle \(PQR\) with any measurement.
Locate \(5\) (the greater of \(2\) and \(5\) in \(\frac{5}{2}\)) points. \(Q_1\), \(Q_2\), \(Q_3\), \(Q_4\) and \(Q_5\) on \(QX\) so that \(QQ_1 = Q_1Q_2 = Q_2Q_3 = Q_3Q_4 = Q_4Q_5\).
Join \(Q_2\) (the 2nd point, 2 being smaller of \(2\) and \(5\) in \(\frac{5}{2}\)) to \(R\) and draw a line through \(Q_5\) parallel to \(Q_2R\), intersecting the extended line segment \(QR\) at \(R'\).