Construct a triangle similar to a given triangle \(PQR\) with its sides equal to \(\frac{7}{3}\) of the corresponding sides of the triangle \(PQR\) (scale factor \(\frac{7}{3} > 1\))
Arrange the steps of constructing in proper order
\(Step-1:\)
\(Step-2:\)
\(Step-3:\)
\(Step-4:\)
\(Step-5:\)
Then, \(P'QR'\) is the required triangle, each of whose side is seven - fourths of the corresponding sides of \(\triangle PQR\).
Answer variants:
Draw a line through \(R'\) parallel to the line \(RP\) to intersect \(QP\) at \(P'\).
Locate \(7\) points (the greater of \(7\) and \(3\) in \(\frac{7}{3}\)). \(Q_1\), \(Q_2\), \(Q_3\), \(Q_4\), \(Q_5\), \(Q_6\) and \(Q_7\) on \(QX\) so that \(QQ_1 = Q_1Q_2 = Q_2Q_3 = Q_3Q_4 = Q_4Q_5 = Q_5Q_6 = Q_6Q_7\).
Draw a ray \(QX\), making an acute angle with \(QR\) on the side opposite to vertex \(P\).
Join \(Q_3\) (the third point, \(3\) being smaller of \(7\) and \(3\) in \(\frac{7}{3}\)) to \(R\) and draw a line through \(Q_7\) parallel to \(Q_3R\), intersecting the extended line segment \(QR\) at \(R'\).
Construct a triangle \(PQR\) with any measurement.