CONSTRUCT A TRIANGLE SIMILAR TO A GIVEN
THE
○ PQR WITH ITS SIDES EQUAL TO 3/5
OF THE CORRESPONDING SIDES OF
TRIANGLE PQR (SCALE FACTOR 315<1')
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Given a triangle PQR, we are required to construct another triangle whose sides 3/5 are of the corresponding sides of the triangle PQR. Steps of construction: (1) Draw any ray QX making an acute angle with QR on the side opposite to the vertex P. (2) Locate 3 (the greater of 2 and 3 in 2/3) points. Q1Q2, Q2 on QX so that QQ1 = Q1Q2 = Q2Q3 (3) Join Q3 R and draw a line through Q2 (the second point, 2 being smaller of 2 and 3 in 2/3) parallel to Q3 R to intersect QR at R’. (4) Draw line through R’ parallel to the line RP to intersect QP at P’. The ΔP’QR’ is the required triangle each of the whose sides is 2/3 of the corresponding sides of 3.
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