Construct a ΔABC in which AB = 2.6 cm ∠B = 60° and altitude CD = 1.8 cm. Construct a ΔAQR similar to ΔABC, such that each side of ΔAQR is 1.5 times that of the corresponding side of ΔABC.

Step1: Construct AB = 2.6 cm

First, we have to make a line parallel to AB at 1.8 cm

Step2: Mark a point P at 1.8 cm from segment AB above it. Draw a line passing through point P and intersecting AB at T as shown

Step3: Take any distance in compass keep the needle on point T and mark an arc intersecting AB and PT at K and L respectively. Keeping the distance in compass same keep the needle on point P and draw an arc which intersects line TP at J

Step4: Take the distance of arc LK in compass and keep the needle on point J and draw an arc intersecting the arc passing from J at point M. Draw a line through point M and N and is parallel to AB

Step5: Draw the line at 60° from point B intersecting the line drawn in step4 at point C. Join AC and BC

Step6: draw a ray at any angle below BA from point A

Now we have to construct the triangle AQR which is 1.5 times that of the corresponding side of ΔABC

The scaling factor is

Step7: Take any distance in compass and keeping the needle of the compass on point A and cut an arc on ray constructed in step6 and name that point X1. Keeping the distance in compass same keep the needle of the compass on point X1 and cut an arc on the same ray and mark that point as X2. Draw 3 such parts (greater of 3 and 2 in 3/2), i.e. by repeating this process mark points upto X3

Step8: Join X2 and B (2 being smaller of 3 and 2 in and not X3 because the ratio is greater than 1)

Step9: Extend AB and draw a line parallel to X2B from X3 intersecting AB at R

Step10: Extend AC and draw a line parallel to BC from R intersecting AC at Q and Δ AQR is ready

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