Q. 134.7( 3 Votes )

# 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.

Answer :

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 X_{1}. Keeping the distance in compass same keep the needle of the compass on point X_{1} and cut an arc on the same ray and mark that point as X_{2}. Draw 3 such parts (greater of 3 and 2 in 3/2), i.e. by repeating this process mark points upto X_{3}

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

Step9: Extend AB and draw a line parallel to X_{2}B from X_{3} 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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