constuct a triangle abc in which ab=5,angle b=60degree,altitude cd=3.construct a triangle aqr similar to triangle abc such that the side of triangle aqr is 1.5 times that of corresponding sodes of triangle acb
Following are the steps of construction : -
1) Construct a line segment AB of length 5 cm.
2) From B, construct an angle of 60° using compass. Name it as ∠ABX.
3) Construct a line EF parallel to line segment AB at a height of 3 cm such that it intersect 
at point C. Join AC.
4) Construct CD perpendicular to AB. CD is the altitude of the triangle ABC of height 3 cm.
5) Construct 
making an acute angle with AB, on the side opposite to vertex C and mark 3 points (Corresponding to the greater of 2 or 3) X1, X2 and X3 on AL such that AX1 = X1X2 = X2X3.
6) Join X2 B (Corresponding to the denominator of the given ratio) and construct a line passing through X3 (Corresponding to the numerator of the given ratio) and parallel to X2B. Let this line intersect the extended line AB at R.
7) Now, construct a line passing through R and parallel to BC, which intersects the extended line AC at point Q.
8) Thus, AQR is the required triangle which is similar to the given triangle ACB and with sides 
of the sides of the given triangle.
