Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1(1/2) times the corresponding sides of the isosceles triangle.

Let us assume that ΔABC is an isosceles triangle having CA and CB of equal lengths, base AB of 8 cm, and AD is the altitude of 4 cm.
A ΔAB’C’ whose sides are 3/2 times of ΔABC can be drawn as follows.
Step 1
Draw a line segment AB of 8 cm. Draw arcs of same radius on both sides of the line segment while taking point A and B as its centre.
Let these arcs intersect each other at O and O’. Join OO’. Let OO’ intersect AB at D.
Step 2
Taking D as centre, Draw an arc of 4 cm radius which cuts the extended line segment OO’ at point C. An isosceles ΔABC is formed, having CD (altitude) as 4 cm and AB (base) as 8 cm.
Step 3
Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C.
Step 4
Locate 3 points (as 3 is greater between 3 and 2) A₁, A₂, and A₃ on AX such that AA₁ = A₁ A₂ = A₂ A₃
Step 5
Join BA₂ and draw a line through A₁ parallel to BA₂ to intersect extended line segment AB at point B’.
Step 6
Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’. ΔAB’C is the required triangle.
Justification
The construction can be justified by proving that
AB’ = 3/2 AB, B’C’ = 3/2 BC, AC’ = 3/2 AC
In ΔABC and ΔAB’C’,
∠ABC and ∠AB’C’ (Corresponding angles)
∠BAC = ∠B’AC’ (Common)
∴ ΔAB’C’ ∼ ΔABC (AA similarity criterion)
⇒ AB’/AB = B’C’/BC = A’C’/AC …(1)
In ΔAA₂B = ΔAA₃B’
∠AA₂B = ∠AA₃B’ (Common)
∠AA₂B = AA₃B’ (Corresponding angles)
∴ ΔAA₂B ∼ AA₃B’ (AA Similarity Criterion)
⇒ AB/AB’ = AA₂/AA₃ ⇒ AB/AB’ = 2/3 …(2)
On comparing equations (1) and (2), we obtain
AB’/AB = B’C’/BC = AC’/AC = 3/2
⇒ AB’ = 3/2 AB,B’C’ = 3/2 BC, AC’ = 3/2 AC
This justifies the construction.