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

Give the justification of the construction

#### Solution

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) A_{1}, A_{2}, and A_{3} on AX such that AA_{1} = A_{1}A_{2} = A_{2}A_{3}.

**Step 5**

Join BA_{2} and draw a line through A_{3} parallel to BA_{2} 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 = ∠AB'C' (Corresponding angles)

∠BAC = ∠B'AC' (Common)

∴ ΔABC ∼ ΔAB'C' (AA similarity criterion)

`=> (AB)/(AB')= (BC)/(B'C') = (AC)/(AC') ....(1)`

In ΔAA_{2}B and ΔAA_{3}B',

∠A_{2}AB = ∠A_{3}AB' (Common)

∠AA_{2}B = ∠AA_{3}B' (Corresponding angles)

∴ ΔAA_{2}B ∼ ΔAA_{3}B' (AA similarity criterion)

`=> (AB)/(AB') = (`

`=>(AB)/(AB') = 2/3 .....2`

On comparing equations (1) and (2), we obtain

`(AB)/(AB')=(BC)/(B'C') = (AC)/(AC') = 2/3`

`=> AB' = 3/2 AB, B'C' = 3/2 BC, AC' = 3/2 AC`

This justifies the construction.