Question

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

Answer 1

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 = A₁A₂ = A₂A₃.

Step 5

Join BA₂ and draw a line through A₃ 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 = ∠AB'C' (Corresponding angles)

∠BAC = ∠B'AC' (Common)

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

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

In ΔAA₂B and ΔAA₃B',

∠A₂AB = ∠A₃AB' (Common)

∠AA₂B = ∠AA₃B' (Corresponding angles)

∴ ΔAA₂B ∼ ΔAA₃B' (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.