Question

Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose side are 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 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

In ΔABC and ΔAB'C',

∠ABC = ∠AB'C' (Corresponding angles)

∠BAC = ∠B'AC' (Common)

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

… (1)

In ΔAA₂B and ΔAA₃B',

∠A₂AB = ∠A₃AB' (Common)

∠AA₂B = ∠AA₃B' (Corresponding angles)

∴ ΔAA₂B ∼ ΔAA₃B' (AA similarity criterion)

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

⇒

This justifies the construction.