Construct an isosceles triangle whose base …

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 aretimes 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.

If base =2x=8

x=4

Side^​​​2​​ =x^{​​​​​​2}+h^{​​​​2}=16+16=32

side=4√2=4×1.41=5.64 cm

AB=AC=5.64 Cm BC=8 cm