Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are of the corresponding sides of the first triangle.

Give the justification of the construction.

Step 1

Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 7 cm radius respectively. Let these arcs intersect each other at point C. ΔABC is the required triangle having length of sides as 5 cm, 6 cm, and 7 cm respectively.

Step 2

Draw a ray AX making acute angle with line AB on the opposite side of vertex C.

Step 3

Locate 7 points, A₁, A₂, A₃, A₄ A₅, A₆, A₇ (as 7 is greater between 5and 7), on line AX such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇.

Step 4

Join BA₅ and draw a line through A₇ parallel to BA₅ to intersect extended line segment AB at point B'.

Step 5

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 ΔAA₅B and ΔAA₇B',

∠A₅AB = ∠A₇AB' (Common)

∠AA₅B = ∠AA₇B' (Corresponding angles)

∴ ΔAA₅B ∼ ΔAA₇B' (AA similarity criterion)

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

⇒

This justifies the construction.