Construction Procedure:

1. Draw a line segment AB =5 cm.

2. Take A and B as centre, and draw the arcs of radius 6 cm and 5 cm respectively.

3. These arcs will intersect each other at point C and therefore ΔABC is the required triangle with the length of sides as 5 cm, 6 cm, and 7 cm respectively.

4. Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C.

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

6. Join the points BA₅ and draw a line from A₇ to BA₅ which is parallel to the line BA₅ where it intersects the extended line segment AB at point B’.

7. Now, draw a line from B’ the extended line segment AC at C’ which is parallel to the line BC and it intersects to make a triangle.

8. Therefore, ΔAB’C’ is the required triangle.

Justification:

The construction of the given problem can be justified by proving that

AB’ = (7/5)AB

B’C’ = (7/5)BC

AC’= (7/5)AC

From the construction, we get B’C’ || BC

∴ ∠AB’C’ = ∠ABC (Corresponding angles)

In ΔAB’C’ and ΔABC,

∠ABC = ∠AB’C (Proved above)

∠BAC = ∠B’AC’ (Common)

∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion)

Therefore, AB’/AB = B’C’/BC= AC’/AC …. (1)

In ΔAA₇B’ and ΔAA₅B,

∠A₇AB’=∠A₅AB (Common)

From the corresponding angles, we get,

∠A A₇B’=∠A A₅B

Therefore, from the AA similarity criterion, we obtain

ΔA A₂B’ and A A₃B

So, AB’/AB = AA₅/AA₇

Therefore, AB /AB’ = 5/7 ……. (2)

From the equations (1) and (2), we get

AB’/AB = B’C’/BC = AC’/ AC = 7/5

This can be written as

AB’ = (7/5)AB

B’C’ = (7/5)BC

AC’= (7/5)AC

Hence, justified.