Draw a ∆ABC in which AB=5 cm, BC=6 cm and ∠ABC=60°. Construct a triangle similar to ABC with scale factorJustify the construction.

Steps of construction:

1. Draw a line segment AB=5 cm.

2. From point B, draw ∠ABY=60°.

To measure angle at B:

a. With B as centre and with any radius, draw another arc cutting the line AB at D.

b. With D as centre and with the same radius, draw an arc cutting the first arc (drawn in step a) at point E.

c. Draw a ray BY passing through E which forms an angle of 60° with the line AB.

3. With B as centre and radius of 6 cm draw an arc intersecting the line BY at C.

4. Join AC, ∆ABC is the required triangle.

5. From A, draw any ray AX downwards making an acute angle.

6. Mark 7 points A₁,A₂,A₃,A₄,A₅,A₆ and A₇ on AX such that AA₁=A₁A₂=A₂A₃=A₃A₄=A₄A₅=A₅A₆=A₆A₇

7. Join A₇B and from A₅ draw A₅M||A₇B intersecting AB at M.

8. From point M draw MN││BC intersecting AC at N. Then, ∆AMN is the required triangle whose sides are equal to of the corresponding sides of ∆ABC.

Justification:

Let AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇ = x

As per the construction A₅M││A₇B

Now,

Also MN || BC,

Therefore rAMC is congruent to rAMN

Thus,

Hence, the new triangle rAMN is similar to the given triangle rABC and its sides are times of the corresponding sides of rABC.