Draw a right ∆ABC in which BC=12 cm, AB=5 cm and ∠B=90°. Construct a triangle similar to it and of scale factor 2/3. Is the new triangle also a right triangle?

Thinking process:

Here, scale factor, i.e, m<n then the triangle to be constructed is smaller than the given triangle. Use this concept and then construct the required triangle.

Steps of construction:

1. Draw a line segment BC=12 cm.

2. From B draw a line which makes a right angle.

a. Now as point B is the initial point as the centre, draw an arc any radius such that, the arc meets the ray BC at point D.

b. With D as centre and with the same radius as before, draw another arc cutting the previous one at point E.

c. Now with E as centre and with the same radius, draw an arc cutting the first arc (drawn in step a) at point F.

d. With E and F as centres, and with a radius more than half the length of FE, draw two arcs intersecting at point G.

e. Join points B and G. The angle formed by GBC is 90°. i.e.

∠ GBC = 90°.

3. From B as centre draw an arc of 5 cm which intersects the line GB at A.

4. Join AC, ABC is the given right triangle.

5. From B draw an acute CBH downwards.

6. On ray BH, mark three point B₁,B₂ and B_3, such that

BB₁=B₁B₂= B₂B₃

7. Join B₃C

8. From point B₂ draw B₂N||B₃C intersect BC at N.

9. From point N draw NM||CA intersect BA at M. ∆MBN is the required triangle. ∆MNB is also a right angled triangle at B.

Justification:

As per the construction, ∆MNB is the required triangle

Let

BB₁=B₁B₂=B₂B₃=x

From triangles ∆BNB₂ and ∆BCB₃, we can say that they are similar

[by AA congruency criteria, ∠ B is common in both triangles and ∠BNB₂ =∠BCB₃ and are corresponding angles of the same transverse as B₃C || B₂N]

---- (i)

Similarly,

∆MBN and ∆ABC are similar.

[by AA congruency criteria, ∠ A is common in both triangles and ∠BAC=∠BMN and are corresponding angles of the same transverse as AC || MN]

Hence

---- (ii)

From (i) and (ii) the constructed triangle ∆MNB is of scale times of the ∆ABC.

Also, as we can clearly see, ∠B = 90° is common in both the triangles ∆ABC and ∆MNB. Hence the constructed triangle ∆MNB is also a right angle triangle.