Draw an isosceles triangle ABC in which AB=AC=6 cm and BC=5 cm. Construct a triangle PQR similar to ∆ABC in which PQ=8 cm. Also justify the construction.

Thinking process:

I. Here, for making two similar triangles with one vertex is same of base, we assume that,

In ∆ABC and ∆PQR: vertex B=vertex Q

So, we get the required scale factor.

II. Now, construct a ∆ABC and then a ∆PBR, similar to ∆ABC whose sides areof the corresponding sides of the ∆ABC.

Let ∆PQR and ∆ABC are similar triangles, and then its scale factor between the corresponding sides is

Steps of construction:

1. Draw a line segment BC=5 cm.

2. Construct OQ the perpendicular bisector of line segment BC meeting BC at P’.

To bisect BC:

a. With B as centre and any radius more than half of the length of BC, draw two arcs on either side of BC.

b. Similarly, with C as centre and any radius more than half of the length of BC; draw two arcs on either side of BC which intersect with the previous arcs at M and N.

c. Join MN to meet the line BC at P’, which is the mid-point.

3. Taking B and C as centres draw two arcs of equal radius 6 cm intersecting each other at A.

4. Join BA and CA. So, ∆ABC is the required isosceles triangle.

5. From B, draw any ray BX making an acute angle, ∠CBX.

6. Locate four points B₁,B₂,B₃ and B₄ on BX such that BB₁=B₁B₂=B₂B₃=B₃B₄

7. Join B₃C and from B₄ draw a line B₄R||B₃C intersecting the extended line segment BC at R.

8. From point R, draw RP||CA meeting the extended line BA at P.

Then, ∆PBR is the required triangles.

Justification:

Let BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = x

As per the construction B₄R││B₃C

Now,

Also, from the construction RP||CA

_{Therefore rABC is congruent to rPBR}

_{(by the AA criteria, as ∠ PBR = ∠ ABC, also as RP||CA, ∠ ACB is corresponding to ∠ PRB so ∠ ACB = ∠ PRB)}

and

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