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.

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

Steps of construction

1. Draw a line BC=5 cm.

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

3. Taking B and C as centers 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. Draw any ray BX making an acute CBX.

6. Locate 4 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. Draw RP||CA meeting BA produced at P.

Then, ∆PBR is the required triangles.

By construction,


BC/CR = 3/1



BR/BC = 4/3





Hence, the new triangle is similar to the given triangle whose sides are 4/3 times of the corresponding sides of the isosceles ∆ABC.