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 B_{1},B_{2},B_{3} and B_{4} on BX such that BB_{1}=B_{1}B_{2}=B_{2}B_{3}=B_{3}B_{4}

7. Join B_{3}C and from B_{4} draw a line B_{4}R||B_{3}C 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,

B_{4}R││B_{3}C

∴ BC/CR = 3/1

Now,

∴

Now,

⇒

⇒

⇒ BR/BC = 4/3

Also,

RP||CA

∴ ΔABC ~ ΔPBR

and

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

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