Question

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.

Answer 1

Let ΔPQR and ΔABC are similar triangles, then its scale factor between the corresponding sides is `(PQ)/(AB) = 8/6 = 4/3`

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’.

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, ΔABC is the required isosceles triangle.

5. From B, draw any ray BX making an acute ∠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 BA produced at P
Then, ΔPBR is the required triangle.

Justification:

∵ B₄R || B₃C 

∴ `(BC) - (CR) = 3/1`

Now, `(BR)/(BC) = (BC + CR)/(BC)`

= `1 + (CR)/(BC) = 1 + 1/3 = 4/3`

Also, RP || CA

∴ ΔABC ∼ ΔPBR

And `(PB)/(AB) = (RP)/(CA) = (BR)/(BC) = 4/3`

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