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 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.
∵ 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.