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.

#### Solution

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 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.** From point R, draw RP||CA meeting BA produced at P

Then, ΔPBR is the required triangle.

**Justification:**

∵ B_{4}R || B_{3}C

∴ `(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.