Question

Two triangles are similar. Smaller triangle’s sides are 4 cm, 5 cm, 6 cm. Perimeter of larger triangle is 90 cm then find the sides of larger triangle.

Answer 1

Given: ΔABC ~ ΔPQR

In ΔABC, AB = 4 cm, BC = 5 cm, AC = 6 cm

In ΔPQR, PQ + QR + PR = 90 cm

To find: PQ, QR and PR

ΔABC ~ ΔPQR   ...[Given]

∴ `(AB)/(PQ) = (BC)/(QR) = (AC)/(PR)`   ...[Corresponding sides of similar triangles]

Let `(AB)/(PQ) = (BC)/(QR) = (AC)/(PR) = k`

∴ `4/(PQ) = 5/(QR) = 6/(PR) = k`   ...[Given]

∴ `4/(PQ) = k, 5/(QR) = k` and `6/(PR) = k`

∴ `PQ = 4/k, QR = 5/k` and `PR = 6/k`   ...(i)

∴ `PQ + QR + PR = 4/k + 5/k + 6/k`

∴ `90 = 15/k`   ...[Given]

∴ `k = 15/90`

= `1/6`

∴ PQ = `4/((1/6))`

= 4 × 6

= 24 cm   ...[From (i)]

QR = `5/((1/6))`

= 5 × 6

= 30 cm   ...[From (i)]

PR = `6/((1/6))` 

= 6 × 6

= 36 cm   ...[From (i)]

∴ The sides of the larger triangle are 24 cm, 30 cm and 36 cm.