Two equal sides of a triangle are each 5 meters less than twice the third side. If the perimeter of the triangle is 55 meters, find the length of its sides?

Let the triangle be Δ ABC.

The triangle has two equal sides, i.e. AB = BC.

Let the length of the third side, ℓ (AC), be x meters. ∴ ℓ (AB) = ℓ (BC) = (2x – 5) meters.

The perimeter of Δ ABC = 55 meters.

According to the given condition,

ℓ (AB) + ℓ (BC) + ℓ (AC) = 55

⇒ 2x – 5 + 2x – 5 + x = 55

5x – 10 = 55

⇒ 5x = 55 + 10 (Transposing 10 to RHS)

5x = 65

⇒ x = (Transposing 5 to RHS)

x = 13

The lengths of the sides of the triangle are

ℓ (AB) = 2x – 5 = 2 (13) – 5 = 26 – 5 = 21 meters.

ℓ (BC) = 2x – 5 = 2 (13) – 5 = 26 – 5 = 21 meters.

ℓ (AC) = x = 13 meters.

∴ The lengths of the sides of Δ ABC are 21 meters, 21 meters and 13 meters.