A rectangular board is to be cut along the diagonal and the pieces rearranged to form an equilateral triangle; and the sides of the triangle must be 50 centimetres. What should be the length and breadth of the rectangle?

Let ABCD be the rectangle and BD be the diagonal and EFG be the derived equilateral triangle. Let EH be perpendicular to FG.

Given,
Δ EFG is an equilateral triangle
⇒ FG = FE = EG = 50 cm.
⇒ FH = HG
(∵ In equilateral triangle median is same as altitude)

⇒ FG = FH + GH = 50 cm
⇒ 2(FH) = 2(GH) = 50 cm
⇒ FH = GH = 25 cm

Also, ∠FEH = 2(∠FEH) = 2(∠GEH) = 60°
⇒ ∠FEH = ∠GEH = 30°
(∵ In equilateral triangle altitude bisects the angle)

We know that sides of any triangle of angles 30°, 60° and 90°
are in the ratio 1:√3:2.

⇒ 25 : EH : 50 = 1:√3:2
⇒ 25 : EH = 1:√3
⇒ EH = 25√3 cm

∴ IN Δ EFG and rect. ABCD
We get,

Δ DBC ≈ Δ EFG

(∵ Δ EFG is same as Δ BDC because both are part of same rectangle divided be diagonal BD.)

∴ AB = CD = FH = HG = 25 cm and BC = EH = 25√3 cm

Length and Breadth of the rectangle is 25√3 cm and 25 cm.