# A rectangular boa

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.

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