Question

Prove that the area of an equilateral triangle described on one side of a side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
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Answer 1

ABCD is a Square,

DB is a diagonal of square,

△DEB and △CBF are Equilateral Triangles.

To Prove:

A(△DEB)

A(△CBF)

=

1/2

Answer 2

Given :

ABCD is a square whose one diagonal is AC. ΔAPC and ΔBQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.

Proof :

Area(ΔBQC) = ½ Area(ΔAPC)

Since, ΔAPC and ΔBQC are both equilateral triangles, as per given,

∴ ΔAPC ~ ΔBQC [AAA similarity criterion]

∴$\frac{ area(ΔAPC)}{area(ΔBQC)}$ = ($\frac{AC²}{BC²}) =$ $\frac{ AC²}{BC²}$

Since, Diagonal = √2 side = √2 BC = AC

($\frac{√2BC}{BC}$)2=2

⇒ area(ΔAPC) = 2 × area(ΔBQC)

⇒ area(ΔBQC) = ½ area(ΔAPC)

$\large\sf\orange{Hence 　 proved ✓}$