Question

Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of an equilateral triangle described on one of its diagonals
(Can the figure be shown below?)​

Answer 1

Answer:

Given:

ABCD is a Square,

DB is a diagonal of square,

△DEB and △CBF are Equilateral Triangles.

To Prove:

$\frac{ar(CBF)}{ar(DEB)} =\frac{1}{2}$  

​

Proof:

Since, △DEB and △CBF are Equilateral Triangles.

∴ Their corresponding sides are in equal ratios.

In a Square ABCD,    $DB=BC \sqrt{2}$              (1)

∴ $\frac{ar(CBF)}{ar(DBE)} = \frac{\sqrt{3}/4*(BC)^{2} }{\sqrt{3}/4*(DB)^{2}}$

∴$\frac{ar(CBF)}{ar(DBE)} = \frac{\sqrt{3}/4*(BC)^{2} }{\sqrt{3}/4*(BC\sqrt{2} )^{2}}$    (From 1)

∴ $\frac{ar(CBF)}{ar(DEB)} =\frac{1}{2}$  

             

See the attachment.

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