## Area of a Square Formed by the Sides of Triangles Inscribed into a Square

The area of the large square is the sum of the area of the smaller square plus four times the area of the triangles ABC and ADC are right angled and similar with scale factor one half since angle

\[\angle BAC= \alpha \rightarrow \angle ABC=90- \alpha \rightarrow \angle CBD= 90-(90- \alpha)= \alpha \rightarrow \angle BCD=90- \alpha\]

.If

\[Ab=x\]

then \[BC=\frac{x}{2}\]

and the area of triangle ABD is \[\frac{1}{2} \times x \times \frac{x}{2}=\frac{x^2}{4}\]

The areas of triangles ABC and BDC are in the ratio

\[(2)^2= 4\]

so Area ABC is \[\frac{4}{1+4} \times \frac{x^2}{4}=\frac{x^2}{5}\]

.Then the area of the green square is

\[x^2 - 4 \times \frac{x^2}{5}=\frac{x^2}{5}\]

.The area of the Green square is one fifth the area of the big square.