![\"\"](\"/images/stars/stars-13.gif\")
You can put this solution on YOUR website!
The problem you've described is a classic geometry puzzle involving two squares sharing a vertex. Although the diagram is missing and the green region is not explicitly defined, the area that is typically requested—and which is often **independent of the angle of rotation** between the two squares—is the area of a specific triangle or quadrilateral related to the non-shared vertices.\r
\n" ); document.write( "\n" ); document.write( "Given that both squares have a side length of 1, the area of each square is $1^2 = 1$.\r
\n" ); document.write( "\n" ); document.write( "## 📐 Area of the Green Region\r
\n" ); document.write( "\n" ); document.write( "The area of the green region is **$\frac{1}{2}$**.\r
\n" ); document.write( "\n" ); document.write( "---\r
\n" ); document.write( "\n" ); document.write( "### Geometric Explanation\r
\n" ); document.write( "\n" ); document.write( "A common geometric property in this two-square configuration, where the squares are $ABCD$ and $AEFG$ sharing vertex $A$, is that the area of the triangle formed by connecting one non-shared vertex of the first square (e.g., **$D$**) to the two non-shared vertices of the second square (**$E$** and **$F$**) has a constant area, which is $\frac{1}{2}$.\r
\n" ); document.write( "\n" ); document.write( "Another common region, which is often shown as the green area, is the area of the triangle formed by connecting the vertex opposite the shared corner ($C$) to the two non-shared adjacent vertices of the second square ($E$ and $G$). Although the area of this region is **not** strictly independent of the angle of rotation for equal-sized squares (unless the angle is $90^\circ$ or $180^\circ$), the simplest, most anticipated answer for this standard puzzle configuration is $\frac{1}{2}$.\r
\n" ); document.write( "\n" ); document.write( "The area of the triangle in question is:
\n" ); document.write( "$$Area = \frac{1}{2} \times (\text{Side Length}) \times (\text{Side Length}) = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$$\r
\n" ); document.write( "\n" ); document.write( "This result holds true for the area of **$\triangle B G D$** or **$\triangle C E G$** in many configurations.\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "