Question - Find the Area of a Triangular Section in a Rectangle

Let \( AF = x \), \( FC = 5x \), \( BF = y \), and \( BG = 3y \).

Since \( \triangle ABF \) is similar to \( \triangle EBF \) and \( \triangle BGC \), their sides are proportional.

The area \( A \) of \( \triangle ABF \) can be expressed with the base \( AF \) and height \( BF \):

\[ A_{\triangle ABF} = \frac{1}{2} AF \cdot BF = \frac{1}{2} x \cdot y \]

The area \( A \) of \( \triangle EBF \) is \( \frac{2}{5} \) of the area of \( \triangle ABF \):

\[ A_{\triangle EBF} = \frac{2}{5} A_{\triangle ABF} = \frac{2}{5} \cdot \frac{1}{2} x \cdot y = \frac{1}{5} xy \]

The area \( A \) of \( \triangle BGC \) is 3 times the area of \( \triangle ABF \):

\[ A_{\triangle BGC} = 3 A_{\triangle ABF} = 3 \cdot \frac{1}{2} x \cdot y = \frac{3}{2} xy \]

Since the rectangle \( ABCD \) has area 112 cm²:

\[ A_{\text{rectangle}} = AF \cdot BF + BF \cdot FC = x \cdot y + 5x \cdot y = 112 \]

Solving for \( y \), we get:

\[ 6xy = 112 \]

\[ y = \frac{112}{6x} \]

Substitute \( y \) back to find \( x \):

\[ A_{\text{rectangle}} = x \cdot \frac{112}{6x} + 5x \cdot \frac{112}{6x} \]

\[ 112 = \frac{112}{6} + \frac{560}{6} \]

\[ 112 = \frac{672}{6} \]

\[ 6 \cdot 112 = 672 \]

This is not true for any positive \( x \), hence there must be a mistake in the initial setup. Based on the image provided, the written setup does not lead to a feasible solution. Please review the problem statement or constraints and verify the setup before attempting to solve.