Finding the Area of a Quadrilateral Within a Rectangle
Let's denote the length of BE as \( x \), so AE is \( 3x \) because BE is one-third of AE.
Given that the area of the rectangle ABCD is \( 112 \, \text{cm}^2 \) and that AE is three times BE, the width of the rectangle is \( 4x \).
Assume the height of the rectangle (BF) is \( h \). From \( 2\text{BF} = 5\text{FC} \), we have \( 2h = \frac{5}{7}(4x) \) because FC is the remainder of the length of the rectangle after removing BF, which is \( \frac{7}{2}h \).
Solving for \( h \), we get \( h = \frac{5}{7} \cdot \frac{4}{2}x = \frac{10}{7}x \).
Now we can find \( x \) from the area of the rectangle:
\[ \text{Area} = 4x \cdot h = 112 \]
\[ 4x \cdot \frac{10}{7}x = 112 \]
\[ \frac{40}{7}x^2 = 112 \]
\[ x^2 = \frac{112 \cdot 7}{40} \]
\[ x^2 = 19.6 \]
\[ x = \sqrt{19.6} \]
\[ x = 4.43 \] (approx)
Now we calculate \( h \):
\[ h = \frac{10}{7} \cdot 4.43 = 6.33 \] (approx)
To find the area of EFGF, we need to find the areas of triangles BEG and CFF and subtract them from the area of the rectangle.
The triangle BEG has a base of x and height h, and the triangle CFF has a base of 3x and height \( \frac{2}{7}h \).
Area of BEG:
\[ \text{Area}_{\text{BEG}} = \frac{1}{2} \cdot x \cdot h = \frac{1}{2} \cdot 4.43 \cdot 6.33 \]
\[ \text{Area}_{\text{BEG}} = 14.01 \] (approx)
Area of CFF:
\[ \text{Area}_{\text{CFF}} = \frac{1}{2} \cdot 3x \cdot \frac{2}{7}h = \frac{1}{2} \cdot 3 \cdot 4.43 \cdot \frac{2}{7} \cdot 6.33 \]
\[ \text{Area}_{\text{CFF}} = 12.15 \] (approx)
Total area of the two triangles:
\[ \text{Area}_{\text{TotalTriangles}} = \text{Area}_{\text{BEG}} + \text{Area}_{\text{CFF}} \]
\[ \text{Area}_{\text{TotalTriangles}} = 14.01 + 12.15 \]
\[ \text{Area}_{\text{TotalTriangles}} = 26.16 \] (approx)
Finally, the area of quadrilateral EFGF is:
\[ \text{Area}_{\text{EFGF}} = 112 - \text{Area}_{\text{TotalTriangles}} \]
\[ \text{Area}_{\text{EFGF}} = 112 - 26.16 \]
\[ \text{Area}_{\text{EFGF}} = 85.84 \, \text{cm}^2 \] (approx)
Please note that the solution is an approximation due to rounding the value of \( x \) to two decimal places.
