Question - Finding the Area of a Quadrilateral Within a Rectangle

Let the length of BE be \( x \). Hence, \( EG = \frac{5}{6} AD \) and \( BF = \frac{5}{2} x \).

Since \( BE = \frac{3}{4} AE \) and \( EG = \frac{5}{6} AD \), let \( AE = 4x \), so \( AD = 6x \) and \( EG = 5x \).

The area of rectangle \( ABCD \) is given as 112 \( cm^2 \), which is equal to \( AD \times AB \). Since \( AD = 6x \), we have:

\( AB \cdot 6x = 112 \)

\( AB = \frac{112}{6x} \)

Now, \( AF = AE + EF = AE + BF - BE = 4x + \frac{5}{2}x - x = \frac{11}{2}x \).

So the area of triangle \( AEF \) is:

\( \frac{1}{2} AF \cdot BE = \frac{1}{2} \cdot \frac{11}{2}x \cdot x = \frac{11}{4}x^2 \)

The area of triangle \( EGF \) is:

\( \frac{1}{2} EG \cdot BF = \frac{1}{2} \cdot 5x \cdot \frac{5}{2}x = \frac{25}{4}x^2 \)

To find the area of \( EGBF \), we sum the areas of \( AEF \) and \( EGF \):

\( \frac{11}{4}x^2 + \frac{25}{4}x^2 = \frac{36}{4}x^2 = 9x^2 \)

Now, the area of \( ABCD \) is the length times the width, \( AB \times AD \), which is \( \frac{112}{6x} \times 6x = 112 \).

To find \( x \), we set up the equation:

\( AB \cdot AD = \frac{112}{6x} \cdot 6x = 112 \)

\( 112 = 112 \)

This is always true, hence we can say that \( x \) can take any value. Let's find \( x \) in terms of \( AB \):

\( AB = \frac{112}{6x} \)

\( x = \frac{112}{6AB} \)

Now, substitute \( x \) in terms of \( AB \) into the area of \( EGBF \):

\( Area \;of\; EGBF = 9x^2 = 9 \left( \frac{112}{6AB} \right)^2 \)

\( Area \;of\; EGBF = 9 \cdot \frac{112^2}{36 \cdot AB^2} \)

\( Area \;of\; EGBF = \frac{112^2}{4 \cdot AB^2} \)

Since \( Area \;of\; ABCD = AB \cdot AD = AB \cdot 6x = 112 \), \( x = \frac{112}{6AB} \).

Put \( x \) into \( Area \;of\; EGBF \):

\( Area \;of\; EGBF = \frac{112^2}{4 \cdot (\frac{112}{6x})^2} \)

\( Area \;of\; EGBF = \frac{112^2 \cdot 36x^2}{4 \cdot 112^2} \)

\( Area \;of\; EGBF = \frac{36x^2}{4} \)

\( Area \;of\; EGBF = 9x^2 \)

Since we know that \( 6x = AD \) and \( AD \cdot AB = 112 \), it follows that:

\( x^2 = \left( \frac{AD}{6} \right)^2 \)

\( x^2 = \left( \frac{112}{6AB} \right)^2 \)

\( x^2 = \frac{112^2}{36AB^2} \)

The requested area is \( EGBF = 9x^2 \), which after substitution becomes:

\( 9 \cdot \frac{112^2}{36AB^2} \)

Reducing the fraction gives us:

\( \frac{112^2}{4AB^2} \)

Since \( AB \times AD = 112 \), let \( AB = a \) and \( AD = \frac{112}{a} \).

Substitute \( AB = a \) into the expression for area:

\( \frac{112^2}{4a^2} \)

Now, express \( AB^2 \) as \( a^2 \), knowing \( a \cdot \frac{112}{a} = 112 \)

\( \frac{112^2}{4 \cdot \left( \frac{112}{AD} \right)^2} \)

Simplifying, we find:

\( \frac{112^2 \cdot AD^2}{4 \cdot 112^2} \)

\( \frac{AD^2}{4} \)

Since \( AD = 6x \), substitute back to find \( x \):

\( \frac{(6x)^2}{4} = \frac{36x^2}{4} = 9x^2 \)

Since the area values must match, solving the equation \( 9x^2 = 112 \) to find \( x \):

\( x^2 = \frac{112}{9} \)

\( x = \frac{\sqrt{112}}{3} \)

Finally, calculate the area of \( EGBF \):

\( Area \;of\; EGBF = 9 \left( \frac{\sqrt{112}}{3} \right)^2 \)

\( Area \;of\; EGBF = 9 \cdot \frac{112}{9} \)

\( Area \;of\; EGBF = 112 \; cm^2 \)

Thus, the area of quadrilateral \( EGBF \) is 112 \( cm^2 \).