Finding the Area of a Quadrilateral Within a Rectangle
<p>Let the length of BE be \( x \). Hence, \( EG = \frac{5}{6} AD \) and \( BF = \frac{5}{2} x \).</p>
<p>Since \( BE = \frac{3}{4} AE \) and \( EG = \frac{5}{6} AD \), let \( AE = 4x \), so \( AD = 6x \) and \( EG = 5x \).</p>
<p>The area of rectangle \( ABCD \) is given as 112 \( cm^2 \), which is equal to \( AD \times AB \). Since \( AD = 6x \), we have:</p>
<p>\( AB \cdot 6x = 112 \)</p>
<p>\( AB = \frac{112}{6x} \)</p>
<p>Now, \( AF = AE + EF = AE + BF - BE = 4x + \frac{5}{2}x - x = \frac{11}{2}x \).</p>
<p>So the area of triangle \( AEF \) is:</p>
<p>\( \frac{1}{2} AF \cdot BE = \frac{1}{2} \cdot \frac{11}{2}x \cdot x = \frac{11}{4}x^2 \)</p>
<p>The area of triangle \( EGF \) is:</p>
<p>\( \frac{1}{2} EG \cdot BF = \frac{1}{2} \cdot 5x \cdot \frac{5}{2}x = \frac{25}{4}x^2 \)</p>
<p>To find the area of \( EGBF \), we sum the areas of \( AEF \) and \( EGF \):</p>
<p>\( \frac{11}{4}x^2 + \frac{25}{4}x^2 = \frac{36}{4}x^2 = 9x^2 \)</p>
<p>Now, the area of \( ABCD \) is the length times the width, \( AB \times AD \), which is \( \frac{112}{6x} \times 6x = 112 \).</p>
<p>To find \( x \), we set up the equation:</p>
<p>\( AB \cdot AD = \frac{112}{6x} \cdot 6x = 112 \)</p>
<p>\( 112 = 112 \)</p>
<p>This is always true, hence we can say that \( x \) can take any value. Let's find \( x \) in terms of \( AB \):</p>
<p>\( AB = \frac{112}{6x} \)</p>
<p>\( x = \frac{112}{6AB} \)</p>
<p>Now, substitute \( x \) in terms of \( AB \) into the area of \( EGBF \):</p>
<p>\( Area \;of\; EGBF = 9x^2 = 9 \left( \frac{112}{6AB} \right)^2 \)</p>
<p>\( Area \;of\; EGBF = 9 \cdot \frac{112^2}{36 \cdot AB^2} \)</p>
<p>\( Area \;of\; EGBF = \frac{112^2}{4 \cdot AB^2} \)</p>
<p>Since \( Area \;of\; ABCD = AB \cdot AD = AB \cdot 6x = 112 \), \( x = \frac{112}{6AB} \).</p>
<p>Put \( x \) into \( Area \;of\; EGBF \):</p>
<p>\( Area \;of\; EGBF = \frac{112^2}{4 \cdot (\frac{112}{6x})^2} \)</p>
<p>\( Area \;of\; EGBF = \frac{112^2 \cdot 36x^2}{4 \cdot 112^2} \)</p>
<p>\( Area \;of\; EGBF = \frac{36x^2}{4} \)</p>
<p>\( Area \;of\; EGBF = 9x^2 \)</p>
<p>Since we know that \( 6x = AD \) and \( AD \cdot AB = 112 \), it follows that:</p>
<p>\( x^2 = \left( \frac{AD}{6} \right)^2 \)</p>
<p>\( x^2 = \left( \frac{112}{6AB} \right)^2 \)</p>
<p>\( x^2 = \frac{112^2}{36AB^2} \)</p>
<p>The requested area is \( EGBF = 9x^2 \), which after substitution becomes:</p>
<p>\( 9 \cdot \frac{112^2}{36AB^2} \)</p>
<p>Reducing the fraction gives us:</p>
<p>\( \frac{112^2}{4AB^2} \)</p>
<p>Since \( AB \times AD = 112 \), let \( AB = a \) and \( AD = \frac{112}{a} \).</p>
<p>Substitute \( AB = a \) into the expression for area:</p>
<p>\( \frac{112^2}{4a^2} \)</p>
<p>Now, express \( AB^2 \) as \( a^2 \), knowing \( a \cdot \frac{112}{a} = 112 \)</p>
<p>\( \frac{112^2}{4 \cdot \left( \frac{112}{AD} \right)^2} \)</p>
<p>Simplifying, we find:</p>
<p>\( \frac{112^2 \cdot AD^2}{4 \cdot 112^2} \)</p>
<p>\( \frac{AD^2}{4} \)</p>
<p>Since \( AD = 6x \), substitute back to find \( x \):</p>
<p>\( \frac{(6x)^2}{4} = \frac{36x^2}{4} = 9x^2 \)</p>
<p>Since the area values must match, solving the equation \( 9x^2 = 112 \) to find \( x \):</p>
<p>\( x^2 = \frac{112}{9} \)</p>
<p>\( x = \frac{\sqrt{112}}{3} \)</p>
<p>Finally, calculate the area of \( EGBF \):</p>
<p>\( Area \;of\; EGBF = 9 \left( \frac{\sqrt{112}}{3} \right)^2 \)</p>
<p>\( Area \;of\; EGBF = 9 \cdot \frac{112}{9} \)</p>
<p>\( Area \;of\; EGBF = 112 \; cm^2 \)</p>
<p>Thus, the area of quadrilateral \( EGBF \) is 112 \( cm^2 \).</p>
