Question - Determining the Area of a Polygon Within a Rectangle
Solution:
Given:
\( [ABCD] = 112 \, \text{cm}^2 \)
\( BE = 3AE, EG = \frac{5}{28} AD, \text{and} 2BF = FC \)
Steps:
1. Find \( AE \) and \( BE \): Since \( AE + BE = AB \) and \( BE = 3AE \), let \( AE = x \) and hence \( BE = 3x \). So, \( x + 3x = AB \) which means \( 4x = AB \). Because \( AB \cdot AD = 112 \), we have \( 4x \cdot AD = 112 \) and \( x = \frac{AD}{4} \). Therefore, \( BE = 3 \times \frac{AD}{4} = \frac{3}{4} AD \).
2. Find \( EG \): \( EG = \frac{5}{28} AD \).
3. Find \( BF \) and \( FC \): Since \( 2BF = FC \) and \( BF + FC = BC = AD \) (as ABCD is a rectangle), let \( BF = y \) and hence \( FC = 2y \). So, \( y + 2y = AD \) which means \( 3y = AD \) and \( y = \frac{AD}{3} \). Therefore, \( FC = 2 \times \frac{AD}{3} = \frac{2}{3} AD \).
4. Calculate the area \( [BEFG] \):
\[\begin{align*} [BEFG] &= [BEC] - [EFGC] \\ &= \frac{1}{2} BE \cdot BC - \frac{1}{2} EG \cdot FC \\ &= \frac{1}{2} \cdot \frac{3}{4} AD \cdot AD - \frac{1}{2} \cdot \frac{5}{28} AD \cdot \frac{2}{3} AD \\ &= \frac{1}{2} \cdot \frac{3}{4} \cdot 112 - \frac{1}{2} \cdot \frac{5}{28} \cdot \frac{2}{3} \cdot 112 \\ &= \frac{3}{8} \cdot 112 - \frac{5}{42} \cdot 112 \\ &= 42 - \frac{5}{42} \cdot 112 \\ &= 42 - \frac{5}{6} \cdot 16 \\ &= 42 - \frac{80}{6} \\ &= 42 - \frac{40}{3} \\ &= 42 - 13\frac{1}{3} \\ &= 28\frac{2}{3} \, \text{cm}^2. \end{align*}\]Thus, the area of \( EFG \) is \( 28\frac{2}{3} \, \text{cm}^2 \).
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