Question - Finding the Area of a Shaded Region in a Geometric Diagram
Solution:
The problem is to find the area of region EGFB within rectangle ABCD. We're given that the area of rectangle ABCD is 112 cm².
First, let's express the lengths BE, EG, and BF in terms of a single variable. Considering that BE is one-third of EG, let's say that EG = 3x, so BE = x. Furthermore, given that BF is half of 5FC and FC is equal to EG (that is, 3x), we have BF = \(\frac{5}{2}\)x.
Next, we can find the side lengths of ABCD by exploiting the fact that the area of the rectangle is the product of its length and width:
- The width of the rectangle is BE + EG, which is x + 3x = 4x.
- The length of the rectangle is BF, which is \(\frac{5}{2}\)x.
Area of ABCD is then:
\[ 4x \times \frac{5}{2}x = 112 \]
\[ 10x^2 = 112 \]
\[ x^2 = \frac{112}{10} \]
\[ x^2 = 11.2 \]
\[ x = \sqrt{11.2} \]
\[ x \approx 3.35 \]
Now that we have x, we can calculate the length of EG (which is 3x):
\[ EG = 3 \times 3.35 \approx 10.05 \]
We also calculate BF:
\[ BF = \frac{5}{2} \times 3.35 \approx 8.38 \]
The area of triangle BEG is half of the rectangle ABCD's area because BEG cuts the rectangle diagonally in half:
\[ \text{Area of BEG} = \frac{\text{Area of ABCD}}{2} \]
\[ \text{Area of BEG} = \frac{112}{2} \]
\[ \text{Area of BEG} = 56 \text{ cm}^2 \]
Finally, we find the area of triangle BFG using its base BF and height x (remember that height is BE):
\[ \text{Area of BFG} = \frac{1}{2} \times BF \times BE \]
\[ \text{Area of BFG} = \frac{1}{2} \times 8.38 \times 3.35 \]
\[ \text{Area of BFG} \approx \frac{1}{2} \times 28.07 \]
\[ \text{Area of BFG} \approx 14.04 \text{ cm}^2 \]
To find the area of the shaded region EGFB, we subtract the area of triangle BFG from the area of triangle BEG:
\[ \text{Area of EGFB} = \text{Area of BEG} - \text{Area of BFG} \]
\[ \text{Area of EGFB} = 56 - 14.04 \]
\[ \text{Area of EGFB} \approx 41.96 \text{ cm}^2 \]
So, the area of EGFB is approximately 41.96 cm².
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