Example Question - shaded region
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Graphical Analysis of Inequalities and Regions
<p>Langkah pertama adalah mengenal pasti persamaan garis dan ketaksamaan yang ditunjukkan dalam graf.</p> <p>Garis yang terdapat dalam graf termasuk:</p> <p>1. \( y = 2 \) : Garis mendatar yang melalui \( y = 2 \).</p> <p>2. \( y = -x + 1 \) : Garis miring dengan gradien -1 yang melalui titik potong y pada \( (0, 1) \).</p> <p>3. \( y = x + 1 \) : Garis miring dengan gradien 1 yang melalui titik potong y pada \( (0, 1) \).</p> <p>Ketaksamaan yang terdapat pada graf:</p> <p>1. \( y \leq 2 \) : Ini berarti bahwa daerah yang diarsir berada pada atau di bawah garis \( y = 2 \).</p> <p>2. \( y \geq -x + 1 \) : Ini berarti bahwa daerah yang diarsir berada pada atau di atas garis \( y = -x + 1 \).</p> <p>3. \( y \leq x + 1 \) : Ini berarti bahwa daerah yang diarsir berada pada atau di bawah garis \( y = x + 1 \).</p> <p>Daerah yang diarsir menunjukkan daerah penyelesaian untuk sistem ketaksamaan yang disajikan oleh garis-garis ini. Daerah ini adalah sempadan yang dibentuk oleh garis-garis ini dan termasuk titik di mana ketiga-tiga garis bertemu, iaitu titik (0.5, 1.5).</p>
Finding the Area of a Shaded Region in a Geometric Diagram
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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