Example Question - polygon
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Calculating the Sum of Interior Angles in a Polygon
<p>Let n be the number of sides (or angles) in the polygon.</p> <p>The formula for the sum of the interior angles in an n-sided polygon is:</p> <p>\[ (n - 2) \times 180^\circ \]</p> <p>We are given that one angle measures \( 50^\circ \). As the polygon appears to be a rhombus, all angles are equal.</p> <p>Since there are four angles in total, we can write:</p> <p>\[ 4 \times 50^\circ = (n - 2) \times 180^\circ \]</p> <p>\[ 200^\circ = (n - 2) \times 180^\circ \]</p> <p>The exact question being asked is not clear from the image, but assuming we need to confirm if the angles provided can form a polygon, we can solve for n:</p> <p>\[ n - 2 = \frac{200^\circ}{180^\circ} \]</p> <p>\[ n - 2 = \frac{20}{18} \]</p> <p>\[ n - 2 = 1\frac{1}{9} \]</p> <p>Given that n must be a whole number, the answer 1\(\frac{1}{9}\) indicates that the values presented do not form a consistent set for a polygon with all equal angle measures. Hence, there seems to be a discrepancy unless the intention was to showcase an irregular polygon, in which case additional information would be needed.</p>
Determining the Area of a Polygon Within a Rectangle
<p>Given:</p> <p>\( [ABCD] = 112 \, \text{cm}^2 \)</p> <p>\( BE = 3AE, EG = \frac{5}{28} AD, \text{and} 2BF = FC \)</p> <p>Steps:</p> <p>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, \( BE = 3 \times \frac{AD}{4} = \frac{3}{4} AD \).</p> <p>2. Find \( EG \): \( EG = \frac{5}{28} AD \).</p> <p>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, \( FC = 2 \times \frac{AD}{3} = \frac{2}{3} AD \).</p> <p>4. Calculate the area \( [BEFG] \):</p> \[\begin{align*} [BEFG] &= [BEC] - [EFGC] \\ &= \frac{1}{2} BE \cdot BC - \frac{1}{2} EG \cdot FC \\ &= \frac{1}{2} \cdot \frac{3}{4} AD \cdot AD - \frac{1}{2} \cdot \frac{5}{28} AD \cdot \frac{2}{3} AD \\ &= \frac{1}{2} \cdot \frac{3}{4} \cdot 112 - \frac{1}{2} \cdot \frac{5}{28} \cdot \frac{2}{3} \cdot 112 \\ &= \frac{3}{8} \cdot 112 - \frac{5}{42} \cdot 112 \\ &= 42 - \frac{5}{42} \cdot 112 \\ &= 42 - \frac{5}{6} \cdot 16 \\ &= 42 - \frac{80}{6} \\ &= 42 - \frac{40}{3} \\ &= 42 - 13\frac{1}{3} \\ &= 28\frac{2}{3} \, \text{cm}^2. \end{align*}\] <p>Thus, the area of \( EFG \) is \( 28\frac{2}{3} \, \text{cm}^2 \).</p>
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