Question

1. find the value of x in the figure below if \\(\overline{mp}\\) is parallel to \\(\overline{ra}\\).

figure is not necessarily drawn to scale.
\\(\bigcirc\\) 18 units
\\(\bigcirc\\) 5 units
\\(\bigcirc\\) 20.5 units
\\(\bigcirc\\) 4.5 units

Explanation:

Step1: Identify the theorem

Since \( \overline{MP} \parallel \overline{RA} \), by the Basic Proportionality Theorem (Thales' theorem), the line \( MP \) divides the sides \( AR \) and \( AA \) (wait, correction: the sides \( AR \) and \( AP \)? Wait, looking at the figure, the triangle is \( ARA \)? No, probably a typo, the vertices are \( R \), \( A \) (lower), and \( A \) (upper)? Wait, no, the figure has points \( R \), \( M \), \( A \) (upper) on the top side, and \( R \), \( A \) (lower), \( P \) on the bottom side? Wait, actually, the triangle is \( A \) (lower) - \( R \) - \( A \) (upper), with \( M \) on \( RA \) (upper) and \( P \) on \( A \) (lower) - \( A \) (upper) side? Wait, no, the correct interpretation is that \( MP \) is parallel to \( RA \), so the triangle has a line segment \( MP \) parallel to \( RA \), so by Thales' theorem, \( \frac{RM}{MA} = \frac{AP}{PA} \)? Wait, no, let's look at the lengths: \( MA = 18 \), \( AP = 5 \), \( PA \) (wait, the side with length 20? Wait, the side from \( P \) to \( A \) (upper) is 20? Wait, no, the figure: \( R \) to \( M \) is \( x \), \( M \) to \( A \) (upper) is 18, \( P \) to \( A \) (upper) is 20, \( A \) (lower) to \( P \) is 5. So the triangle is \( A \) (lower) - \( R \) - \( A \) (upper), with \( M \) on \( RA \) (upper) and \( P \) on \( A \) (lower) - \( A \) (upper) side. So \( MP \parallel RA \), so by Thales' theorem, \( \frac{RM}{MA} = \frac{AP}{PA} \)? Wait, no, the segments: \( AP = 5 \), \( PA \) (wait, the entire side from \( A \) (lower) to \( A \) (upper) is \( AP + PA = 5 + 20 = 25 \)? Wait, no, the side from \( A \) (lower) to \( P \) is 5, and from \( P \) to \( A \) (upper) is 20, so total length \( AA = 5 + 20 = 25 \). Then \( MP \) is parallel to \( RA \), so in triangle \( ARA \) (wait, no, the triangle is \( A \) (lower) - \( R \) - \( A \) (upper), with \( M \) on \( RA \) (upper) and \( P \) on \( A \) (lower) - \( A \) (upper) side. So by Thales' theorem, \( \frac{RM}{MA} = \frac{AP}{PA} \)? Wait, no, the correct ratio is \( \frac{RM}{MA} = \frac{AP}{PA} \)? Wait, no, Thales' theorem states that if a line is drawn parallel to one side of a triangle, intersecting the other two sides, then it divides those sides proportionally. So the two sides are \( AR \) (from \( A \) (lower) to \( R \)) and \( AA \) (from \( A \) (lower) to \( A \) (upper)). Wait, no, the sides are \( A \) (lower) - \( R \) and \( A \) (lower) - \( A \) (upper). Then \( MP \) is parallel to \( R \) - \( A \) (lower)? No, \( MP \) is parallel to \( RA \) (where \( RA \) is from \( R \) to \( A \) (lower)). So the triangle is \( A \) (lower) - \( R \) - \( A \) (upper), with \( M \) on \( R \) - \( A \) (upper) and \( P \) on \( A \) (lower) - \( A \) (upper). Then \( MP \parallel RA \) ( \( RA \) is from \( R \) to \( A \) (lower) ). So by Thales' theorem, \( \frac{RM}{MA} = \frac{AP}{PA} \). Wait, \( RM = x \), \( MA = 18 \), \( AP = 5 \), \( PA = 20 \)? Wait, no, \( AP \) is 5, \( PA \) is 20? Wait, the side from \( A \) (lower) to \( P \) is 5, and from \( P \) to \( A \) (upper) is 20, so the ratio of \( AP \) to \( PA \) is \( 5:20 = 1:4 \)? No, wait, the correct ratio is \( \frac{RM}{MA} = \frac{AP}{PA} \)? Wait, no, Thales' theorem: if a line is drawn parallel to one side of a triangle, intersecting the other two sides, then it divides those sides proportionally. So the two sides are \( A \) (lower) - \( R \) and \( A \) (lower) - \( A \) (upper). The line \( MP \) intersects \( A \) (lower) - \( R \) at \( M \)? No, \( M \) is on \( R \) - \( A \) (upper), and \( P \) is on \( A \) (lower) - \( A \) (upper). So the triangle is \( A \) (lower) - \( R \) - \( A \) (upper), with \( M \) on \( R \) - \( A \) (upper) and \( P \) on \( A \) (lower) - \( A \) (upper). Then \( MP \parallel RA \) ( \( RA \) is from \( R \) to \( A \) (lower) ). So the segments: \( R \) - \( M \) is \( x \), \( M \) - \( A \) (upper) is 18, \( A \) (lower) - \( P \) is 5, \( P \) - \( A \) (upper) is 20. So by Thales' theorem, \( \frac{RM}{MA} = \frac{AP}{PA} \)? Wait, \( AP = 5 \), \( PA = 20 \), so \( \frac{x}{18} = \frac{5}{20} \)? No, that would be \( x = 18 \times \frac{5}{20} = 4.5 \). Wait, that makes sense. So the ratio is \( \frac{RM}{MA} = \frac{AP}{PA} \), where \( RM = x \), \( MA = 18 \), \( AP = 5 \), \( PA = 20 \). So \( \frac{x}{18} = \frac{5}{20} \).

Step2: Solve for \( x \)

Simplify \( \frac{5}{20} = \frac{1}{4} \), so \( x = 18 \times \frac{1}{4} = 4.5 \).

Answer:

4.5 units