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 \(\overline{MP} \parallel \overline{RA}\), we can use the Basic Proportionality Theorem (Thales' theorem), which 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.

Step2: Set Up the Proportion

Let's consider triangle \(ARA'\) (assuming the lower vertex is \(A\) and the side with length \(20\) and \(5\) is part of the triangle). The line \(MP\) is parallel to \(RA\), so we have the proportion: \(\frac{x}{18}=\frac{5}{20}\) (Wait, no, let's correct. Wait, the segments: The side with length \(20\) is split into \(MP\) related? Wait, actually, the triangle has sides: the side from the lower \(A\) to the upper \(A\) has length \(20\), and the segment from \(P\) to the lower \(A\) is \(5\)? Wait, no, let's re - examine the figure. The figure has points \(R\), \(M\), \(A\) on the top line, with \(RM = x\), \(MA=18\). The lower side has \(A\) (lower), \(P\), and the other \(A\) (upper)? Wait, maybe it's a triangle with a line parallel to one side. Let's assume that the triangle has a base with length \(x + 18\) and a line \(MP\) parallel to \(RA\), and the other side of the triangle has length \(20\) with a segment of length \(5\) from the vertex. So by the Basic Proportionality Theorem (Thales' theorem), \(\frac{x}{18}=\frac{5}{20}\)? No, that would be wrong. Wait, actually, the correct proportion should be \(\frac{x}{18}=\frac{5}{20}\)? Wait, no, let's do it properly. If \(MP\parallel RA\), then the triangles are similar? Wait, no, maybe the segments are such that \(\frac{x}{18}=\frac{5}{20}\) is incorrect. Wait, let's think again. The length from \(P\) to the lower \(A\) is \(5\), and from \(P\) to the upper \(A\) is \(20 - 5=15\)? No, the given length is \(20\) as the side from \(P\) to upper \(A\), and \(5\) from lower \(A\) to \(P\). Wait, maybe the correct proportion is \(\frac{x}{18}=\frac{5}{20}\) is wrong. Wait, let's set up the proportion correctly. Let's say that in triangle \(ARA\) (the big triangle), the line \(MP\) is parallel to \(RA\), so the ratio of the segments on one side is equal to the ratio on the other side. So \(\frac{x}{18}=\frac{5}{20}\) is incorrect. Wait, actually, the length of the side with \(x\) and \(18\) is parallel to the side with \(5\) and \(20\)? No, \(MP\) is parallel to \(RA\), so the triangles are similar. So the ratio of corresponding sides is equal. So if \(MP\parallel RA\), then \(\frac{x}{18}=\frac{5}{20}\) is wrong. Wait, let's solve \(\frac{x}{18}=\frac{5}{20}\), cross - multiply: \(20x=18\times5\), \(20x = 90\), \(x=\frac{90}{20}=4.5\). Ah, that makes sense. So the proportion is set up as the ratio of \(x\) to \(18\) is equal to the ratio of \(5\) to \(20\) (because \(MP\) is parallel to \(RA\), so the line \(MP\) divides the two sides of the triangle proportionally).

Step3: Solve for \(x\)

We have the proportion \(\frac{x}{18}=\frac{5}{20}\). Cross - multiply: \(20x=18\times5\). Calculate \(18\times5 = 90\). Then \(x=\frac{90}{20}=4.5\).

Answer:

4.5 units