Question

given lines ( l ), ( m ), and ( n ) are parallel and cut by two transversal lines, find the value of ( x ). round your answer to the nearest tenth if necessary.

Explanation:

Step1: Apply the Basic Proportionality Theorem (Thales' theorem) for parallel lines cut by transversals.

When three parallel lines are cut by two transversals, the segments are proportional. So, we set up the proportion: $\frac{22}{7} = \frac{30}{x}$ (assuming the segments are corresponding). Wait, actually, the correct proportion should be based on the similar triangles formed. Let's re - check. The segments on one transversal are 22 and 7, and on the other transversal are 30 and $x$. So the proportion is $\frac{22}{7}=\frac{30}{x}$? Wait, no, maybe it's $\frac{7}{22}=\frac{x}{30}$. Let's think again. If we consider the two transversals intersecting the three parallel lines, the ratios of the corresponding segments are equal. So if the length between $l$ and $m$ on one transversal is 22, between $m$ and $n$ is 7, and on the other transversal, between the top two intersections is 30, and between the bottom two (where $x$ is) is $x$, then the ratio of the segments on the first transversal (between $l - m$ and $m - n$) should equal the ratio of the segments on the second transversal (between the top two and $x$). So $\frac{7}{22}=\frac{x}{30}$.

Step2: Solve the proportion for $x$.

Cross - multiply: $22x = 7\times30$.
Calculate $7\times30 = 210$. So $22x=210$.
Then $x=\frac{210}{22}\approx9.545\approx9.5$ (rounded to the nearest tenth). Wait, no, wait. Maybe I mixed up the proportion. Let's re - establish the proportion. Let's say the two transversals create segments: on the first transversal (the one with 22 and 7), the total length from $l$ to $n$ is $22 + 7=29$? No, that's not the right way. Wait, the correct approach is that when three parallel lines are cut by two transversals, the ratio of the lengths of the segments on one transversal is equal to the ratio of the lengths of the corresponding segments on the other transversal. So if we have segments $a = 22$ (between $l$ and $m$), $b = 7$ (between $m$ and $n$) on one transversal, and segments $c = 30$ (between the top two intersections) and $d=x$ (between the bottom two intersections) on the other transversal, then $\frac{a}{b}=\frac{c}{d}$, so $\frac{22}{7}=\frac{30}{x}$. Wait, but if we solve $\frac{22}{7}=\frac{30}{x}$, then $x=\frac{30\times7}{22}=\frac{210}{22}\approx9.5$? Wait, no, that gives a small number. Wait, maybe the proportion is $\frac{7}{22}=\frac{x}{30}$. Let's solve that: $22x = 7\times30=210$, so $x=\frac{210}{22}\approx9.5$. Wait, but let's check the diagram again. The segment of length 22 is between $l$ and $m$, 7 is between $m$ and $n$, and 30 is between the top two lines, and $x$ is between the bottom two lines. So the ratio of the distance between $m - n$ to $l - m$ should be equal to the ratio of $x$ to 30. So $\frac{7}{22}=\frac{x}{30}$. Solving for $x$: $x=\frac{7\times30}{22}=\frac{210}{22}\approx9.5$.

Answer:

$\approx9.5$