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Question
quiz 4: proport if ab = 35, bc = 15, and ef = 60, then the value of de is: 140. 180. 200. 210.
Step1: Identify the theorem
This is a problem about parallel lines and transversals, so we can use the Basic Proportionality Theorem (Thales' theorem) or the concept of similar triangles/parallel line segment ratios. The segments \(AB\), \(BC\) are on one transversal, and \(DE\), \(EF\) are on another transversal, with the lines containing \(AD\), \(BE\), \(CF\) being parallel. So, \(\frac{AB}{BC}=\frac{DE}{EF}\)? Wait, no, wait. Wait, actually, the transversal with \(AB\) and \(BC\) and the transversal with \(DE\) and \(EF\) – wait, maybe the correct ratio is \(\frac{AB + BC}{BC}=\frac{DE}{EF}\)? Wait, no, let's check the lengths. Wait, \(AB = 35\), \(BC = 15\), so the total length from \(A\) to \(C\) is \(AB + BC=35 + 15 = 50\)? Wait, no, maybe the lines are such that \(\frac{AB}{BC}=\frac{DE}{EF}\) is not correct. Wait, actually, the three parallel lines (the ones with arrows) cut the transversals \(AC\) (with \(A\), \(B\), \(C\)) and \(DF\) (with \(D\), \(E\), \(F\)). So by the theorem of parallel lines cutting transversals proportionally, \(\frac{AB}{BC}=\frac{DE}{EF}\)? Wait, no, that would be if \(BE\) and \(CF\) are parallel, but actually, the three lines (the ones with arrows) are parallel, so the ratio of the segments on one transversal is equal to the ratio on the other. Wait, let's re-express: Let the three parallel lines be \(l_1\) (through \(A\), \(D\)), \(l_2\) (through \(B\), \(E\)), \(l_3\) (through \(C\), \(F\)). Then the transversal \(AC\) has segments \(AB = 35\), \(BC = 15\), and transversal \(DF\) has segments \(DE\) (what we need to find) and \(EF = 60\). Wait, no, that can't be. Wait, maybe the correct proportion is \(\frac{AB}{BC}=\frac{DE}{EF}\)? Wait, no, let's calculate the ratio. Wait, maybe I mixed up the segments. Wait, actually, the correct proportion is \(\frac{AB}{BC}=\frac{DE}{EF}\) is incorrect. Wait, let's think again. The lines \(AD\), \(BE\), \(CF\) are parallel (the ones with arrows), so the transversals \(AC\) (with \(A\), \(B\), \(C\)) and \(DF\) (with \(D\), \(E\), \(F\)) are cut by these parallel lines. So by the theorem, \(\frac{AB}{BC}=\frac{DE}{EF}\)? Wait, no, that would mean \(\frac{35}{15}=\frac{DE}{60}\), but that would give \(DE=\frac{35\times60}{15}=140\), but that's one of the options. Wait, but wait, maybe the segments are \(AB\) and \(AC\) vs \(DE\) and \(DF\). Wait, no, let's check the answer options. The options include 140, 180, 200, 210. Wait, if we use \(\frac{AB}{BC}=\frac{DE}{EF}\), then \(DE=\frac{AB\times EF}{BC}=\frac{35\times60}{15}=140\). But wait, maybe the correct ratio is \(\frac{AB + BC}{BC}=\frac{DE + EF}{EF}\)? Wait, no, that would be \(\frac{50}{15}=\frac{DE + 60}{60}\), then \(50\times60 = 15\times(DE + 60)\), \(3000 = 15DE + 900\), \(15DE = 2100\), \(DE = 140\). Wait, same result. Wait, maybe the problem is that \(AB\) and \(BC\) are on one transversal, and \(DE\) and \(EF\) are on another, with the parallel lines, so the ratio of \(AB\) to \(BC\) is equal to the ratio of \(DE\) to \(EF\). So:
Step2: Set up the proportion
Given \(AB = 35\), \(BC = 15\), \(EF = 60\). Let \(DE = x\). By the proportionality of parallel lines cutting transversals, \(\frac{AB}{BC}=\frac{DE}{EF}\). Wait, no, actually, the correct proportion is \(\frac{AB}{AC}=\frac{DE}{DF}\), but \(AC = AB + BC = 35 + 15 = 50\), \(DF = DE + EF = x + 60\). So \(\frac{35}{50}=\frac{x}{x + 60}\). Solving this: \(35(x + 60)=50x\), \(35x + 2100 = 50x\), \(15x = 2100\), \(x = 140\). Wait, that's the same as before. So either way, we get \(DE = 140\)? Wait, but let's check the options. The first o…
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