Question

the diagram shows the intersections of several straight roads. the avenues run parallel to each other. first ave. main 280 ft second ave. 140 ft oak a b 113 ft amana walks along oak from point a to b. to the nearest foot, how far does she walk? 75 ft 226 ft 307 ft 347 ft

Explanation:

Step1: Identify Similar Triangles

Since avenues are parallel, triangles formed are similar. So, $\frac{AB}{113}=\frac{280 + 140}{140}$ (by basic proportionality or similar triangles property).

Step2: Solve for AB

Simplify the right - hand side: $\frac{420}{140}=3$. Then $AB = 113\times3=339$? Wait, maybe I misapplied. Wait, another approach: Let the distance from A to B be $x$. The two parallel lines (First and Second Ave) cut the transversals (Main and Oak). So by similar triangles, $\frac{x}{113}=\frac{280}{140}$. Wait, $280$ is the segment between First and Second on Main, $140$ between Second and the end on Main, and $113$ is the segment after B on Oak. Wait, correct proportion: The ratio of the segments on Main is $\frac{280}{140}=2$, so the ratio on Oak should be the same. Wait, no, maybe the triangles are similar with sides $280, 140$ and $x, 113$. Wait, actually, the two triangles (one with base $280 + 140$ and height related, and the other with base $140$ and height $113$? No, better: The lines First Ave and Second Ave are parallel, so the triangles are similar. So the ratio of corresponding sides is equal. Let’s denote $AB = x$. Then $\frac{x}{113}=\frac{280}{140}$. Wait, $280\div140 = 2$, so $x=113\times2 = 226$. Yes, that makes sense. Because the segment between First and Second Ave on Main is $280$, and between Second Ave and the end is $140$, so the ratio of those two is $280/140 = 2$. So the segment on Oak from A to B should be twice the segment from B to the end, which is $113\times2 = 226$.

Answer:

226 ft (corresponding to the option "226 ft")