Question

1. the figure shows the roads connecting towns a, b, and c. a music company has a store in each town and a distributive warehouse w at the midpoint of road \\(\overline{xy}\\).

a. what is the distance from the warehouse to point x?
b. a truck starts at the warehouse, delivers instruments to the stores in towns a, b, and c (in this order) and then returns to the warehouse, assuming the driver takes the shortest possible route. what is the total length of the trip, route?

Explanation:

Step1: Identify Midline Theorem

In triangle \(ABC\), \(X\) and \(Y\) are midpoints (since marks indicate midpoints), so \(XY\) is a midline. By Midline Theorem, \(XY=\frac{1}{2}AB\). \(AB = 9\) mi, so \(XY=\frac{9}{2}=4.5\) mi? Wait, no, wait. Wait, \(W\) is midpoint of \(XY\)? Wait, the problem is distance from warehouse \(W\) to \(X\). Wait, first, let's check the triangle. \(C\) to \(X\) is 3.5 mi, \(X\) to \(A\) is equal to \(C\) to \(Y\) (midpoints). Wait, maybe \(XY\) is parallel to \(AB\) and half its length. But \(W\) is the midpoint of \(XY\), so distance from \(W\) to \(X\) is half of \(XY\). Wait, no, maybe I misread. Wait, the triangle has \(CA\) with \(C\) to \(X\) as 3.5 mi, \(X\) to \(A\) as equal (midpoint), so \(CA = 7\) mi? Wait, no, the side \(AB\) is 9 mi, \(BY = 4\) mi, so \(Y\) is midpoint of \(BC\) (since \(BY = YC\)? Wait, the marks: \(C\) to \(Y\) and \(Y\) to \(B\) have the same mark, so \(Y\) is midpoint of \(BC\). Similarly, \(X\) is midpoint of \(CA\) (since \(C\) to \(X\) and \(X\) to \(A\) have same mark? Wait, \(C\) to \(X\) is 3.5 mi, so \(X\) to \(A\) is also 3.5 mi, so \(CA = 7\) mi. Then \(XY\) is midline, so \(XY=\frac{1}{2}AB=\frac{9}{2}=4.5\) mi. Then \(W\) is midpoint of \(XY\), so \(WX=\frac{1}{2}XY=\frac{4.5}{2}=2.25\) mi? Wait, no, maybe the triangle is \(ABC\) with \(AB = 9\), \(BC = 8\) (since \(BY = 4\), so \(BC = 8\)), \(CA = 7\) (3.5*2). Then \(XY\) connects midpoints of \(CA\) and \(BC\), so \(XY\parallel AB\) and \(XY = \frac{1}{2}AB = 4.5\) mi. Then \(W\) is midpoint of \(XY\), so distance from \(W\) to \(X\) is half of \(XY\), so \(4.5/2 = 2.25\) mi? Wait, maybe I made a mistake. Wait, the problem says "distance from the warehouse \(W\) to point \(X\)". Let's re-express:

Wait, \(X\) is midpoint of \(CA\), \(Y\) is midpoint of \(BC\), so \(XY\) is midline, length \(XY = \frac{1}{2}AB = 4.5\) mi. \(W\) is midpoint of \(XY\), so \(WX = \frac{1}{2}XY = 2.25\) mi.

Step2: For part b, total route: Warehouse \(W\) to \(A\), \(A\) to \(B\), \(B\) to \(C\), \(C\) to \(W\)? Wait, no, the truck starts at warehouse, delivers to \(A\), \(B\), \(C\), then returns to warehouse. Wait, shortest possible route: \(W\) to \(X\) to \(A\), \(A\) to \(B\), \(B\) to \(Y\) to \(C\), \(C\) to \(X\) to \(W\)? No, better: \(W\) to \(X\) (2.25 mi), \(X\) to \(A\) (3.5 mi), \(A\) to \(B\) (9 mi), \(B\) to \(Y\) (4 mi), \(Y\) to \(C\) (4 mi), \(C\) to \(X\) (3.5 mi), \(X\) to \(W\) (2.25 mi)? No, that's not right. Wait, maybe the route is \(W\) to \(X\) (2.25), \(X\) to \(A\) (3.5), \(A\) to \(B\) (9), \(B\) to \(Y\) (4), \(Y\) to \(C\) (4), \(C\) to \(X\) (3.5), \(X\) to \(W\) (2.25)? No, that's too long. Wait, maybe the shortest route is \(W\) to \(X\) (2.25), \(X\) to \(A\) (3.5), \(A\) to \(B\) (9), \(B\) to \(Y\) (4), \(Y\) to \(C\) (4), \(C\) to \(X\) (3.5), \(X\) to \(W\) (2.25)? No, that's not. Wait, maybe the triangle's perimeter and the midline. Wait, no, let's do part a first correctly.

Wait, maybe I misread the triangle. Let's look at the labels: \(C\) to \(X\) is 3.5 mi, \(X\) to \(A\) is same (midpoint), so \(CA = 7\) mi. \(Y\) to \(B\) is 4 mi, \(Y\) to \(C\) is same (midpoint), so \(BC = 8\) mi. \(AB = 9\) mi. Then \(XY\) is midline, so \(XY = \frac{1}{2}AB = 4.5\) mi. \(W\) is midpoint of \(XY\), so \(WX = \frac{1}{2} \times 4.5 = 2.25\) mi. So part a: 2.25 mi.

For part b: Truck starts at \(W\), goes to \(A\) (via \(X\): \(WX + XA = 2.25 + 3.5 = 5.75\) mi), then \(A\) to \(B\) (9 mi), then \(B\) to \(C\) (8 mi), then \(C\) to \(W\) (via \(Y\): \(YC + YW = 4 + 2.25 = 6.25\) mi)? Wait, no, \(YW\) is same as \(WX\) (2.25 mi). Wait, \(C\) to \(Y\) is 4 mi, \(Y\) to \(W\) is 2.25 mi, so \(C\) to \(W\) is 4 + 2.25 = 6.25 mi. Then total route: \(W\) to \(A\): 5.75, \(A\) to \(B\): 9, \(B\) to \(C\): 8, \(C\) to \(W\): 6.25. Total: 5.75 + 9 + 8 + 6.25 = 29 mi? Wait, no, maybe the shortest route is \(W\) to \(X\) (2.25), \(X\) to \(A\) (3.5), \(A\) to \(B\) (9), \(B\) to \(Y\) (4), \(Y\) to \(C\) (4), \(C\) to \(X\) (3.5), \(X\) to \(W\) (2.25). Wait, that's 2.25 + 3.5 + 9 + 4 + 4 + 3.5 + 2.25 = 29 mi. Alternatively, maybe the triangle's perimeter plus \(XY\)? No, midline is inside. Wait, maybe I made a mistake in part a. Let's re-express:

Wait, the problem says "the figure shows the roads connecting towns A, B, and C. A music company has a store in each town and a distributive warehouse W at the midpoint of road XY." So \(XY\) is a road, \(W\) is midpoint of \(XY\). So \(X\) is on \(CA\), \(Y\) is on \(BC\), \(XY\) is a road. So \(CX = 3.5\) mi, so \(XA = 3.5\) mi (since \(X\) is midpoint? Wait, the marks: \(CX\) and \(XA\) have the same mark, so \(X\) is midpoint of \(CA\). Similarly, \(BY\) and \(YC\) have same mark, so \(Y\) is midpoint of \(BC\). Then \(XY\) is midline, so \(XY = \frac{1}{2}AB = \frac{9}{2} = 4.5\) mi. Then \(W\) is midpoint of \(XY\), so \(WX = \frac{1}{2} \times 4.5 = 2.25\) mi. So part a: 2.25 miles.

For part b: The truck starts at \(W\), goes to \(A\) (route: \(W\) to \(X\) to \(A\): \(2.25 + 3.5 = 5.75\) mi), then \(A\) to \(B\) (9 mi), then \(B\) to \(C\) (route: \(B\) to \(Y\) to \(C\): \(4 + 4 = 8\) mi), then \(C\) to \(W\) (route: \(C\) to \(X\) to \(W\): \(3.5 + 2.25 = 5.75\) mi). Wait, but that's \(5.75 + 9 + 8 + 5.75 = 28.5\) mi? Wait, no, \(C\) to \(X\) is 3.5, \(X\) to \(W\) is 2.25, so 5.75. \(W\) to \(X\) to \(A\) is 5.75. \(A\) to \(B\) is 9. \(B\) to \(Y\) to \(C\) is 8. So total: 5.75 + 9 + 8 + 5.75 = 28.5 mi. Wait, maybe I messed up the direction. Alternatively, maybe the shortest route is \(W\) to \(X\) (2.25), \(X\) to \(A\) (3.5), \(A\) to \(B\) (9), \(B\) to \(Y\) (4), \(Y\) to \(W\) (2.25), \(Y\) to \(C\) (4), \(C\) to \(X\) (3.5), \(X\) to \(W\) (2.25)? No, that's longer. Wait, no, the truck delivers to \(A\), \(B\), \(C\) in order, then returns to warehouse. So order: \(W \to A \to B \to C \to W\). So distance \(W\) to \(A\): \(WX + XA = 2.25 + 3.5 = 5.75\). \(A\) to \(B\): 9. \(B\) to \(C\): \(BY + YC = 4 + 4 = 8\). \(C\) to \(W\): \(CX + XW = 3.5 + 2.25 = 5.75\). Total: \(5.75 + 9 + 8 + 5.75 = 28.5\) mi. Wait, but maybe \(XY\) is parallel to \(AB\), but the road lengths: \(CA = 7\) (3.5*2), \(BC = 8\) (4*2), \(AB = 9\). Then the perimeter of triangle \(ABC\) is \(7 + 8 + 9 = 24\) mi. But the warehouse is inside, so the route through \(W\) would be adding the midline? No, maybe the shortest route is the perimeter of the triangle plus the length of \(XY\)? No, that doesn't make sense. Wait, maybe I made a mistake in part a. Let's check again.

Wait, the problem says "distance from the warehouse \(W\) to point \(X\)". \(W\) is midpoint of \(XY\), so \(WX = \frac{1}{2}XY\). \(XY\) is midline, so \(XY = \frac{1}{2}AB = 4.5\), so \(WX = 2.25\) mi. That seems right.

Answer:

a. The distance from \(W\) to \(X\) is \(\boldsymbol{2.25}\) miles.
b. The total length of the trip is \(\boldsymbol{28.5}\) miles (or depending on correct calculation, maybe 29, but likely 28.5).

(Note: For part b, if we consider \(W\) to \(A\) as \(WX + XA = 2.25 + 3.5 = 5.75\), \(A\) to \(B = 9\), \(B\) to \(C = 8\), \(C\) to \(W = CX + XW = 3.5 + 2.25 = 5.75\), total \(5.75 + 9 + 8 + 5.75 = 28.5\) miles.)