Question

cg, eh, fj are midsegments of △abd, △gcd, and △ghe, respectively. find each measure.

1. cg
2. eh
3. m∠dcg
4. m∠fjh
5. 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,

Answer:

Problem 9: Find \( CG \) (Midsegment in \( \triangle ABD \))

Step 1: Recall Midsegment Theorem

The midsegment of a triangle is parallel to the third side and half its length. In \( \triangle ABD \), \( CG \) is a midsegment (since \( C, G, H, E, F, J \) are midpoints implied by midsegment labels). The third side is \( AB = 33 \).

Step 2: Apply Midsegment Formula

Midsegment length \( = \frac{1}{2} \times \) third side length. So \( CG = \frac{1}{2} \times AB = \frac{1}{2} \times 33 = 16.5 \).

Problem 10: Find \( EH \) (Midsegment in \( \triangle GCD \))

Step 1: Identify Third Side for \( \triangle GCD \)

In \( \triangle GCD \), the third side is \( CD \) (or \( AB \)? Wait, \( CD \) is related to \( BC = 22 \)? Wait, \( \triangle GCD \): midsegment \( EH \) should be half of \( CD \). Wait, \( CD \) is equal to \( AB \)? No, \( BC = 22 \), maybe \( CD = 22 \)? Wait, midsegment \( EH = \frac{1}{2} \times CD \). If \( CD = 22 \), then \( EH = \frac{1}{2} \times 22 = 11 \).

Problem 11: Find \( m\angle DCG \) (Wait, \( m\angle FJH \))

Step 1: Corresponding Angles (Midsegments Parallel)

Since midsegments are parallel to sides, \( \angle FJH \) corresponds to \( \angle B = 57^\circ \) (by corresponding angles, as midsegments are parallel to \( AB \) or \( BD \)). So \( m\angle FJH = 57^\circ \).

Problem 13a: Distance from Warehouse \( W \) to \( X \) (Midsegment in \( \triangle ABC \))

Step 1: Midsegment Theorem in \( \triangle ABC \)

\( XY \) is a midsegment (since \( X, Y \) are midpoints, \( W \) is midpoint of \( XY \)? Wait, \( AC = 3.5 \times 2 = 7 \)? Wait, \( XY \) is midsegment, so \( XY = \frac{1}{2} \times AB \)? Wait, \( AB = 9 \) mi? Wait, \( X \) is midpoint of \( AC \), \( Y \) is midpoint of \( BC \). So \( XY \) is midsegment, length \( XY = \frac{1}{2} \times AB = \frac{1}{2} \times 9 = 4.5 \) mi. Then \( W \) is midpoint of \( XY \), so \( WX = \frac{1}{2} \times XY = \frac{1}{2} \times 4.5 = 2.25 \) mi? Wait, no: \( AC = 3.5 \times 2 = 7 \) mi, \( BC = 4 \times 2 = 8 \) mi? Wait, maybe \( X \) is midpoint of \( AC \), so \( CX = 3.5 \) mi, so \( AC = 7 \) mi. \( XY \) is midsegment, so \( XY \parallel AB \), \( XY = \frac{1}{2} AB = \frac{1}{2} \times 9 = 4.5 \) mi. \( W \) is midpoint of \( XY \), so \( WX = \frac{1}{2} XY = 2.25 \) mi.

Problem 13b: Total Trip Length (Warehouse \( W \) to \( A, B, C \) and back)

Step 1: Shortest Route (Triangle Inequality, Midsegments)

The shortest route is \( W \to X \to A \to B \to Y \to C \to W \)? Wait, no: \( W \) is midpoint of \( XY \), \( X \) midpoint of \( AC \), \( Y \) midpoint of \( BC \). So \( WX = 2.25 \) mi (from 13a), \( XA = 3.5 \) mi, \( AB = 9 \) mi, \( BY = 4 \) mi, \( YC = 3.5 \) mi, \( CW = WX + XC = 2.25 + 3.5 = 5.75 \) mi? Wait, no, better: total length is \( WX + XA + AB + BY + YC + CW \)? Wait, no, the truck starts at \( W \), goes to \( A \), \( B \), \( C \), then back to \( W \). Using midsegments: \( WA = WX + XA = 2.25 + 3.5 = 5.75 \) mi, \( WB = WY + YB = 2.25 + 4 = 6.25 \) mi, \( WC = 3.5 + 2.25 = 5.75 \) mi? No, simpler: the perimeter of the triangle formed by \( W, A, B, C \)? Wait, \( \triangle ABC \) has \( AC = 7 \) mi, \( BC = 8 \) mi, \( AB = 9 \) mi. \( W \) is midpoint of \( XY \), so \( WX = 4.5/2 = 2.25 \), \( XY = 4.5 \). The trip is \( W \to A \to B \to C \to W \). \( WA = XA + WX = 3.5 + 2.25 = 5.75 \), \( AB = 9 \), \( BC = 8 \), \( CW = CX + XW = 3.5 + 2.25 = 5.75 \). Total length: \( 5.75 + 9 + 8 + 5.75 = 28.5 \) mi? Wait, no, maybe using midsegment for total: \( W \) to \( A \): \( 3.5 + 2.25 = 5.75 \), \( A \) to \( B \): 9, \( B \) to \( C \): 8, \( C \) to \( W \): \( 3.5 + 2.25 = 5.75 \). Total: \( 5.75 + 9 + 8 + 5.75 = 28.5 \) mi.

Final Answers (Sample for Key Problems):

• 9. \( \boldsymbol{16.5} \)
• 10. \( \boldsymbol{11} \)
• 11. \( \boldsymbol{57^\circ} \)
• 13a. \( \boldsymbol{2.25} \) mi
• 13b. \( \boldsymbol{28.5} \) mi

(Note: Answers depend on precise midsegment identification; adjust if triangle labels differ.)