Question

1. (03.06 mc)

the diagram below models the layout at a carnival where g, r, p, c, b, and e are various locations on the grounds. grpc is a parallelogram.

part a: identify a pair of similar triangles. (2 points)

part b: explain how you know the triangles from part a are similar. (4 points)

part c: find the distance from b to e and from p to e. show your work. (4 points)

Explanation:

Part A:

Step 1: Analyze the parallelogram

Since \( GRPC \) is a parallelogram, \( GR \parallel CP \) and \( GC \parallel RP \). Also, \( \angle GCB \) and \( \angle EPB \) are vertical angles? Wait, no, let's look at the triangles. Let's consider \( \triangle GCB \) and \( \triangle EPB \)? Wait, no, maybe \( \triangle GBC \) and \( \triangle EBP \)? Wait, actually, since \( GRPC \) is a parallelogram, \( GC \parallel RP \), so \( \angle GCB=\angle EPB \) (corresponding angles) and \( \angle GBC=\angle EBP \) (vertical angles). Wait, another approach: \( GRPC \) is a parallelogram, so \( GC = RP \) and \( GR = CP \). Wait, the triangles: \( \triangle GBC \) and \( \triangle EBP \)? Wait, no, let's check the sides. \( CB = 350 \), \( BP = 250 \), \( GB = 450 \)? Wait, no, \( GC = 400 \), \( GB = 450 \), \( CB = 350 \). Wait, actually, since \( GR \parallel CP \), then \( \angle GBC=\angle EBP \) (vertical angles) and \( \angle BGC=\angle BEP \) (alternate interior angles) because \( GC \parallel RP \)? Wait, maybe \( \triangle GCB \sim \triangle EPB \)? Wait, no, let's see: \( GRPC \) is a parallelogram, so \( GC \parallel RP \), so \( \angle GCB = \angle EPB \) (corresponding angles) and \( \angle CGB = \angle PEB \) (alternate interior angles). Also, \( \angle GBC = \angle EBP \) (vertical angles). So by AA similarity, \( \triangle GBC \sim \triangle EBP \). Wait, but maybe \( \triangle GCB \) and \( \triangle EPB \)? Wait, actually, the correct pair is \( \triangle GBC \) and \( \triangle EBP \) (or \( \triangle GCB \) and \( \triangle EPB \)). Alternatively, \( \triangle GBC \sim \triangle EBP \) by AA similarity.

Part B:

Step 1: Identify angles

Since \( GRPC \) is a parallelogram, \( GC \parallel RP \). Therefore, \( \angle GCB=\angle EPB \) (corresponding angles, as \( GC \) and \( RP \) are parallel and \( CP \) is a transversal). Also, \( \angle GBC=\angle EBP \) (vertical angles, since \( CB \) and \( BP \) are on a straight line, so \( \angle GBC \) and \( \angle EBP \) are vertical angles).

Step 2: Apply AA similarity

By the Angle - Angle (AA) similarity criterion, if two angles of one triangle are congruent to two angles of another triangle, the triangles are similar. Here, \( \angle GCB=\angle EPB \) and \( \angle GBC=\angle EBP \), so \( \triangle GBC \sim \triangle EBP \).

Part C:

Step 1: Set up the proportion

Since \( \triangle GBC \sim \triangle EBP \), the ratios of corresponding sides are equal. So \( \frac{CB}{BP}=\frac{GB}{EB}=\frac{GC}{EP} \). We know \( CB = 350 \) ft, \( BP = 250 \) ft, \( GC = 400 \) ft, \( GB = 450 \) ft.

Step 2: Find \( EB \)

Using \( \frac{CB}{BP}=\frac{GB}{EB} \), substitute the values: \( \frac{350}{250}=\frac{450}{EB} \). Cross - multiply: \( 350\times EB = 250\times450 \). Then \( EB=\frac{250\times450}{350}=\frac{112500}{350}=\frac{2250}{7}\approx321.43 \) ft? Wait, no, wait, maybe I mixed up the sides. Wait, \( \triangle GBC \sim \triangle EBP \), so corresponding sides: \( CB \) corresponds to \( BP \), \( GB \) corresponds to \( EB \), \( GC \) corresponds to \( EP \). Wait, no, \( \angle GCB=\angle EPB \), \( \angle GBC=\angle EBP \), so \( CB \) corresponds to \( BP \), \( GB \) corresponds to \( EB \), \( GC \) corresponds to \( EP \). So \( \frac{CB}{BP}=\frac{GB}{EB}=\frac{GC}{EP} \). So \( \frac{350}{250}=\frac{450}{EB} \). Solving for \( EB \): \( EB=\frac{450\times250}{350}=\frac{112500}{350}=\frac{2250}{7}\approx321.43 \) ft. Then for \( EP \): \( \frac{350}{250}=\frac{400}{EP} \), so \( EP=\frac{400\times250}{350}=\frac{100000}{350}=\frac{2000}{7}\approx285.71 \) ft. Wait, but maybe the similarity ratio is different. Wait, maybe \( \triangle GBC \) and \( \triangle EBP \) have a ratio of \( CB:BP = 350:250 = 7:5 \). So \( GB:EB = 7:5 \), so \( EB=\frac{5}{7}\times450=\frac{2250}{7}\approx321.43 \) ft, and \( GC:EP = 7:5 \), so \( EP=\frac{5}{7}\times400=\frac{2000}{7}\approx285.71 \) ft.

Answer:

Part A:

A pair of similar triangles is \( \boldsymbol{\triangle GBC} \) and \( \boldsymbol{\triangle EBP} \) (or \( \triangle GCB \) and \( \triangle EPB \)).

Part B:

1. Since \( GRPC \) is a parallelogram, \( GC \parallel RP \). Thus, \( \angle GCB=\angle EPB \) (corresponding angles, as \( GC \) and \( RP \) are parallel and \( CP \) is a transversal).
2. \( \angle GBC=\angle EBP \) (vertical angles, formed by the intersection of \( CB \) and \( BP \)).
3. By the AA (Angle - Angle) Similarity Criterion, if two angles of one triangle are congruent to two angles of another triangle, the triangles are similar. Since \( \angle GCB=\angle EPB \) and \( \angle GBC=\angle EBP \), \( \triangle GBC \sim \triangle EBP \).

Part C:

Step 1: Determine the similarity ratio

From \( \triangle GBC \sim \triangle EBP \), the ratio of corresponding sides is \( \frac{CB}{BP}=\frac{350}{250}=\frac{7}{5} \).

Step 2: Find \( EB \)

Using the proportion \( \frac{GB}{EB}=\frac{7}{5} \) (where \( GB = 450 \) ft):

$$ \frac{450}{EB}=\frac{7}{5} $$

Cross - multiply: \( 7\times EB = 450\times5 \)

$$ EB=\frac{2250}{7}\approx321.43 \text{ ft} $$

Step 3: Find \( EP \)

Using the proportion \( \frac{GC}{EP}=\frac{7}{5} \) (where \( GC = 400 \) ft):

$$ \frac{400}{EP}=\frac{7}{5} $$

Cross - multiply: \( 7\times EP = 400\times5 \)

$$ EP=\frac{2000}{7}\approx285.71 \text{ ft} $$