Question

a triangle has an area of 6 square inches and a perimeter of 12 inches. suppose it is dilated by some scale factor, an
the area and perimeter of the image are calculated.
match each graph with the relationship it represents.

1: scale factor is the ( x )-value; perimeter is the ( y )-value
2: scale factor is the ( x )-value; area is the ( y )-value
3: perimeter is the ( x )-value; scale factor is the ( y )-value
4: area is the ( x )-value; scale factor is the ( y )-value

Explanation:

To solve this, we analyze the relationships between scale factor, perimeter, and area of a dilated triangle:

Key Relationships:

• Perimeter and Scale Factor: Perimeter is a linear measurement. If the scale factor is \( k \), the new perimeter \( P' = k \times \) original perimeter. This is a linear relationship (straight line through the origin, \( y = mx \)).
• Area and Scale Factor: Area is a two - dimensional measurement. If the scale factor is \( k \), the new area \( A' = k^{2} \times \) original area. This is a quadratic relationship (curved graph, \( y = ax^{2} \)).

Matching Graphs to Relationships:

1. Relationship 1: Scale factor (\( x \)) → Perimeter (\( y \))

Perimeter and scale factor have a linear relationship (\( y = kx \), since perimeter scales linearly with the scale factor). Graph B is a straight line through the origin, so it matches Relationship 1.

1. Relationship 2: Scale factor (\( x \)) → Area (\( y \))

Area and scale factor have a quadratic relationship (\( y = kx^{2} \), since area scales with the square of the scale factor). Graph D is a curved (parabola - like) graph, so it matches Relationship 2.

1. Relationship 3: Perimeter (\( x \)) → Scale factor (\( y \))

Solving \( x = k_{\text{original}} \times y \) (since perimeter \( x = \) original perimeter \( \times y \) (scale factor)), we get \( y=\frac{x}{k_{\text{original}}} \), which is a linear relationship. Graph C is a straight line through the origin, so it matches Relationship 3.

1. Relationship 4: Area (\( x \)) → Scale factor (\( y \))

Solving \( x = k_{\text{original}} \times y^{2} \) (since area \( x=\) original area \( \times y^{2} \) (scale factor squared)), we get \( y = \sqrt{\frac{x}{k_{\text{original}}}} \), which is a square - root (curved) relationship. Graph A is a curved graph (square - root - like), so it matches Relationship 4.

Final Matches:

• Graph A ↔ Relationship 4
• Graph B ↔ Relationship 1
• Graph C ↔ Relationship 3
• Graph D ↔ Relationship 2

(If you need to present the answer as per the matching structure in the problem, the pairs are:
A - 4, B - 1, C - 3, D - 2)

Answer:

To solve this, we analyze the relationships between scale factor, perimeter, and area of a dilated triangle:

Key Relationships:

• Perimeter and Scale Factor: Perimeter is a linear measurement. If the scale factor is \( k \), the new perimeter \( P' = k \times \) original perimeter. This is a linear relationship (straight line through the origin, \( y = mx \)).
• Area and Scale Factor: Area is a two - dimensional measurement. If the scale factor is \( k \), the new area \( A' = k^{2} \times \) original area. This is a quadratic relationship (curved graph, \( y = ax^{2} \)).

Matching Graphs to Relationships:

1. Relationship 1: Scale factor (\( x \)) → Perimeter (\( y \))

Perimeter and scale factor have a linear relationship (\( y = kx \), since perimeter scales linearly with the scale factor). Graph B is a straight line through the origin, so it matches Relationship 1.

1. Relationship 2: Scale factor (\( x \)) → Area (\( y \))

Area and scale factor have a quadratic relationship (\( y = kx^{2} \), since area scales with the square of the scale factor). Graph D is a curved (parabola - like) graph, so it matches Relationship 2.

1. Relationship 3: Perimeter (\( x \)) → Scale factor (\( y \))

Solving \( x = k_{\text{original}} \times y \) (since perimeter \( x = \) original perimeter \( \times y \) (scale factor)), we get \( y=\frac{x}{k_{\text{original}}} \), which is a linear relationship. Graph C is a straight line through the origin, so it matches Relationship 3.

1. Relationship 4: Area (\( x \)) → Scale factor (\( y \))

Solving \( x = k_{\text{original}} \times y^{2} \) (since area \( x=\) original area \( \times y^{2} \) (scale factor squared)), we get \( y = \sqrt{\frac{x}{k_{\text{original}}}} \), which is a square - root (curved) relationship. Graph A is a curved graph (square - root - like), so it matches Relationship 4.

Final Matches:

• Graph A ↔ Relationship 4
• Graph B ↔ Relationship 1
• Graph C ↔ Relationship 3
• Graph D ↔ Relationship 2

(If you need to present the answer as per the matching structure in the problem, the pairs are:
A - 4, B - 1, C - 3, D - 2)