Question

Determine the scale factor for each pair of similar triangles. The perimeter of \(\triangle R S T\) is 57 feet. If \(\triangle R S T \sim \triangle H K N\) and the scale factor of \(\triangle R S T\) to \(\triangle H K N\) is \(\frac{3}{2}\), find the perimeter of \(\triangle H K N\).

Short answer

The perimeter of \( \triangle H K N \) is 38 feet.

Step-by-step solution

Understand Given Information

We are given that the perimeter of \( \triangle R S T \) is 57 feet, and it is similar to \( \triangle H K N \) with a scale factor of \( \frac{3}{2} \) from \( \triangle R S T \) to \( \triangle H K N \).

Define Scale Factor Relationship

The scale factor \( \frac{3}{2} \) means each side and the perimeter of \( \triangle H K N \) is \( \frac{2}{3} \) of the corresponding length in \( \triangle R S T \). This is because the scale factor described for transformation from \( \triangle R S T \) to \( \triangle H K N \) implies that each dimension of \( \triangle H K N \) is \( \frac{3}{2} \) times that of \( \triangle R S T \), inversely it means the smaller triangle (\( \triangle R S T \)) is \( \frac{2}{3} \) times smaller than the larger triangle (\( \triangle H K N \)).

Calculate Perimeter of \( \triangle H K N \)

To find the perimeter of \( \triangle H K N \), apply the inverse of the given scale factor to the perimeter of \( \triangle R S T \). Since the scale factor from \( \triangle R S T \) to \( \triangle H K N \) is \( \frac{2}{3} \), the perimeter of \( \triangle H K N \) is \( 57 \times \frac{2}{3} = 38 \) feet.

Key concepts

Understanding Scale Factor

When we talk about the 'scale factor' in geometry, we're referring to how much we need to multiply dimensions of a shape to obtain a corresponding shape. In the case of similar triangles, this scale factor describes the ratio of corresponding sides between two triangles.
For instance, if we have two similar triangles and we say the scale factor is \( \frac{3}{2} \), it means each side of the larger triangle is 1.5 times the length of the sides of the smaller triangle.

• If multiplying the dimensions of Triangle A by \( \frac{3}{2} \) gives us Triangle B, then Triangle B is larger than Triangle A.
• Conversely, if we consider Triangle B as our starting triangle, the scale factor from B to A would then be \( \frac{2}{3} \).

In our exercise, this concept of scaling is applied to the perimeter as well. So, understanding this factor is the key first step to solving problems involving similar triangles.

Calculating Triangle Perimeter

The perimeter of a triangle is simply the total length around it, which means adding up the lengths of all its sides. For similar triangles, the perimeter of one is directly related to the perimeter of the other through their scale factor.
So if we know the perimeter of one triangle and the scale factor, we can figure out the perimeter of the other similar triangle.

• Take Triangle RST with a perimeter of 57 feet.
• We need to find the perimeter of Triangle HKN, knowing the scale factor is \( \frac{3}{2} \).
• Realizing that, by proportion, Triangle RST is the smaller triangle, we multiply its perimeter by the inverse of the scale factor, \( \frac{2}{3} \).
• This means Triangle HKN has a perimeter of \( 57 \times \frac{2}{3} = 38 \) feet.

By manipulating the scale factor correctly, we accurately determine the perimeter of the similar triangle.

Simplifying Geometry Problem Solving

Solving geometry problems can feel challenging, but breaking them into clear steps makes it manageable. Here's a simple strategy to use:

• Start by understanding all given information. Identify what's known and what needs to be found.
• Identify any relationships or formulas, such as the scale factor, that can connect different pieces of information.
• Apply these relationships to calculate the unknowns. Be systematic to avoid errors.

In our problem, we started by noting the given perimeters and the scale factor. We then used the relationship defined by the scale factor to find the unknown perimeter.
Tackle geometry using this step-by-step method, and you'll find that even complex problems become much simpler to solve!