# Lesson 7 Homework Practice Area And Perimeter Of Similar Figures Answer Key

## Lesson 7 Homework Practice Area and Perimeter of Similar Figures Answer Key

In this lesson, you will learn how to find the area and perimeter of similar figures. Similar figures are figures that have the same shape but not necessarily the same size. They have corresponding angles that are congruent and corresponding sides that are proportional.

To find the area and perimeter of similar figures, you can use the following formulas:

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The ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding sides. That is, if two figures are similar with a scale factor of k, then $$\fracA_1A_2 = k^2$$ where $A_1$ and $A_2$ are the areas of the two figures.

The ratio of the perimeters of two similar figures is equal to the ratio of their corresponding sides. That is, if two figures are similar with a scale factor of k, then $$\fracP_1P_2 = k$$ where $P_1$ and $P_2$ are the perimeters of the two figures.

Let's look at some examples of how to apply these formulas to find the area and perimeter of similar figures.

## Example 1

For each pair of similar figures, find the perimeter of the second figure.

Solution:

The first pair of triangles are similar with a scale factor of $$\frac47$$. To find the perimeter of the second triangle, we can use the formula $$\fracP_1P_2 = k$$ where $k = \frac47$. We know that $P_1 = 24$ m, so we can solve for $P_2$ by cross-multiplying: $$\frac24P_2 = \frac47 \implies P_2 = \frac24 \times 74 = 42$$ m. Therefore, the perimeter of the second triangle is 42 m.

The second pair of triangles are similar with a scale factor of $$\frac912$$. To find the perimeter of the second triangle, we can use the same formula as before: $$\fracP_1P_2 = k$$ where $k = \frac912$. We know that $P_1 = 8$ ft, so we can solve for $P_2$ by cross-multiplying: $$\frac8P_2 = \frac912 \implies P_2 = \frac8 \times 129 = 10.67$$ ft. Therefore, the perimeter of the second triangle is 10.67 ft.

## Example 2

For each pair of similar figures, find the area of the second figure.

Solution:

The first pair of rectangles are similar with a scale factor of $$\frac35$$. To find the area of the second rectangle, we can use the formula $$\fracA_1A_2 = k^2$$ where $k = \frac35$. We know that $A_1 = 48$ cm, so we can solve for $A_2$ by cross-multiplying: $$\frac48A_2 = \left(\frac35\right)^2 \implies A_2 = \frac48 \times 259 = 133.33$$ cm. Therefore, the area of the second rectangle is 133.33 cm.

The second pair of rectangles are similar with a scale factor of $$\frac14081$$. To find the area of the second rectangle, we can use the same formula as before: $$\fracA_1A_2 = k^2$$ where $k = \frac14081$. We know that $A_1 = 9$ in, so we can solve for $A_2$ by cross-multiplying: $$\frac9A_2 = \left(\frac14081\right)^2 \implies A_2 = \frac9 \times 81^2140^2 = 3.15$$ in. Therefore, the area of the second rectangle is 3.15 in.

For more practice and answer keys, you can check out the following resources:

[Answered: Lesson 7 Skills Practice - Area and bartleby]

[Perimeter and area of similar figures practice Flashcards Quizlet]

[Area and Perimeter of Similar Figures Flashcards Quizlet]