High School

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The perimeter of Pentagon A is 15 inches, and its area is 30 in\(^2\). The perimeter of Pentagon B is 25 inches. What is the area of Pentagon B, assuming these pentagons are similar?

Answer :

Final answer:

To find the area of pentagon B, we can use the fact that similar polygons have their corresponding sides proportional.

Explanation:

To find the area of pentagon B, we can use the fact that similar polygons have their corresponding sides proportional. If the perimeter of pentagon A is 15 in and the perimeter of pentagon B is 25 in, we can set up the proportion:

perimeter of pentagon B / perimeter of pentagon A = corresponding side lengths of pentagon B / corresponding side lengths of pentagon A

To find the corresponding side lengths of pentagon B, we can multiply the corresponding side lengths of pentagon A by the ratio of the perimeters:

corresponding side lengths of pentagon B = corresponding side lengths of pentagon A * (perimeter of pentagon B / perimeter of pentagon A)

Once we have the corresponding side lengths of pentagon B, we can use the formula for the area of a regular pentagon: Area = (1/4) * sqrt(5 * (5 + 2 * sqrt(5))) * s^2, where s is the length of a side. Calculate the area of pentagon B using the corresponding side lengths.

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Rewritten by : Barada

Answer:

The area of pentagon B is [tex]83\frac{1}{3}\ in^{2}[/tex]

Step-by-step explanation:

step 1

Find the scale factor

we know that

If two figures are similar, then the ratio of its perimeters is equal to the scale factor

Let

z-----> the scale factor

x----> perimeter pentagon B

y----> perimeter pentagon A

[tex]z=\frac{x}{y}[/tex]

substitute the values

[tex]z=\frac{25}{15}[/tex]

Simplify

[tex]z=\frac{5}{3}[/tex] ----> scale factor

step 2

Find the area of pentagon B

we know that

If two figures are similar, then the ratio of its areas is equal to the scale factor squared

Let

z-----> the scale factor

x----> area pentagon B

y----> area pentagon A

[tex]z^{2}=\frac{x}{y}[/tex]

we have

[tex]z=\frac{5}{3}[/tex]

[tex]y=30\ in^{2}[/tex]

substitute and solve for x

[tex](\frac{5}{3})^{2}=\frac{x}{30}[/tex]

[tex](\frac{25}{9})=\frac{x}{30}[/tex]

[tex]x=30*(\frac{25}{9})=83.33\ in^{2}[/tex]

convert to mixed number

[tex]83.33=83\frac{1}{3}\ in^{2}[/tex]