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. SCULPTURE An artist creates two metal sculptures in the shape of regular octagons. A side of the larger sculpture is 3.5 times longer than a side of the smaller octagon. The area of the smaller octagon is 19.28 square inches.
a. What is the area of the larger octagon?
b. The artist is going to pack the sculptures in a cylindrical box to take them to an art show. To the nearest inch, what is the diameter of the smallest box he can use? Explain your reasoning. (Hint: Find the radius of the larger octagon.)

Sagot :

facundo3141592

By using what we know about octagons, we will see that:

  • a) The area is 244.89 in^2
  • b) The diameter is 18.58 in

What is the area of the larger octagon?

For an octagon of side length S, the area is:

A = 2*(1 + √2)*S^2

In this case, the area of the smaller octagon is:

19.28 in^2 =  2*(1 + √2)*S^2

Solving for S we get:

S = √( 19.28 in^2/(2*(1 + √2))) = 2.03 in

If we apply an enlargement of a factor 3.5, the new side length is:

S' = 3.5*2.03 in = 7.11 in

Then the area of the larger octagon is:

A' = 2*(1 + √2)*(7.11 in)^2 = 244.89 in^2

b) For an octagon of side length S, the radius is:

R = S*√(1 + 1/√2)

Then the radius of the larger octagon is:

R' = (7.11 in)*√(1 + 1/√2) = 9.29 in

Then the diameter of the box must be:

D = 2*R' = 2*9.29 in = 18.58 in

If you want to learn more about octagons, you can read:

https://brainly.com/question/1592456

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