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Bunuel
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POE method:

Circle has a circumference of \(16\pi\) = \(2*pi*R\) => \(R=8\). Area of circle = \(3.14*8^2\)= 192
Now let's eliminate options 192<C<D<E, since inscribed square has to have area less than the circle's one. And we have to elemenate A, because \(8 \sqrt{2}\) = 11, which is the way too small (6% of the area of our circle).

(A) \(8 \sqrt{2}\)
(B) 128
(C) 256
(D) \(128\pi\)
(E) \(256\pi\)

Answer (B)
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Bunuel


In the figure, the circle circumscribed about square ABCD has a circumference of \(16\pi\). The length of a side of square ABCD is s. What is the square’s area?


(A) \(8 \sqrt{2}\)

(B) 128

(C) 256

(D) \(128\pi\)

(E) \(256\pi\)


Attachment:
2018-04-24_1708.png


Since the circumference is 16π and C = πd , we see that the circle’s diameter = 16, which is also the length of the diagonal of the square.

Since diagonal = s√2, we have:

s√2 = 16

s = 16/√2

Thus, the area of the square is s^2 = (16/√2)^2 = 256/2 = 128.

Answer: B
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The length of the circumference (perimeter) is 16π.

The diameter of the circle is the diagonal of the circumscribed square.

Let us remember that:

16π = Diameter x π

So the diameter is 16, and the diagonal of the circumscribed square is also 16.

Let us remember that:

The area of a square of side S is Sexp2

In a square of side "A", the diagonal raised to 2 of the square is equal to 2 times the side squared.

In a square of side "S", 2 x Sexp2 = 16exp2

Then:

The area of the square with side S is:

Sexp2 = (16 x16)/2
= 8x16

A positive integer, whose unit is 8: ....8

Answer B
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