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In the figure above, if the area of the inscribed rectangula
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17 Dec 2016, 07:52
17 Dec 2016, 07:52
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In the figure above, if the area of the inscribed rectangular region is 32, then the circumference of the circle is
(A) \(20 \pi\)
(B) \(4 \pi \sqrt{5}\)
(C) \(4 \pi \sqrt{3}\)
(D) \(2 \pi \sqrt{5}\)
(E) \(2 \pi \sqrt{3}\)
Re: In the figure above, if the area of the inscribed rectangula
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29 Sep 2017, 07:49
29 Sep 2017, 07:49
2
The area of the rectangular is \(x*2x = 32\), from which we get x = 4 (we exclude x = -4 because a side of a rectangular cannot have a negative length.
Then, using Pitagora's theorem we can find the length of the hypotenuse of the triangle, which is half of the rectangular and we get it equal to \(\sqrt(80) = 4sqrt(5)\).
Finally, the circumference of the circle is given by \(2r*\pi\) where 2r is the diameter that in this case equals \(4sqrt(5)\).
Thus, the circumference is equal to \(4\pi\sqrt(5)\). Answer B!
Then, using Pitagora's theorem we can find the length of the hypotenuse of the triangle, which is half of the rectangular and we get it equal to \(\sqrt(80) = 4sqrt(5)\).
Finally, the circumference of the circle is given by \(2r*\pi\) where 2r is the diameter that in this case equals \(4sqrt(5)\).
Thus, the circumference is equal to \(4\pi\sqrt(5)\). Answer B!
Re: In the figure above, if the area of the inscribed rectangula
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01 Jan 2022, 22:43
01 Jan 2022, 22:43
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