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Updated on: 14 Jul 2019, 21:26
Originally posted by bumblebee14 on 01 Mar 2015, 08:31.
Last edited by Bunuel on 14 Jul 2019, 21:26, edited 2 times in total.
Last edited by Bunuel on 14 Jul 2019, 21:26, edited 2 times in total.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
82% (01:06) correct
18%
(01:20)
wrong
based on 782
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In the figure below, a square is inscribed in a circle. If the area of the square region is 16, what is the area of the circular region?
(A) 2π
(B) 4π
(C) 8π
(D) 12π
(E) 16π
Show SpoilerAnswer
Attachment:
square-in-circle.png [ 19.04 KiB | Viewed 67771 times ]
Area of a square = (Diagonal)^2 /2
16 = (Diagonal)^2 /2
32 = (Diagonal)^2
Diagonal = 4 (2)^(1/2)
Diagonal = Diameter in the given figure
Radius = Diameter/2
Radius = 2(2)^(1/2)
Area of a circle = πr^2
= π [2(2)^(1/2)]^2
= 8π
Answer is C
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sashiim20
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07 Jul 2019, 08:59
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bumblebee14
Let side of Square be \(= a\)
Radius of Circle \(= r\)
Area of square \(= a^2= 16\)
Side of square \(= a = 4\)
Diagonal of square \(= 4\sqrt{2}\)
Diagonal of square \(=\) Diameter of circle
\(4\sqrt{2} = 2r\)
\(r = \frac{4\sqrt{2}}{2} = 2\sqrt{2}\)
Area of circle \(= \pi r^2 = \pi (2\sqrt{2})^2 = 8\pi\)
Answer C
27 Jul 2019, 14:35
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I made the same error as you did. Pythagoras does work here, but 16π is a wrong from calculation error.
\(4^2 + 4^2 = c^2\)
\(16 + 16 = c^2\) (--> THIS DOES NOT EQUATE TO C = 16!)
\(32 = c^2\)
\(c = 4 \sqrt{2}\)
c = the diameter (d) of this circle. Need the radius which is \(\frac{d}{2}\).
\(r = \frac{4 \sqrt{2}}{2}\)
\(r = 2 \sqrt{2}\)
Now plug in our radius amount into the circle area formula \(A = π r^2\)
\(A = π (2 \sqrt{2})^2\)
\(= π (4 * 2)\)
= 8π
\(4^2 + 4^2 = c^2\)
\(16 + 16 = c^2\) (--> THIS DOES NOT EQUATE TO C = 16!)
\(32 = c^2\)
\(c = 4 \sqrt{2}\)
c = the diameter (d) of this circle. Need the radius which is \(\frac{d}{2}\).
\(r = \frac{4 \sqrt{2}}{2}\)
\(r = 2 \sqrt{2}\)
Now plug in our radius amount into the circle area formula \(A = π r^2\)
\(A = π (2 \sqrt{2})^2\)
\(= π (4 * 2)\)
= 8π
hanutsingh
