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12 Feb 2018, 00:15
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
76% (01:56) correct
24%
(02:14)
wrong
based on 88
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In the square above with side 4, the ratio (area of shaded region)/(area of unshaded region)
(A) (2 + x)/4
(B) (4 + x)/8
(C) 2
(D) (4 + x)/(4 - x)
(E) 2x
Attachment:
2018-02-12_1108.png [ 11.51 KiB | Viewed 6821 times ]
12 Feb 2018, 01:56
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Bunuel
We'll show two approaches:
The first one is Logical and relies on the maximal and minimal values:
The shaded area is at least half of the total area and at most the whole area.
Therefore our ratio is between half:half = 1 to everything:nothing = infinity.
So we need to find an answer that has value 1 for x = 0 and value infinity for x = 4
Only (D) fits.
The standard, Precise approach is to just calculate:
The shaded area is a trapezoid has bases of length 4 and x and height of length 4.
Therefore its area is (4+x)/2 * 4 = 2(4+x)
The unshaded area is a right triangle with sides 4 and 4 - x
Therefore its area is 4(4 - x)/2 = 2(4 - x)
So the ratio is 2(4+x)/2(4-x) = (4+x)/(4 - x)
Once again, (D) is our answer.
12 Feb 2018, 02:01
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Attachment:
2018-02-12_1108.png [ 11.43 KiB | Viewed 5819 times ]
The right angled triangle has a base of 4 and height of 4-x.
The area of this right angled triangle is \(\frac{1}{2}*4*(4-x) = 2(4-x)\)
Area of the shaded portion is \(4x + 8-2x = 2(x+4)\)
Area of the unshaded portion is 16 - 2(x+4) = 8-2x = 2(4-x)
Therefore, the ratio of the shaded region to the unshaded region is \(\frac{2(4+x)}{2(4-x)} = \frac{(4+x)}{(4-x)}\) (Option D)
