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ABCD is a square .........

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ABCD is a square ......... [#permalink] New post 27 Aug 2016, 23:49
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#GREpracticequestion ABCD is a square and AEC.gif
#GREpracticequestion ABCD is a square and AEC.gif [ 2.19 KiB | Viewed 118 times ]


ABCD is a square and AEC and AFC are one fourth of the circumference of the circle whose radius is equal to the length of the side of the square ABCD. Ratio of unshaded to shaded region?


(A) 22 : 3
(B) 4 : 3
(C) 3 : 4
(D) 7 : 4
(E) cannot be determined
[Reveal] Spoiler: OA

Last edited by Carcass on 27 Jan 2019, 03:52, edited 3 times in total.
corrected the OA
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Re: ABCD is a square ......... [#permalink] New post 13 Apr 2018, 00:02
I would say c
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Re: ABCD is a square ......... [#permalink] New post 30 Apr 2018, 01:07
Me too c
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Re: ABCD is a square ......... [#permalink] New post 21 Jan 2019, 17:43
Answer should be C.

Let side of Square = \(a\)

If AEC is 1/4 of circumference it means that Angle ADC is 90 as the circle is cut into 4 quarters. (also evident from the fact that ADC is the angle of a square)
Since AD is the radius as well, we can find the area of the arc AEC. Similarly we can find the area of the Arc AFC. Now if you add the two areas up, you notice that the shaded region is counted twice. To remove this double counting and excess area you subtract the area of the square. This gives us the area of the shaded region.

Area of Shaded portion = \(2*\frac{\pi*a^2}{4}-a^2\)\(=\) \(\frac{\pi*a^2}{2}-a^2 =\frac{\pi*a^2-2*a^2}{2}\)

Area of Unshaded portion = Area of Square - Area of Shaded portion

= \(a^2 - (2*\frac{\pi*a^2}{4}-a^2)\)\(=\) \(2*a^2 - \frac{\pi*a^2}{2}=\frac{4*a^2 - \pi*a^2}{2}\)

\(\frac{Area of Unshaded portion}{Area of Shaded portion}=\frac{4*a^2 - \pi*a^2}{\pi*a^2-2*a^2}=\frac{4 - \pi}{\pi-2}\)

Putting \(\pi = \frac{22}{7}\) in above ratio , we get \(\frac{Area of Unshaded portion}{Area of Shaded portion}= \frac{(28-22)}{(22-14)} =\frac{6}{8}=\frac{3}{4}\)

Answer C
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Re: ABCD is a square ......... [#permalink] New post 23 Jan 2019, 21:07
AE wrote:
Answer should be C.

Let side of Square = \(a\)

If AEC is 1/4 of circumference it means that Angle ADC is 90 as the circle is cut into 4 quarters. (also evident from the fact that ADC is the angle of a square)
Since AD is the radius as well, we can find the area of the arc AEC. Similarly we can find the area of the Arc AFC. Now if you add the two areas up, you notice that the shaded region is counted twice. To remove this double counting and excess area you subtract the area of the square. This gives us the area of the shaded region.

Area of Shaded portion = \(2*\frac{\pi*a^2}{4}-a^2\)\(=\) \(\frac{\pi*a^2}{2}-a^2 =\frac{\pi*a^2-2*a^2}{2}\)

Area of Unshaded portion = Area of Square - Area of Shaded portion

= \(a^2 - (2*\frac{\pi*a^2}{4}-a^2)\)\(=\) \(2*a^2 - \frac{\pi*a^2}{2}=\frac{4*a^2 - \pi*a^2}{2}\)

\(\frac{Area of Unshaded portion}{Area of Shaded portion}=\frac{4*a^2 - \pi*a^2}{\pi*a^2-2*a^2}=\frac{4 - \pi}{\pi-2}\)

Putting \(\pi = \frac{22}{7}\) in above ratio , we get \(\frac{Area of Unshaded portion}{Area of Shaded portion}= \frac{(28-22)}{(22-14)} =\frac{6}{8}=\frac{3}{4}\)

Answer C


Now if you add the two areas up, you notice that the shaded region is counted twice. To remove this double counting and excess area you subtract the area of the square

Area of Shaded portion = \(2*\frac{\pi*a^2}{4}-a^2\)\(=\) \(\frac{\pi*a^2}{2}-a^2 =\frac{\pi*a^2-2*a^2}{2}\)


Please can you simplify this,it's quite confusing.
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Re: ABCD is a square ......... [#permalink] New post 23 Jan 2019, 22:36
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Expert's post
Sonalika42 wrote:
ABCD is a square and AEC and AFC are one fourth of the circumference of the circle whose radius is equal to the length of the side of the square ABCD. Ratio of unshaded to shaded region? image is attached below



(A) 22 : 3
(B) 4 : 3
(C) 3 : 4
(D) 7 : 4
(E) cannot be determined

Join AC, it divides the square into 2 equal parts..
Let us take one part ABFC, ABC is isosceles right angled triangle with sides, say 2, so area = (1/2)*2*2=2
Area of ABFC.. ABFC is 1/4th of a circle with side 2, so area = \(\frac{\pi*2^2}{4}=\pi\)
so shaded portion = area of ABFC - area of ABC = \(\pi-2\)
But there are two shade portion, one each side of AC, so area = 2(\(\pi-2\))=\(2\pi-4\)

therefore, the area of unshaded portion = area of square - area of shaded portion = \(2*2-2(\pi-2)=4-2\pi+4\)=\(8-2\pi\)

Ratio unshaded to shaded = \(8-2\pi\):\(2\pi-4\)=\(4-\pi\):\(\pi-2\)=\(\frac{6}{7}:\frac{8}{7}=6:8=3:4\)
C
Attachments

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_________________

Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

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Re: ABCD is a square ......... [#permalink] New post 23 Jan 2019, 23:00
Many thanks! @chetan2u
Re: ABCD is a square .........   [#permalink] 23 Jan 2019, 23:00
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