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ABCD is a square and AEC and AFC are one fourth of the circumference

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ABCD is a square and AEC and AFC are one fourth of the circumference  [#permalink]

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New post 03 Jul 2018, 09:25
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A
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C
D
E

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  95% (hard)

Question Stats:

35% (03:28) correct 65% (02:43) wrong based on 40 sessions

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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

Attachment:
Image50.gif
Image50.gif [ 2.19 KiB | Viewed 836 times ]

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Re: ABCD is a square and AEC and AFC are one fourth of the circumference  [#permalink]

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New post 03 Jul 2018, 12:44
OA: C

Let side of Square = \(a\)

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 \(\frac{Area of Unshaded portion}{Area of Shaded portion}= \frac{(28-22)}{(22-14)} =\frac{6}{8}=\frac{3}{4}\)
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Re: ABCD is a square and AEC and AFC are one fourth of the circumference  [#permalink]

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New post 03 Jul 2018, 13:03
Image

Let's assume the side of the square to be 7. The area of this square is 49.

Similarly, as the length of the sector is also 7, the area of one sector \(\frac{1}{4}*\frac{22}{7}*7^2 = \frac{77}{2}\)
There are 2 such sectors AEC and AFC each having the same area, making the total area \(77\)

We use the following diagram to find the area of the shaded region
Attachment:
Image50.gif


Area of AEC = 1 + 2 | Area of AFC = 2 + 3 | Area of Square = 1 + 2 + 3

The area of shaded region = Area of AEC + Area of AFC - Area of Square = 77 - 49 = 28
Area of unshaded region = Area of square - Area of shaded region = 49 - 28 = 21

Therefore, the ratio of the unshaded to the shaded region is 21:28 or 3:4(Option C)
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Re: ABCD is a square and AEC and AFC are one fourth of the circumference  [#permalink]

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New post 03 Jul 2018, 13:14
Bunuel wrote:
Image

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

Attachment:
Image50.gif


Area of unshaded region=Area of square-2*Area of segment AEC or AFC

Let the side of the square be 'a' unit.

Area of segment= Area of sector-Area of triangle=\(\frac{1}{2}\pi*a-\frac{1}{2} \sqrt{2}a*\frac{a}{\sqrt{2}}\)
=\(\frac{\pi a^2}{4}-\frac{a^2}{2}\)
=\(\frac{a^2(\pi-2)}{4}\)

Now, the required ratio=\(\frac{A_{shaded}}{A_{unshaded}}\)=\(\frac{Area of square}{2*Area of segments AEC or AFC}-1\)
=\(\frac{a^2}{a^2(\pi-2)/2}-1\)
=\(\frac{2}{\pi-2}-1\)
=\(\frac{4-\pi}{\pi-2}\)
=\(\frac{3}{4}\)

Ans. C
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Re: ABCD is a square and AEC and AFC are one fourth of the circumference  [#permalink]

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New post 04 Jul 2018, 12:51
Princ wrote:
OA: C

Let side of Square = \(a\)

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 \(\frac{Area of Unshaded portion}{Area of Shaded portion}= \frac{(28-22)}{(22-14)} =\frac{6}{8}=\frac{3}{4}\)


I'm confused how you get the first equation. Could you please elaborate? It says AEC and AFC are 1/4 of circumference i.e. 1/4*2*pi*a

I fail to see how this came up: \(2*\frac{\pi*a^2}{4}-a^2\)
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ABCD is a square and AEC and AFC are one fourth of the circumference  [#permalink]

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New post 04 Jul 2018, 13:07
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Quote:
I'm confused how you get the first equation. Could you please elaborate? It says AEC and AFC are 1/4 of circumference i.e. 1/4*2*pi*a

I fail to see how this came up: \(2*\frac{\pi*a^2}{4}-a^2\)


Hi,

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.

Go through the solution posted by Princ now, should make things clearer. :grin:
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Re: ABCD is a square and AEC and AFC are one fourth of the circumference  [#permalink]

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New post 05 Jul 2018, 11:34
1
rajudantuluri

munzuto has already explained how I got to first equation.
In this question, Double counting concept is similar to two overlapping set problem.
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Re: ABCD is a square and AEC and AFC are one fourth of the circumference &nbs [#permalink] 05 Jul 2018, 11:34
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