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In the figure shown, ABCD is a square with side length 5 sqr

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In the figure shown, ABCD is a square with side length 5 sqr  [#permalink]

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New post Updated on: 13 Jul 2013, 08:27
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A
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In the figure shown, ABCD is a square with side length 5 sqrt 2. Two quarter-circles BED and DFB are centered at A and C, respectively. What is the area of the shaded region?

A. 50pi-25
B. 25pi−50
C. 50pi/3
D. 25pi/3
E. 25pi

Originally posted by sudhir18n on 06 Jul 2011, 09:30.
Last edited by Bunuel on 13 Jul 2013, 08:27, edited 1 time in total.
Edited the question.
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Re: Adv GMAT- Manhattan  [#permalink]

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New post 06 Jul 2011, 09:55
4
2
sudhir18n wrote:
In the figure shown, ABCD is a square with side length 5 sqrt 2. Two quarter-circles BED and DFB are centered at A and C, respectively. What is the area of the shaded region?


A.50pi-25
B.25pi−50
C.50pi/3
D.25pi/3
E.25pi

I got the answer.. but whats the fastest way to do this?


There are a lot of ways to do this. You could add the two areas of the quarter-circles ABD and BCD: when you do that, you'll be adding the area of ADFB once, the area of CDEB once, and the area of the shaded region twice. That's just the area of the entire square plus the area of the shaded region. So if you then subtract the area of the square, you'll have the area of the shaded region (this would be easier to demonstrate by drawing diagrams, but hopefully what I mean is clear - it's very much like a Venn diagram question, in which you add the numbers in each of two overlapping groups; in that case you're counting the overlap twice).

The area of each quarter-circle is (1/4)*Pi*(5sqrt(2))^2 = 25*Pi/2. Adding the two quarter-circles' areas gives 25*Pi. Then subtracting the area of the square to get the area of the shaded region gives 25*Pi - (5*sqrt(2))^2 = 25*Pi - 50.

You might also be able to pick the right answer here without really doing any work. The area of the square is 50, and the area of the shaded region looks to be about half the area of the square. That alone gets you to answers B or D immediately. Since you can pretty reliably guess that you'll need to subtract one area from another to get the answer, B almost has to be correct.
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Re: Adv GMAT- Manhattan  [#permalink]

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New post 06 Jul 2011, 09:56
radius of each quarter = 5sqrt2.
ABCD = DFBC + ADFB
DFBC = PI*(SIDE)*SIDE/4 = 25PI/2
ADFB = ABCD - DFBC = 50 - 25PI/2
DFBE = ADEB - ADFB = 25PI/2 - 50 + 25PI/2 = 25PI - 50. B
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Re: Adv GMAT- Manhattan  [#permalink]

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New post 06 Jul 2011, 10:47
1
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sudhir18n wrote:
In the figure shown, ABCD is a square with side length 5 sqrt 2. Two quarter-circles BED and DFB are centered at A and C, respectively. What is the area of the shaded region?


A.50pi-25
B.25pi−50
C.50pi/3
D.25pi/3
E.25pi

I got the answer.. but whats the fastest way to do this?


Solved the same way Ian suggested:

\((5\sqrt{2})^2=\frac{\pi (5\sqrt{2})^2}{4}+\frac{\pi (5\sqrt{2})^2}{4}-Region\)

\(Region=25\pi\) − \(50\)

Ans:"B"
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Re: Adv GMAT- Manhattan  [#permalink]

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New post 08 Jul 2011, 11:23
sudhir18n wrote:
In the figure shown, ABCD is a square with side length 5 sqrt 2. Two quarter-circles BED and DFB are centered at A and C, respectively. What is the area of the shaded region?


A.50pi-25
B.25pi−50
C.50pi/3
D.25pi/3
E.25pi

I got the answer.. but whats the fastest way to do this?



The fastest way according to me is as below

Half of the required area is = the area of quarter of the circle centering at C- Area of the triangle BCD
= pi * \(Radius ^2\)* 1/2 - 1/2*Height of the triangle BCD * Base of the triangle BCD
=pi * \({5sqrt2} ^2\) -1/2 *\(5sqrt2\) *\(5sqrt2\)
= (25*pi )/2-25

so the required area is 25pi-50........ :) :)
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Re: Adv GMAT- Manhattan  [#permalink]

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New post 08 Jul 2011, 13:34
i solved the Q in one of the longer ways possible..i felt Ian's solution is the quickest way to reach the ans..thanks Ian
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Re: In the figure shown, ABCD is a square with side length 5 sqr  [#permalink]

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New post 27 Apr 2016, 03:17
for me easiest way was:

half Required area = sector area - area of triangle
=( (central angle*pi*r^2)/2*pi) - ((1/2 )* b*h)

Required area = 2 * half Required area

hence answer is B
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Re: In the figure shown, ABCD is a square with side length 5 sqr  [#permalink]

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New post 27 Apr 2016, 19:14
area of the square 50
area of 1/4 of circle: 25pi/2
now...from 50 we deduct 25pi/2 = and get value of 1 un-shaded portion.
(50-25pi/2)*2 = 100-25pi
from the whole area: 50, we deduct sum of the both un-shaded regions:
50-(100-25pi) = 25pi-50.
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Re: In the figure shown, ABCD is a square with side length 5 sqr  [#permalink]

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New post 01 Jun 2017, 17:24
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sudhir18n wrote:
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The attachment 166-1.gif is no longer available
In the figure shown, ABCD is a square with side length 5 sqrt 2. Two quarter-circles BED and DFB are centered at A and C, respectively. What is the area of the shaded region?

A. 50pi-25
B. 25pi−50
C. 50pi/3
D. 25pi/3
E. 25pi


This question appeared on an elementary school exam in Singapore, the leading country in mathematics as of now. When the teachers were asked how they were able to explain such a complex problem to children the answer was "using shapes and puzzle pieces."
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Re: In the figure shown, ABCD is a square with side length 5 sqr  [#permalink]

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Re: In the figure shown, ABCD is a square with side length 5 sqr &nbs [#permalink] 05 Aug 2018, 14:41
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