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In the figure below, rectangles CABD and PMNQ are inscribed in a circl

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In the figure below, rectangles CABD and PMNQ are inscribed in a circl [#permalink]

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New post 28 Aug 2015, 02:23
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Question Stats:

63% (03:59) correct 38% (04:12) wrong based on 56 sessions

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In the figure below, rectangles CABD and PMNQ are inscribed in a circle with center O as shown, such that segment CD lies on segment MN, and points A, B, P, and Q lie on the circle. If the radius of the circle is 25, rectangles CABD and PMNQ have the same dimensions, and the shortest distance from O to segment MN is 6, what is the area of the shaded region?

Image


(A) \(625\pi-420\)
(B) \(625\pi-500\)
(C) \(625\pi-640\)
(D) \(625\pi-840\)
(E) \(625\pi-960\)



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[Reveal] Spoiler:
Attachment:
JD_NONSBD_Hard004.GIF
JD_NONSBD_Hard004.GIF [ 4.72 KiB | Viewed 1483 times ]
[Reveal] Spoiler: OA

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Re: In the figure below, rectangles CABD and PMNQ are inscribed in a circl [#permalink]

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New post 28 Aug 2015, 19:31
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Let the width of rectangle NQ = x
Length of rectangle PQ =y
(y/2)^2+(x+6)^2 = 25

Considering Pythagorean triples (3,4,5) => (3*5,4*5,5*5)
y/2 = 15 => y = 30
x+6 = 20 => x = 14

Area or rectangle = 30*14 = 420

Area of shaded region = Area of circle - 2*area of rectangle
= 625*pi - 840

Answer (D)
Attachments

shadedarea.png
shadedarea.png [ 12.28 KiB | Viewed 1160 times ]

Kudos [?]: 25 [2], given: 31

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Re: In the figure below, rectangles CABD and PMNQ are inscribed in a circl [#permalink]

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New post 30 Aug 2015, 00:53
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pluto82 wrote:
Let the width of rectangle NQ = x
Length of rectangle PQ =y
(y/2)^2+(x+6)^2 = 25

Considering Pythagorean triples (3,4,5) => (3*5,4*5,5*5)
y/2 = 15 => y = 30
x+6 = 20 => x = 14

Area or rectangle = 30*14 = 420

Area of shaded region = Area of circle - 2*area of rectangle
= 625*pi - 840

Answer (D)



To determine the area of shaded region : {(Area of circle)- (Area of ABDC +Area of MNQP)}

We are given the radius (25 units).

Let R be mid point of MN such that OR =6 units (the shortest distance from O to segment MN is 6)

Let T be a point on PQ such that RT = x units

OT =OR + RT = 6+x units

Let TQ = y units.

OTQ is a right angled triangled at T

Using Pythagorean Principle :

OT ^2 +TQ ^2 = OQ
(6+x)^2 + y ^2 = 25 ^2

The triplets could be (15, 20) or (20 ,15). So length & breadth of the rectangles can be (9, 20) or (14,15).

Case 1:(9, 20*2)
Area of the rectangle : 360 units
Area of 2 rectangle : 2 * 360 : 720 units

Area of shaded region = area of circle - area of 2 rectangles.
========= 625 pie - 720

Case 2:(14 , 15*2)
Area of rectangle : 420 units
Area of 2 rectangles : 840 units

Area of shaded region :area of circle - area of 2 rectangles.
= 625 pie - 840.

Case 2 matches with option D

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Re: In the figure below, rectangles CABD and PMNQ are inscribed in a circl [#permalink]

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New post 04 Sep 2017, 07:46
Ans is D:

HERE I HAVE SOLVED BY ASSUMPTION METHOD ASSUMING THAT THEY ARE FORMING TRIPLET TAKING INTO ACCOUNT THAT L AND B ARE INTEGERS

I tried solving by other methods also but could not solve.

Although by assumption i got 2 values of shaded area of which one is matching my answer, that is option D .

Apart from that we could have once more possible answer , Luckily that is not available
i have shown both answers in my image.
Attachments

IMG_6886.JPG
IMG_6886.JPG [ 1.39 MiB | Viewed 153 times ]


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Re: In the figure below, rectangles CABD and PMNQ are inscribed in a circl   [#permalink] 04 Sep 2017, 07:46
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In the figure below, rectangles CABD and PMNQ are inscribed in a circl

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