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805+ (Hard)|   Geometry|      
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Bunuel
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In the figure below, rectangles CABD and PMNQ are inscribed in a circl 28 Aug 2015, 02:23
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95% (hard)

Question Stats:

44% (03:47) correct 56% (03:31) wrong based on 55 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?




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



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D
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In the figure below, rectangles CABD and PMNQ are inscribed in a circl 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)
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In the figure below, rectangles CABD and PMNQ are inscribed in a circl 30 Aug 2015, 00:53
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pluto82
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)
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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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In the figure below, rectangles CABD and PMNQ are inscribed in a circl 04 Sep 2017, 07:46
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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.
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In the figure below, rectangles CABD and PMNQ are inscribed in a circl 18 Aug 2019, 06:25
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Let the length of the rectangle be \(x\) and breadth be \(y\).

From the attached diagram, we can draw two equations

\((x-6)^2+(\frac{y}{2})^2=25^2\) ---------------- (I)
\((\frac{x}{2})^2+(y+6)^2=25^2\)---------------- (II)

Equating (I) and (II)

\(x^2-12x+36+\frac{y^2}{4}=\frac{x^2}{4}+y^2+12y+36\)
\(x^2-\frac{x^2}{4}+\frac{y^2}{4}-y^2=12x+12y\)

\(\frac{3x^2}{4}-\frac{3y^2}{4}=12(x+y)\)
\(\frac{3(x^2-y^2)}{4}=12(x+y)\)

\(\frac{(x+y)(x-y)}{4}=4(x+y)\)
\(x-y=16\) and so

\(x=y+16\)

Substitute for \(x\) in (I)

\((y+16-6)^2+(\frac{y}{2})^2=25^2\)
\((y+10)^2+\frac{y^2}{4}=625\)

\(y^2+20y+100+\frac{y^2}{4}=625\)
\(\frac{5y^2}{4}+20y-525=0\)

\(\frac{y^2}{4}+4y-105=0\)
\(y^2+16y-420=0\)

\((y+30)(y-14)=0\)

Therefore, \(y=14\) and so \(x=16+14=30\)

Area of rectangle is \(14*30=420\)
So combined area of the two rectangles is \(420*2=840\)

Therefore area of the shaded region is \(625\pi-840\)

Answer is (D)
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In the figure below, rectangles CABD and PMNQ are inscribed in a circl 29 Oct 2020, 04:00
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