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# In the figure above, AB and CD are diameters of the circle with centre

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Manager
Joined: 15 Dec 2015
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GMAT 1: 660 Q46 V35
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In the figure above, AB and CD are diameters of the circle with centre [#permalink]

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04 Feb 2018, 12:41
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In the figure above, AB and CD are diameters of the circle with centre as O, AEB is arc of circle with centre as D and AFB is an arc of the circle with centre as C. If AB = 20cm, what is the area of the shaded region?

(A) 50
(B) 100
(C) 150
(D) 200
(E) 250
[Reveal] Spoiler: OA

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2018-02-05 00_59_51-Microsoft Office 2010.png [ 311.91 KiB | Viewed 353 times ]

Math Expert
Joined: 02 Aug 2009
Posts: 5780
In the figure above, AB and CD are diameters of the circle with centre [#permalink]

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04 Feb 2018, 19:17
DHAR wrote:
In the figure above, AB and CD are diameters of the circle with centre as O, AEB is arc of circle with centre as D and AFB is an arc of the circle with centre as C. If AB = 20cm, what is the area of the shaded region?

(A) 50
(B) 100
(C) 150
(D) 200
(E) 250

lets find dia of arc AEB..
Dia of circle is 20 and triangle ADB will be right isosceles triangle with angle D as 90..
therefore AD or BD will be $$\frac{20}{\sqrt{2}}=10\sqrt{2}$$, this is same as RADIUS of the arc ADB and arc AFB

Now area of unshaded portion AEBO..
area of AEBO = area of segment AEBD - area of triangle ABD..
$$pi*(10\sqrt{2})^2*\frac{90}{360}-\frac{1}{2}*10\sqrt{2}*10\sqrt{2}=50pi-100$$..
so the total unshaded portion's area = 2*(50pi-100)=100pi-200

total area of circle = $$pi*10^2=100pi$$

so area of shaded region = 100pi-(100pi-200)=200

D
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2018-02-05 00_59_51-Microsoft Office 2010.png [ 260.77 KiB | Viewed 319 times ]

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In the figure above, AB and CD are diameters of the circle with centre [#permalink]

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05 Feb 2018, 01:57
DHAR wrote:
In the figure above, AB and CD are diameters of the circle with centre as O, AEB is arc of circle with centre as D and AFB is an arc of the circle with centre as C. If AB = 20cm, what is the area of the shaded region?

(A) 50
(B) 100
(C) 150
(D) 200
(E) 250

Good Question... Kudos to you.. This question tests many Concepts.

For the arc AEB with Centre D, AD = ED = R (Radius of Bigger Circle in for which AEB is an arc).
r = Radius of smaller circle with Centre O.
Now, as AB is the Diameter of Circle with centre O, it will subtend 90 Degree at point C and D.
From Triangle BOD, we have -
$$R^2 = r^2 + r^2$$ -----> $$R = \sqrt{2}r.$$

Now, lets find the Area of AEBOA = Area of SECTOR with Centre D - Area of Triangle ADB
Area of AEBFA = $$\frac{90}{360}*pi*R^2 - \frac{1}{2}*2r*r = pi*\frac{r^2}{2} - r^2$$

Now, multiply this area with 2 to get the total area of arc AEBFA = $$pi*r^2 - 2r^2$$
So, area of shaded region = Area of circle with Centre O - Area of unshaded region AEBFA = $$pi*r^2 - (pi*r^2 - 2r^2)$$
Area of shaded region = $$2r^2$$ = 200.
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In the figure above, AB and CD are diameters of the circle with centre   [#permalink] 05 Feb 2018, 01:57
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