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04 Feb 2018, 12:41
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
55% (02:51) correct
45%
(02:55)
wrong
based on 88
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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
(A) 50
(B) 100
(C) 150
(D) 200
(E) 250
Attachments
2018-02-05 00_59_51-Microsoft Office 2010.png [ 311.91 KiB | Viewed 10965 times ]
04 Feb 2018, 19:17
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DHAR
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 10837 times ]
rahul16singh28
Joined: 31 Jul 2017
Last visit: 09 Jun 2020
Posts: 428
Given Kudos: 752
Location: Malaysia
Schools: INSEAD Jan '19
GPA: 3.95
WE:Consulting (Energy)
05 Feb 2018, 01:57
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DHAR
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.
20 Mar 2020, 04:17
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Hi
I am able to visualise 2 semi hemispheres here. But unable to split and solve them because of the intersection with each other.
Can someone help?
Thank you.
I am able to visualise 2 semi hemispheres here. But unable to split and solve them because of the intersection with each other.
Can someone help?
Thank you.
