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28 Aug 2015, 02:23
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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) \(625\pi-960\)
Kudos for a correct solution.
JD_NONSBD_Hard004.GIF [ 4.72 KiB | Viewed 5330 times ]
(A) \(625\pi-420\)
(B) \(625\pi-500\)
(C) \(625\pi-640\)
(D) \(625\pi-840\)
(E) \(625\pi-960\)
Kudos for a correct solution.
Attachment:
JD_NONSBD_Hard004.GIF [ 4.72 KiB | Viewed 5330 times ]
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)
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 [ 12.28 KiB | Viewed 4787 times ]
rohitt911
Joined: 17 Feb 2015
Last visit: 29 Sep 2017
Posts: 14
Given Kudos: 51
Schools: Alberta"19 Degroote McMaster"19 HEC Montreal '20 Ivey '19 Schulich Jan'19 Rotman '19 Desautels '19
WE:Sales (Consulting)
30 Aug 2015, 00:53
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pluto82
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
