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# In the below diagram, points A and B lie on the circle with center O,

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Math Expert
Joined: 02 Sep 2009
Posts: 43304

Kudos [?]: 139252 [0], given: 12781

In the below diagram, points A and B lie on the circle with center O, [#permalink]

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31 Aug 2015, 09:27
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Question Stats:

61% (03:07) correct 39% (03:22) wrong based on 93 sessions

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Knewton Brutal Challenge

In the below diagram, points A and B lie on the circle with center O, and rectangle ABCD and triangle ODC have the same area. If the area of the circle with center O is pi times the area of rectangle ABCD, what is the ratio of the length of the radius of the circle to the length of segment DC?

(A) 2 : 1
(B) $$2 : \sqrt{2}$$
(C) 1 : 1
(D) $$1 : \sqrt{2}$$
(E) 1 : 2

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:

Brutal.GIF [ 3.16 KiB | Viewed 1702 times ]
[Reveal] Spoiler: OA

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Kudos [?]: 139252 [0], given: 12781

Manager
Joined: 29 Jul 2015
Posts: 159

Kudos [?]: 202 [1], given: 59

Re: In the below diagram, points A and B lie on the circle with center O, [#permalink]

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31 Aug 2015, 16:00
1
KUDOS
Bunuel wrote:

Knewton Brutal Challenge

In the below diagram, points A and B lie on the circle with center O, and rectangle ABCD and triangle ODC have the same area. If the area of the circle with center O is pi times the area of rectangle ABCD, what is the ratio of the length of the radius of the circle to the length of segment DC?

(A) 2 : 1
(B) $$2 : \sqrt{2}$$
(C) 1 : 1
(D) $$1 : \sqrt{2}$$
(E) 1 : 2

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
Brutal.GIF

Let OF be the altitude on DC cutting AB at E.
Here AE=EB and DF=FC (perpendicular drawn from the centre to the chord bisects the chord.)
Let R b the radius. OA=OB=R

It is given that area of triangle ODC = area of rectangle ABCD
So,
1/2 ( OF x DC ) = BC x DC

or OF = 2 BC
or OE = BC

In triangle OEB,
OE^2 + EB^2 = OB^2
or BC^2 + (1/2 DC)^2 = R^2
or BC^2 + 1/4 DC^2 = R^2 ...(1)

It is also given that Area of circle is equal to pi times area of ABCD
So,
pi x R x R = pi x BC x DC

or R^2 = BC x DC ...(2)

From (1) & (2)

BC x DC = BC^2 +1/4 DC^2
or 4 x BC x DC = 4 x BC^2 + DC^2
or DC^2 - 4 x BC x DC + 4 x BC^2 = 0
or (DC - 2BC)^2=0
or DC = 2BC
or BC = DC/2

putting this value in (2)

R^2 = (DC^2) / 2
or (R^2) / (DC^2) =1/2
or R/DC = 1/ $$\sqrt{2}$$

Kudos [?]: 202 [1], given: 59

Director
Joined: 13 Mar 2017
Posts: 560

Kudos [?]: 145 [0], given: 64

Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
In the below diagram, points A and B lie on the circle with center O, [#permalink]

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04 Aug 2017, 21:54
Bunuel wrote:

Knewton Brutal Challenge

In the below diagram, points A and B lie on the circle with center O, and rectangle ABCD and triangle ODC have the same area. If the area of the circle with center O is pi times the area of rectangle ABCD, what is the ratio of the length of the radius of the circle to the length of segment DC?

(A) 2 : 1
(B) $$2 : \sqrt{2}$$
(C) 1 : 1
(D) $$1 : \sqrt{2}$$
(E) 1 : 2

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
Brutal.GIF

Let the length of recangle be l and breadth be b
Area of rectangle = lb

Join the points OA . Let radius of circle be r . So, OA = r
Lets draw a line perpendicular from O to DC cutting AB at M and DC at N
So OM = $$\sqrt{r^2-(l^2)/4}$$
MN = b
ON = b+$$\sqrt{r^2-(l^2)/4}$$

Area ODC = 1/2 * (b+$$\sqrt{r^2-(l^2)/4}$$)* l = Area ABCD = lb
b = $$\sqrt{r^2-(l^2)/4}$$
$$b^2 = r^2 - l^2/4$$

Area of circle = pi * area of rectangle
pi * r^2 = pi * lb
r^2 = lb = l* $$\sqrt{r^2-(l^2)/4}$$
$$r^4 = l^2 * (r^2 - l^2/4)$$
$$r^4 = r^2*l^2 -l^4/4$$
$$(2r^2 - l^2)^2 = 0$$
$$r^2/l^2 = 1/2$$
r/l = 1/$$\sqrt{2}$$

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Kudos [?]: 145 [0], given: 64

In the below diagram, points A and B lie on the circle with center O,   [#permalink] 04 Aug 2017, 21:54
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