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# geometry question

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Joined: 11 Apr 2011
Posts: 91

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17 May 2011, 07:29
00:00

Difficulty:

(N/A)

Question Stats:

50% (07:43) correct 50% (02:19) wrong based on 2 sessions

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In the xy-coordinate system, rectangle ABCD is inscribed within a circle having the equation x*x + y*y =25 . Line segment AC is a diagonal of the rectangle and lies on the x-axis. Vertex B lies in quadrant II and vertex D lies in quadrant IV. If side BC lies on line y= 3x +5 , what is the area of rectangle ABCD?

(A) 15
(B) 30
(C) 40
(D) 45
(E) 50

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17 May 2011, 08:34
Concept : area of the quadrilateral ABCD = area of triangle ABC + area of triangle CDA

Area of triangle ABC = 1/2 * AC * (altitude = BE) E be the point of intersection of perpendicular from B on AC.

1 BC intersects the circle at B and C.
Coordinates of B have to be found out.

y= 3x + 5 x^2 + y^2 = 25

gives 10 x^2 + 30x = 0 meaning x = 0 or -3

y = 3 * (-3) + 5 = 4 = length of BE = altitude,

Base = AC = 5+5 = 10.

hence area = 2* [ 1/2 * 4 * 10] = 40. C

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Re: geometry question &nbs [#permalink] 17 May 2011, 08:34
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# geometry question

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