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Let ABCD be a rhombus with AC = 16 and BD = 30. Let N be a point on AB

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Joined: 02 Sep 2009
Posts: 59589
Let ABCD be a rhombus with AC = 16 and BD = 30. Let N be a point on AB  [#permalink]

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20 Mar 2019, 23:32
00:00

Difficulty:

65% (hard)

Question Stats:

41% (02:01) correct 59% (03:44) wrong based on 17 sessions

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Let ABCD be a rhombus with AC = 16 and BD = 30. Let N be a point on AB, and let P and Q be the feet of the perpendiculars from N to AC and BD, respectively. Which of the following is closest to the minimum possible value of PQ?

(A) 6.5
(B) 6.75
(C) 7
(D) 7.25
(E) 7.5

Attachment:

c22d4d20155faaf8bd0cf51f26a5795d14b32322.png [ 14.98 KiB | Viewed 449 times ]

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Re: Let ABCD be a rhombus with AC = 16 and BD = 30. Let N be a point on AB  [#permalink]

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21 Mar 2019, 01:28
1
Let AC and BD intersect at O.
ABCDis a rhombus, then AC and BD are perpendicular bisectors.
angle Q = 90, so OPNQ is a rectangle. Since the diagonals of a rectangle are of equal length, PQ = ON, ON Perpendicular AB.

area of rhombus = 1/2 * 16* 30 = 240 each side = 60 each
so
1/2
AB = √8^2+15^2/ 2= 17/2
17*ON/2 = 60
ON = 120/ 17
IMO C ~7
Bunuel wrote:

Let ABCD be a rhombus with AC = 16 and BD = 30. Let N be a point on AB, and let P and Q be the feet of the perpendiculars from N to AC and BD, respectively. Which of the following is closest to the minimum possible value of PQ?

(A) 6.5
(B) 6.75
(C) 7
(D) 7.25
(E) 7.5

Attachment:
c22d4d20155faaf8bd0cf51f26a5795d14b32322.png
Re: Let ABCD be a rhombus with AC = 16 and BD = 30. Let N be a point on AB   [#permalink] 21 Mar 2019, 01:28
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