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13 May 2022, 01:30
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C
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Difficulty:
75%
(hard)
Question Stats:
33% (01:47) correct
67%
(02:59)
wrong
based on 18
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
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05 Jun 2022, 18:34
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Bunuel
If you like this question, you're a geometry nerd...and I like this question a LOT!!
Let's make the point in the center of the rhombus E.
PNQE is a rectangle.
Since the two diagonals of a rectangle are equal, the minimum value of PQ will occur when we also have the minimum value for NE, and that is when NE is perpendicular to AB.
If we look at triangle ABE, we can find the area in a two different ways. One of those ways is 1/2*AE*BE and the second is 1/2*AB*NE. AE=8, BE=15, and AB=17. Let's set the two areas equal to each other and plug in the values we already know:
\(\frac{1}{2}*8*15 = \frac{1}{2}*17*NE\)
\(60 = 8.5*NE\)
\(NE = \frac{60}{8.5}\)
NE = 7 and a teeny bit
Answer choice C.
