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Logical Geometry Problem

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Logical Geometry Problem

by Anchit 777 » Fri Feb 26, 2010 7:37 am
A parallelogram ABCD has diagonal AC intersected by segment BM at P, such that M is the midpoint of CD. If AP = 65 and PM= 30, then find the largest possible integral value of AB:


A> 124 B> 120 C> 119 D>118
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by sanju09 » Sat Feb 27, 2010 2:15 am
Anchit 777 wrote:A parallelogram ABCD has diagonal AC intersected by segment BM at P, such that M is the midpoint of CD. If AP = 65 and PM= 30, then find the largest possible integral value of AB:


A> 124 B> 120 C> 119 D>118
Drawn the figure as directed :?:

Find, ∆APB ~ ∆CPM. How?

We can now have, AP/CP = PB/PM, or CP × PB = 65 × 30, and we further have

AP/AB = CP/CM, or

65/AB = CP/ (½ AB), or

CP = 65/2, and hence PB = 60.

Now, in ∆APB, AP = 65, and PB = 60, the third side AB CANNOT get to the sum of AP and PB. Why? In other words, AB < 125, Hence, [spoiler]124[/spoiler] is the greatest integer AB.

[spoiler]A[/spoiler]
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