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]