+177 A rectangle measures 6 centimeters by 25 centimeters. The probability that a point randomly chosen inside the rectangle is closer to a side of length 25 than a side of length 6 is \(\frac p q\), where \(p\) and \(q\) are relatively prime positive whole numbers. What is the sum of \(p\) and \(q\)?
+128475 I AM NOT totally sure about this one, TGO, but it's an interesting problem......here's my best attempt
Look at the image below :
Note that any point lying within triangles ADE and BCF will be CLOSER to a side of 6 than to a side of 25
Each of these triangles has a base of 6 and a height of 3
So....their total area is 6 * 3 =18
And the total area of the rectangle = 25 * 6 =150
So....the region that a point could fall into and be closer to a side of 25 than a side of 6 has an area of : 150 -18 = 132
So.... p / q = 132 /150 = 22 / 25
So..... p + q = 22 + 25 = 47
