+1439 In the diagram, $ABCD$ is a square. Find $PR.$
M, N, O, L are midpoints of sides.
AB = 12
-482 To find PR, we can break down the problem into smaller steps:
Find the lengths of LM and NO: Since L and M are midpoints of AB and AD, respectively, LM is half the length of AB, which is 12/2 = 6.
Similarly, NO is half the length of BC, which is also 12/2 = 6.
Find the length of MO: Diagonal BD bisects square ABCD, creating two right triangles ABD and CBD. Since AB = BC = 12, both triangles are isosceles right triangles with hypotenuses of length 12.
Using the Pythagorean theorem, we can find the length of BD, the common leg for both triangles: AB^2 + BD^2 = AD^2 = 12^2 + BD^2 = 144.
Solving for BD gives BD = 12√2. MO, being half of diagonal BD, has length 12√2/2 = 6√2.
Find the length of PR: Triangle LMN is a right triangle with right angle at M. LM and NO are legs, and we already found their lengths (6 and 6).
Using the Pythagorean theorem again: LM^2 + NO^2 = PR^2 = 6^2 + 6^2 = 72. Taking the square root of both sides gives PR = 6√2.
Therefore, the length of segment PR is 6*sqrt(2).
