1) The document describes a geometry problem where a square is drawn with labeled line segments of lengths AP=7, BP=35, CP=49, and the variable DP=x. Altitudes are drawn from point P to each side. 2) Using the Pythagorean theorem across various triangles formed by the lines and altitudes sets up four equations. 3) Combining and subtracting the equations isolates x2, allowing x to be solved for as 35.

1. Jillian Fluet’s POTW solution

2. A B C D E F G H Draw a square and label the line segments so AP=7 BP=35 CP=49 and DP=x. Then drop altitudes from point P to each side of the square. Label the point on AB, E, the point on BC, F, the point on CD, G, and the point on DA, H. P EP, AH, BF have a length of a GP, DH, CF have a length of s-a EP, BE, CG have a length of b HP, AE, DG have a length of s-b 1 2 3 4

3. Then use the Pythagorean Theorem to solve for x. Triangle 1: a ²+b² =(35)² Triangle 2: (s-a) ²+b² =49² Triangle 3: a ²+(s-b)² =7² Triangle 4: (s-a) ²+(s-b)² =x² Add the 2 nd and 3 rd triangles together and subtract the sum of the 1 st and 4 th triangles. Then you get: 0= 49²+7²-35²-x² Solve: 0= 49²+7²-35²-x² x ² =49²+7²-35² x² =35² x=35 or x= (­35) x=35 because you can’t have a negative length