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
