You had the right idea using the Pythagorean theorem to solve for b.
Problem is for that triangle to work, the 5 and the 2√2 would have to switch places. The length of a leg cannot be larger than the length of the hypotenuse for it to truly be a right triangle.
Pythagorean theorem only works for the right triangles. Only way to "solve this problem would be to bring in complex numbers.
5² + b² = (2√2)²
25 + b² = 2²(√2)²
25 + b² = 4(2)
25 + b² = 8
b² = 8 - 25
b² = - 17
b = √-17
b= (√17i)
Then the problem with THIS is a measurement/distance cannot be negative... which goes against exactly what that complex number i is.
If a quadrilateral has exactly 2 lines of symmetry and both are angle bisectors then the figure must be a rhombus because it has exactly 2 pairs of congruent angles.
