设直角三角形直角边长为a,b.
ab/2=a+b+√(a^2+b^2)
≥2√(ab)+√(2ab),
ab-(4+2√2)√(ab)≥0,
√(ab)≥4+2√2,
ab/2≥12+8√2,
a=b时,等号成立,所以面积最小值为12+8√2.