Suppose ABCD is a rectangle. Find AB and AD if point M is the midpoint of BC,AM ⊥ MD , and the perimeter of ABCD is 34 in.. Someone answered this question but Ad is not 4 and AB is not 9. . Thank you

we know that
Perimeter=2AB+2AD=34 in-----> AB+AD=17-----> AB=17-AD-----> equation 1

MA=MD
MA²=AB²+(AD/2)²
for the triangle AMD
AD²=[MA²+MD²]----> AD²=2*[MA²]----> AD²=2*[AB²+(AD/2)²]---> equation 2

I substitute 1 in 2 AD²=2*[(17-AD)²+(AD/2)²]----> AD²=2*[289-34AD+AD²+0.25AD²] AD²=578-68AD+2.50AD²--------> 1.50AD²-68AD+578 1.50AD²-68AD+578=0  using a graph tool to solve the quadratic equation see the attached figure AD1=11.33 in AD2=34 in----------is not solution because (AB+AD=17) Solution is AD=11.33 in AB=17-11.33--------> 17-11.33-----> AB=5.67 in  the answer is AD=11.33 in AB=5.67 in