I will give brainiest to whoever answers correctly !!
1 answer:
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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
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
<span>AB=5.67 in</span>