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>
Answer:
Step-by-step explanation: 8*2x, 8*-14+13=4x-27
16x-112+13=4x-27
16x-99=4x-27
-4x -4x
12x-99=-27
+99 +99
12x = 72
/12 /12
x=6
x/2+2 = 0
x+4/2 = 0
x+4 = 0