The given figure
There is a circle of radius of 10
Then the hypotenuse of the right triangle is the radius of the circle = 10
Since one leg of the triangle is 6
Then we can use the Pythagoras Theorem to find AE, then multiply it by 2 to find AB
Since < AEM = 90 degrees, then
[tex]AM^2=AE^2+EM^2[/tex]M is the center of the circle
Since AM = 10, ME = 6, then
[tex]\begin{gathered} 10^2=AE^2+6^2 \\ 100=AE^2+36 \end{gathered}[/tex]Subtract 36 from both sides
[tex]\begin{gathered} 100-36=AE^2+36-36 \\ 64=AE^2 \end{gathered}[/tex]Take a square root for both sides
[tex]\begin{gathered} \sqrt[]{64}=\sqrt[]{AE^2} \\ 8=AE \end{gathered}[/tex]Since DC is perpendicular to AB and passing through the center of the circle, then
DC bisects AB, which means E is the mid-point of AB
AE = EB = 8
AB = 8 + 8 = 16
AB = 16
