From the given figure, we can see the AB and CD are perpendicular to each other. Let the point of intersection of AB and CD be O as shown in the attached image.
Join AC and BC
Consider the triangle COA,
By Pythagoras theorem, which states
[tex] (Hypotenuse)^{2}=(Base)^{2}+(Perpendicular)^{2} [/tex]
[tex] (AC)^{2}=(CO)^{2}+(OA)^{2} [/tex] (Equation 1)
CO=OD= 3 in (If two congruent circles intersect at two points, then their centers lie on the perpendicular bisector of the common chord)
AO=OB=4 in
From equation 1, we get
[tex] (AC)^{2}=(3)^{2}+(4)^{2} [/tex]
[tex] (AC)^{2}=25 [/tex]
[tex] AC= \sqrt{25}= 5 [/tex] in.
