$Circles..$

$Solution:$

Let's say, $E$ is the intersection of $AB$ and $CD$ . We know that $\angle CEB$ = $\frac{1}{2}*(Arc CB + Arc AD)$

--> $Arc CB + Arc AD = 180 degrees$ .

This means, the length of $Arc CB + Arc AD$ is simply $half$ of the circumference. $C = (7)(2)(\pi) = 14\pi$ .

Therefore, $\boxed{Arc CB + Arc AD = C/2 = 7\pi units -> 7*\frac{22}{7} -> 22 units}$

$-----------------$

$Note:$

A whole $circumference$ varies directly as the $whole$ sum of the angles in the circle or in short,

$C = 360k$

In our situation, $C = 14\pi --> 14\pi = 360k$

In the other hand, $C = 180k$

by using ratio and proportion, we obtain $C = 7\pi$ or in conclusion, $\boxed{7\pi} -> 7*\frac{22}{7} -> 22 units$ is the $sum$ of the lengths of Arc CB & Arc AY.

we can also do this by visualising , first let the chords perpendicular be the diameters the the length of the sum of arc of the two opposite quadrants formed will be half of the full circle and if we move the point of intersection away from the centre or vice versa , we will se that the length of the opposite arcs formed changes proportionately , one decreases and one increases but the sum remains a constant .