Two circles with different radii have chords AB and CD, such that AB is congruent to CD. Are the arcs intersected by these chords also congruent? Explain.Consider the type of triangle that may be drawn by connecting the endpoints of a chord to the center of a circle.Compare the triangles made by two circles with different radii.
Solution. Let R1 and R2 be the corresponding radii of circles, and O, T be the centers of the circles. Notice that it is implicitly assumed that assuming that each of the AOB and CTD is less of equal to pi. Then we have a isosceles triangles AOB and CTD such that OA=OB=R1, TC=TD=R2. For definiteness assume that R1<R2. We claim that the arc A^B intersected by AB is greater than C^D, the arc intersected by CD: (*) A^B > C^D Indeed, A^B = R1 angle(AOB), C^D = R2 angle(CTD). So instead of (*) we have to prove that (**) angle(AOB) > angle(CTD). Moreover, since each of these angles is less of equal to pi, and so angle(AOB)/2, angle(CTD)/2 is less of equal to pi/2, it suffices to show that (***) tan(AOB/2) > tan(CTD/2). From triangle AOB we get that tan(AOB/2) = AB/(2*R1) tan(CTD/2) = CD/(2*R2) Since AB=CD, and R1<We obtain (***), and therefore (*).
Dear Lear. If two circles have different radii, then congruence of arcs does not follow from the congruence of chords. It is also clear from the formula a=2Rsin(alpha/2), where a is the chord length, alpha is the central angle. The measure of an arc (in radians or degrees) is the measure of the central angle subtended by this arc.
This reply is very confusing. When question just wants to know yes or no if the arcs intercepted by this congruent