Question #4126

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.

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Expert's answer

Let R

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=R

TC=TD=R

For definiteness assume that

R

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*R

tan(CTD/2) = CD/(2*R

Since AB=CD, and R

## Comments

Assignment Expert12.07.20, 21:47Dear 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.

Lear10.07.20, 20:46This reply is very confusing. When question just wants to know yes or no if the arcs intercepted by this congruent

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