Two rings with the same chord have the same area

Question

A ring (annulus) is the region between two circles with the same centre. Draw a chord of the outer circle that just touches the inner circle. Call its length L. Two different rings are drawn: one thin with large circles, one fat with small circles. It happens that their tangent chords have the same length. Prove that the two rings have the same area, and express that area using only L.

Accepted Answer

Both rings have area pi times (L/2) squared, the area of a circle whose diameter is the chord. Set up. Let the outer radius be R and the inner radius r. The chord of length L touches the inner circle at its midpoint M (the radius to the point of tangency is perpendicular to the chord, and a perpendicular from the centre bisects any chord). So the centre O, the midpoint M, and one end P of the chord form a right-angled triangle with OM = r, MP = L/2, and OP = R. Pythagoras. R squared = r squared + (L/2) squared, so R squared minus r squared = (L/2) squared. Area of the ring. The ring's area is…

Two rings with the same chord have the same area

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