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(B) Let the radii of the two circles be $r_1 = 5 \ cm$ and $r_2 = 12 \ cm$. Since the circles intersect at $90^o$,the distance between their centers $

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$\frac{120}{13}$

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(B) Let the radii of the two circles be $r_1 = 5 \ cm$ and $r_2 = 12 \ cm$. Since the circles intersect at $90^o$,the distance between their centers $d$ is given by $d = \sqrt{r_1^2 + r_2^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \ cm$.Let the common chord have length $2x$. The common chord is perpendicular to the line joining the centers.In the triangle formed by the two centers and one of the intersection points,the altitude to the hypotenuse is $x$.Using the area of the triangle: $	ext{Area} = \frac{1}{2} 	imes r_1 	imes r_2 = \frac{1}{2} 	imes d 	imes x$.$\frac{1}{2} 	imes 5 	imes 12 = \frac{1}{2} 	imes 13 	imes x$.$60 = 13x \implies x = \frac{60}{13}$.The length of the common chord is $2x = 2 	imes \frac{60}{13} = \frac{120}{13} \ cm$.

If the angle of intersection at a point where the two circles with radii $5 \ cm$ and $12 \ cm$ intersect is $90^o$,then the length (in $cm$) of their common chord is

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