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?
- $\dfrac{60}{13}$
- $\dfrac{120}{13}$
- $\dfrac{13}{2}$
- $\dfrac{13}{5}$
For two circles with radii r1=5 and r2=12 intersecting at 90 degrees, the distance between centers d = sqrt(5^2 + 12^2) = 13. The length of the common chord is (2 * r1 * r2) / d = (2 * 5 * 12) / 13 = 120 / 13.
The angle of intersection of the two circles is 90 degrees, which means the radii drawn to the point of intersection are perpendicular to each other. We use the Pythagorean theorem on the triangle formed by the two radii and the line joining the centers to find the distance between centers, d = sqrt(5^2 + 12^2) = sqrt(25 + 144) = 13 cm. The line joining the centers is the perpendicular bisector of the common chord, splitting the isosceles triangle of radii into two identical right-angled triangles. Let the length of the common chord be L; then half of this chord is L/2. In one of these right-angled triangles with hypotenuse 5 and another side as 12, we use the area equivalence to find the altitude to the hypotenuse, but more directly we calculate the chord using similar triangles or trigonometric relations. The area of the 5-12-13 triangle is 0.5 * 5 * 12 = 30. Let h be the perpendicular distance from the intersection point to the line of centers, then 0.5 * 13 * h = 30, so h = 60/13. This perpendicular h is exactly half the length of the common chord. Therefore, the full length of the common chord is 2 * (60/13) = 120/13. The result is 120/13.