A chord of length 48 cm is at a distance of 7 cm from the centre of the circle. What is the length of the chord of the same circle which is at a distance of 15 cm from the centre of the circle?

40 cm

MP = PN = \(\frac{MN}{2}\) = \(\frac{48}{2}\) = 24 cm and

The distance from the center as d1 = 15 cm.

From the pyhtagoras theorem the radius r of the circle is

= r = √(\( {a }^{ 2} \) + \( {d1 }^{ 2} \))

= √(\( {24 }^{ 2} \) + \( {7 }^{ 2} \))

= \(\sqrt {625 }\)

= 25 cm.

For the unknown chord, let's denote half of its length as b, and the known distance from the center as d2 = 15 cm.

Again using the pythagoras theorem, we find \(\frac{x}{2}\) as,

= \(\frac{x}{2}\) = √(\( {r }^{ 2} \) + \( {d2 }^{ 2} \))

= √(\( {25 }^{ 2} \) - \( {15 }^{ 2} \))

= \(\sqrt {400 }\)

= 20 cm.

Full length of chord = 2 x 20 = 40 cm.

Therefore, the length of the chord of the same circle which is at a distance of 15 cm from the center of the circle is 40 cm.