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?
Answer & explanation
Correct answer: option 1

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.