From a point on a circle, two perpendicular chords are drawn so that they are respectively 6 cm and 3 cm away from the center. Find the lengths of these chords.
Let O be the center of the circle, and let A and B be the points where the perpendicular chords intersect the circle, as shown in the diagram below:
```
A ---- 6 cm ---- O ---- 3 cm ---- B
```
Since the two chords are perpendicular, they intersect at the center O, dividing each other into two segments of equal length. Let x be the length of each segment of the longer chord AO and y be the length of each segment of the shorter chord OB.
By using the Pythagorean theorem for right triangles OAB and OBA, we have:
OA^2 = x^2 + 6^2
OB^2 = y^2 + 3^2
Since OA = OB (both radii of the circle), we have:
x^2 + 36 = y^2 + 9
x^2 - y^2 = 27
(x + y)(x - y) = 27
Since x and y are the lengths of the segments of the chords, the lengths of the chords AO and OB are 2x and 2y, respectively.
To find the lengths of the chords, we need to solve for x and y. By trying different pairs of factors of 27, we find that x = 9 and y = 6 satisfy the conditions.
Therefore, the lengths of the chords are:
Chord AO = 2x = 2(9) = 18 cm
Chord OB = 2y = 2(6) = 12 cm.