In a circle, two parallel chords on the same side of a diameter have lengths $4$ cm and $6$ cm. If the distance between these chords is $1$ cm, then the radius of the circle, in cm, is

Let's visualize this problem step by step. We have a circle with two parallel chords on the same side of a diameter, and we need to find the radius.

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When we draw a perpendicular line from the center of a circle to any chord, it always bisects that chord.

This means:

The 6 cm chord gets divided into two equal parts of 3 cm each

The 4 cm chord gets divided into two equal parts of 2 cm each

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Let's define our variables:

Let the radius of the circle = $r$ cm

Let the distance from center to the 6 cm chord = $x$ cm

Since the chords are 1 cm apart, the distance from center to the 4 cm chord = $(x + 1)$ cm

Since both chords are on the same side of the diameter, and the 4 cm chord is farther from the center than the 6 cm chord, we add the distance between them.

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For any chord in a circle, we can form a right triangle where:

Hypotenuse = radius of the circle

One leg = distance from center to chord

Other leg = half the length of the chord

For the 6 cm chord:

$r^2 = x^2 + 3^2$

$r^2 = x^2 + 9$

For the 4 cm chord:

$r^2 = (x+1)^2 + 2^2$

$r^2 = (x+1)^2 + 4$

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Since both expressions equal $r^2$:

$x^2 + 9 = (x+1)^2 + 4$

$x^2 + 9 = x^2 + 2x + 1 + 4$

$x^2 + 9 = x^2 + 2x + 5$

$9 = 2x + 5$

$4 = 2x$

$x = 2$

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Now we can substitute $x = 2$ back into either equation:

$r^2 = x^2 + 9$

$r^2 = 2^2 + 9$

$r^2 = 4 + 9$

$r^2 = 13$

Therefore: $r = \sqrt{13}$ cm

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The radius of the circle is $\sqrt{13}$ cm.