There are two parallel chords measuring 16 cm and 12 cm, both situated on the same side of the center of a circle. The space between the two chords is 2 cm. What is the radius of the circle?

Correct option is B

Given :

Length of first chord = 16 cm
Length of second chord = 12 cm
Distance between the two parallel chords = 2 cm
Both chords are on the same side of the center

Formula Used :
For a circle of radius ( r ), distance of a chord of length ( l ) from the center is
d $= \sqrt{r^2 - \left(\frac{l}{2}\right)^2}$

Solution :

Let the distances of the chords from the center be ( $d_1$ ) and ( $d_2$ ).

For chord of length 16 cm:
$d_1 = \sqrt{r^2 - 8^2} = \sqrt{r^2 - 64}$

For chord of length 12 cm:
$d_2 = \sqrt{r^2 - 6^2} = \sqrt{r^2 - 36}$

Given that the distance between the two chords is 2 cm:
$d_2 - d_1 = 2$

$\sqrt{r^2 - 36} - \sqrt{r^2 - 64} = 2$

Square both sides:
r $^2 - 36 + r^2 - 64 - 2\sqrt{(r^2 - 36)(r^2 - 64)} = 4$

$2r^2 - 100 - 4 = 2\sqrt{(r^2 - 36)(r^2 - 64)}$

$2r^2 - 104 = 2\sqrt{(r^2 - 36)(r^2 - 64)}$

$r^2 - 52 = \sqrt{(r^2 - 36)(r^2 - 64)}$

Square again:
( $r^2 - 52)^2 = (r^2 - 36)(r^2 - 64)$

$r^4 - 104r^2 + 2704 = r^4 - 100r^2 + 2304$

$-104r^2 + 2704 = -100r^2 + 2304$

$4r^2 = 400$

$r^2 =$ 100
r = 10 cm

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There are two parallel chords measuring 16 cm and 12 cm, both situated on the same side of the center of a circle. The space between the two chords is 2 cm. What is the radius of the circle?

8 cm

10 cm

12 cm

15 cm

Correct option is B

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