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