Question

in the given figure O is the centre of the circle chord ab is parallel to code CD AB equal to 64 cm is equal to 48 cm and radius of circle is equal to 40 cm find the distance between the two chords​

Answer 1

Distance between the two chords is 56 cm.

Step-by-step explanation:

Given:

Here O is the centre of the circle.

Chord AB is parallel to chord CD.

Let radius OA = OC = 40 cm

Chord AB = 64 cm and CD = 48 cm

Draw OP and OQ

Therefore,

$AP = \frac{1}{2}AB$

$AP = \frac{1}{2}\times 64$

AP = 32 cm.

$CQ = \frac{1}{2} CD$

$CQ = \frac{1}{2} \times 48$

CQ = 24 cm.

In ΔOAP,

$OP^2+ AP^2 = OA^2$

$OP^2 = OA^2 - AP^2$

$OP^2 = 40^2 - 32^2$       [given]

$OP^2 = 1600 -1024$

$OP^2=576$

$OP = \sqrt{576}$

OP = 24 cm.

In ΔOQD,

$OQ^2 + CQ^2 = OC^2$

$OQ^2 = OC^2 - CQ^2$

$OQ^2 = 40^2+ 24^2$

$OQ^2=1600-576$

$OQ^2=1024$

$OQ = \sqrt{1024}$

OQ = 32 cm.

Now,

PQ = OP + OQ = 24 + 32 = 56 cm.

Hence, distance between the two chords is 56 cm.