Question

AB = 14 cm and CD=6 cm are two parallel chords of a circle with centre O. Find the distance between the chords AB and CD. ​

Answer 1

Answer:

The radius of the circle is 10 cm.

Step-by-step explanation:

Given,

AB and CD are two parallel chords of a circle with center O.

Length of AB = 16 cm

Length of CD = 12 cm

We have to find out the radius of the circle.

Solution,

We have drawn the circle for your reference.

And also given length of MN = 14.

Let the radius of the circle be 'r'.

And let the length of ON be 'x'.

AN=

$\frac{ab}{ 2} = \frac{16}{2} = 8cm$

AGAIN,CM=

$\frac{cd}{2} = \frac{12}{2} = 6cm$

Now, In ΔANO,

By Pythagoras theorem, square of the hypotenuse is equal to the sum of squares of the other two sides of triangle.

${an}^{2} + {no}^{2} = {oa}^{2}$

On putting the values, we get;

${8}^{2} + {x}^{2} = {r}^{2} \: equation \: \: 1$

Again,In ∆CMO,

${cm}^{2} + {om}^{2} = {oc}^{2}$

$since \: on = x$

•°•

$om = 14 - x$

${6}^{2} + (14 - x) = {r}^{2} \: \: \: \: equation \: 2$

Equation 1 = Equation 2 (due to radius)

${8}^{2} + {x}^{2} = {6}^{2} + (14 - x)$

Now we solve the equation to get the value of 'x'.

$64 + {x}^{2} = 36 + 196 + {x}^{2} - 28x$

$28x + {x}^{2} - {x}^{2} = 232 - 64$

$28x = 168$

$x = \frac{168}{28} = 6$

Now putting the value of 'x' in equation 1, we get;

$64 + 36 = {r}^{2}$

${r}^{2} = 100$

$r = \sqrt{100} = 10cm$

Hence The radius of the circle is 10 cm.