Question

AB and CD are two parallel chords of a circle which are on opposite sides of the centre such that AB = 10 cm, CD = 24 cm and the distance between AB and CD is 17 cm. Find the radius of the circle.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

AB and CD are two parallel chords of a circle which are on opposite sides of the centre such that AB = 10 cm, CD = 24 cm and the distance between AB and CD is 17 cm.

Construction :-

From centre O, draw OE and OF perpendiculars on AB and CD respectively such that EF = 17 cm. Join OA and OC

We know, perpendicular drawn from centre of the circle bisects the chord.

It means, AE = EB = 5 cm

and

CF = DF = 12 cm

Let assume that OE = x cm, so OF = 17 - x cm

Let further assume that radius of circle be r cm.

Now, In right angle triangle OAE

By using Pythagoras Theorem, we have

$\rm \: {OA}^{2} = {OE}^{2} + {AE}^{2}$

$\rm \: {r}^{2} = {x}^{2} + {5}^{2}$

$\rm \: {r}^{2} = {x}^{2} + 25 - - - - (1)$

Now, In right-angle triangle OCF

By using Pythagoras Theorem, we have

$\rm \: {OC}^{2} = {OF}^{2} + {CF}^{2}$

$\rm \: {r}^{2} = {(17 - x)}^{2} + {12}^{2}$

On substituting the value from equation (1), we get

$\rm \: {x}^{2} + 25= 289 + {x}^{2} - 34x + 144$

$\rm \: 25= 433 - 34x$

$\rm \: 34x= 433 -25$

$\rm \: 34x= 408$

$\rm\implies \:x = 12 \: cm$

On substituting the value of x in equation (1), we get

$\rm \: {r}^{2} = {12}^{2} + 25$

$\rm \: {r}^{2} = 144 + 25$

$\rm \: {r}^{2} = 169$

$\rm \: {r}^{2} = {13}^{2}$

$\rm\implies \:r \: = \: 13 \: cm$

$\rule{190pt}{2pt}$

Additional Information :-

1. Equal chords subtends equal angles at the centre.

2. Equal chords are equidistant from the centre.

3. If a line is perpendicular bisector of the chord, it passes through the centre of circle.

4. If angle subtended at the centre by chords are equal, then chords are also equal.

5. One and only one circle passes through 3 non collinear points.

Answer 2

Answer:

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