Question

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In the figure, given below, AB and CD are two parallel chords and O is the center. If the radius of the circle is 15cm , find the distance MN between the two chords of lengths 24cm and 18cm respectively.
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Answer 1

Given – AB = 24 cm, cd = 18 cm

⇒ AM = 12 cm, CN = 9 cm Also, OA = OC = 15 cm

Let MO = y cm, and ON = x cm In right angled ∆AMO

(OA)2=(AM)2+(OM)2

⇒ 152=122+y2

⇒ y2=152-122

⇒ y2=225-144

⇒y2=81

⇒ y = 9 cm

In right angled ΔCON

(OC)2=(ON)2+(CN)2

⇒

152=x2+92

⇒

x2=152-92

⇒

x2=225-81

⇒ x2=144

 ⇒ y = 12 cm

Now, MN = MO + ON

= y + x

= 9 cm +12 cm

= 21 cm

hope it's help u

Answer 2

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

Given that,

• O be the center of circle having radius, r = 15 cm.

• AB and CD are two parallel chords on the either side of the center of circle such that AB = 24 cm and CD = 18 cm

• OM and ON are perpendicular drawn from center O on AB and CD respectively.

Now, we have to find distance between two parallel chord.

Now,

We know, perpendicular drawn from centre bisects the chord.

So, AM = MB = $\rm \: \dfrac{1}{2} AB$ = 12 cm

Also, CN = ND = $\rm \: \dfrac{1}{2} CD$ = 9 cm

Now, Join OA and OC.

So, OA = OC = r = 15 cm

Now, In $\triangle$ OAM,

By using Pythagoras Theorem, we have

$\rm \: {OA}^{2} = {AM}^{2} + {OM}^{2}$

$\rm \: {15}^{2} = {12}^{2} + {OM}^{2}$

$\rm \: 225 = 144 + {OM}^{2}$

$\rm \: {OM}^{2} = 225 - 144$

$\rm \: {OM}^{2} = 81$

$\rm \: {OM}^{2} = {9}^{2}$

$\bf\implies \:OM \: = \: 9 \: cm$

Now, In $\triangle$ OCN

By using Pythagoras Theorem, we have

$\rm \: {ON}^{2} + {CN}^{2} = {OC}^{2}$

$\rm \: {ON}^{2} + {9}^{2} = {15}^{2}$

$\rm \: {ON}^{2} + 81 = 225$

$\rm \: {ON}^{2} = 225 - 81$

$\rm \: {ON}^{2} = 144$

$\rm \: {ON}^{2} = {12}^{2}$

$\bf\implies \:ON \: = \: 12 \: cm$

Now,

$\rm \: MN$

$\rm \: = \: OM + ON$

$\rm \: = \: 9 + 12$

$\rm \: = \: 21 \: cm$

Hence,

$\bf\implies \:MN \: = \: 21 \: cm$