Question

For two concentric circle of radius 25 cm and 7 cm the chord of the larger circle that touches the smaller circle find the length ?

Answer 1

Given:-

• Two conentric circle with center O of radii 25 cm and 7cm.
• A chord(AB) of larger circle touches smaller circle.

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To Find:-

• Length of chord (AB)

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Theoream Used:-

• Pythagorean Theorem
• The line perpendicular from the centre of a circle to the chord bisects the chord.

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Construction:-

• Join OA
• Join OB
• Draw perpendicular bisector from the chord to the centre of a circle (OC) .

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Solution:-

➥ In ∆AOC;

AO(hypotenuse) = Radii of large circle = 25 cm

OC(perpendicular) = Radii of small circle = 7cm

∠AOC = 90°

➥ According to pythagorean theorem in ∆AOC

Hypotenuse² = Perpendicular² + Base²

➟ (AO)² = (OC)² + (AC)²

➟ 25² = 7² + (AC)²

➟ 625 = 49 + AC²

➟ AC² = 625 - 49

➟ AC = √576

➟ AC = √24 × 24

∴AC = 24cm

➥ As we know,

The line perpendicular from the centre of a circle to the chord bisects the chord.

➥ Hence,

AB = 2AC

= 2 × 24

= 48cm

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Answer:-

Length of chord is 48cm.

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Answer 2

${\pink{\tt{\underline{\underline{Given}}}}}$

Two concentric circles with radii 7cm and 25cm .

${\pink{\tt{\underline{\underline{To\ Find}}}}}$

The length of the cord of the larger circle which touches the smaller circle.

${\pink{\tt{\underline{\underline{Solution }}}}}$

On applying Pythagoras theorem in triangle AOC

$\bb{{(AO) }^{2} = {(AC)}^{2}+{(OC)}^{2}}$

$\bb{{(25)}^{2} = {(AC)} ^{2}+{(7)}^{2}}$

$\bb{625 = {(AC)}^{2}+49}$

$\bb{(AC)}^{2} = 625-49}$

$\bb{AC = \sqrt{576}}$

AC = 24

Similarly, if we apply Pythagoras theorem in triangle OCB,

We get CB=24cm

Therefore, AB = AC+CB

AB = 24+24

AB = 48cm

Hence, the length of the cord of the larger circle which touches the smaller circle is 48cm

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