Question

At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is :

Answer 1

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A chord CD is drawn which is parallel to XY and at a distance of 8cm from A.

As we know that tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴∠OAY=90°

As sum of cointerior angle is 180°.

$\texttt{Therefore,}$

$\textsf{∠OAY+∠OED=180°}$

$\textsf{⇒∠OED=90°}$

$\textsf{AE=8cm(From fig.)}$

Now in △OEC, by pythagoras theorem, OC²= OE² + EC²

$\textsf{⇒EC²= OC²− OE²}$

$\textsf{⇒EC² = (5)² − (3)²}$

$\textsf{⇒EC= 25−9 = 4}$

Therefore,

Length of chord CD = 2×CE (∵perpendicular from centre to the chord bisects the chord)

$\textsf{⇒CD = 2×4 = 8cm}$

Hence the length of the chord CD is 8cm.

Answer 2

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A chord CD is drawn which is parallel to XY and at a distance of 8cm from A.

As we know that tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴∠OAY=90°

As sum of cointerior angle is 180°.

$\texttt{Therefore,}$

$\textsf{∠OAY+∠OED=180°}$

$\textsf{⇒∠OED=90°}$

$\textsf{AE=8cm(From fig.)}$

Now in △OEC, by pythagoras theorem, OC²= OE² + EC²

$\textsf{⇒EC²= OC²− OE²}$

$\textsf{⇒EC² = (5)² − (3)²}$

$\textsf{⇒EC= 25−9 = 4}$

Therefore,

Length of chord CD = 2×CE (∵perpendicular from centre to the chord bisects the chord)

$\textsf{⇒CD = 2×4 = 8cm}$

Hence the length of the chord CD is 8cm.