Question

Two concentric circles are of radii 21 cm and 29cm. Find the length of the chord of the larger circle which touches the smaller circle.​

Answer 1

Answer:

The length of the chord of the larger circle which touches the smaller circle is 40 cm.

Step-by-step explanation:

Let PQ be the chord of the larger circle which touches the smaller circle at the point L . Since PQ is tangent at the point L . Since PQ is tangent at the point L to the smaller circle with centre O.Thus

OL = 21 cm and OP = 29 cm

Therefore, OL is perpendicular to PQ

Since PQ is a chord of the bigger circle and OL is perpendicular PQ.

Therefore, PQ = 2PL

In right angled triangle OPL.

PL =√(OP²-OL²)

$= \sqrt{ {29}^{2} - {21}^{2} } \\ = \sqrt{400} \\ = 20 \: cm$

Therefore, chord PQ = 2 PL = 2×20 cm = 40 cm

So, length of the chord PQ is 40 cm.

This is a problem of circle.

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Answer 2

Answer:

The length of the chord of the larger circle which touches the smaller circle is 40 cm.

Step-by-step explanation:

Let PQ be the chord of the larger circle which touches the smaller circle at the point L . Since PQ is tangent at the point L . Since PQ is tangent at the point L to the smaller circle with centre O.Thus

OL = 21 cm and OP = 29 cm

Therefore, OL is perpendicular to PQ

Since PQ is a chord of the bigger circle and OL is perpendicular PQ.

Therefore, PQ = 2PL

In right angled triangle OPL.

PL =√(OP²-OL²)

\begin{gathered} = \sqrt{ {29}^{2} - {21}^{2} } \\ = \sqrt{400} \\ = 20 \: cm\end{gathered}

=

29

2

−21

2

=

400

=20cm

Therefore, chord PQ = 2 PL = 2×20 cm = 40 cm

So, length of the chord PQ is 40 cm.