Question

7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the
larger circle which touches the smaller circle.
8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that
AB+CD=AD+BC​

Answer 1

${\tt{\blue{\underline{\underline{\huge{Question \: : }}}}}}$

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

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Let two concentric circles be centered at point O.

Let PQ be the chord of the larger circle which touches the smaller circle at point A.

Therefore, PQ is tangent to the smaller circle.

Now, OA Ʇ PQ [Since OA is the radius of the circle]

Apply Pythagoras theorem in ΔOAP, we get

OA2 + AP2 = OP2

=> 32 + AP2 = 52

=> 9 + AP2 = 25

=> AP2 = 25 - 9

=> AP2 = 16

=> AP = √16

=> AP = 4

In ΔOAP,

Since OA Ʇ PQ,

=> AP = AQ [Perpendicular from the center of the circle bisects the chord]

So, PQ = 2 * AP = 2 * 4 = 8

Hence, the length of the chord of the larger circle is 8 cm.