Question

two concentric circles are of radii 26 cm and 10 cm find the length of the chord of the bigger circle which touches the smaller circle...

Answer 1

Answer:

$\boxed{\sf \: The\:length\:of\:chord\:is\:48\:cm \: }$

Step-by-step explanation:

Let assume that C(O, r) and C(O, R) be two concentric circles with centre O and such that R = 26 cm and r = 10 cm.

Let further assume that AB be a chord of larger circle touches the small circle at C.

Construction :- Join OC

We know, radius and tangent are perpendicular to each other.

So, OC is perpendicular to AB.

Now, as OC is perpendicular to AB and we know, perpendicular drawn from centre bisects the chord.

So, AC = CB

Now, In right-angle triangle OAC

By using Pythagoras Theorem, we have

$\sf \: {OA}^{2} = {OC}^{2} + {AC}^{2}$

$\sf \: {R}^{2} = {r}^{2} + {AC}^{2}$

$\sf \: {26}^{2} = {10}^{2} + {AC}^{2}$

$\sf \: 676 = 100 + {AC}^{2}$

$\sf \: {AC}^{2} = 676 - 100$

$\sf \: {AC}^{2} = 576$

$\bf\implies \:AC = 24\: cm$

Now,

$\bf\implies \:AB = 2AC = 48 \: cm$

Hence,

$\implies\sf \: \boxed{\sf \: The\:length\:of\:chord\:is\:48\:cm \: }$