Question

Given only info : length t of the tangent inside the ring.

Find the area inside the ring ie. Enclosed between the two circles.

Answer 1

Let Radius of the outer circle be PN=b and inner circle be PO=a
given MN = t
now PO radius is drawn..so the radius makes a right angle because the line is tangent ..so the radius bisects the tangent line..so half of the tangent line be ON (c)=t/2
now from Pythagoras theorem..
${b}^{2} = {a}^{2} + {c}^{2} \\ or \: {c}^{2} = {b}^{2} - {a}^{2}$

$now \: area \: of \: outer \: circle \: \pi {b}^{2} \: and .. area \: of \: inner \: circle \: \pi {a}^{2}$

now area inside the ring
i.e. enclosed between the two circles
$\pi {b}^{2} - \pi {a}^{2} = \pi( {b}^{2} - {a}^{2} ) = \pi ({ \frac{t}{2} })^{2} = \frac{ {t}^{2} }{4} \pi$

Answer 2

ANSWER.


$\frac{\pi {t}^{2} }4{}$


FORMULA USED ☞

$\pi {r}^{2} \: (area \: of \: circle)$


SOLUTION.


Kindly look in attachment