Question

$the \: difference \: between \: areas \\ of \: two \: circles \: is \: 462 \: sq \: units \\ and \: the \: sum \: of \: their \: \\ circumferences \: is \: 132 \: sq \: units \\ find \: the \: radius \: of \: each \: circle$
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Answer 1

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$The \: difference \: between \: areas \: of \: two \\ circles \: is \: 462 \: sq \: units \: and \: the \: sum \: of \\ their \: circumferences \: is \: 132 \: units. \: Find \\ the \: radius \: of \: each \: circle$

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$\huge{\bigstar}{ \green{ \mathfrak{ \underline{ \underline{Answer}}}}}{\bigstar}$

《☆Let the radius are $r_1$and $r_2$☆》

$\boxed{\red{\bold{Given \ area \ relation:}}}$

$\purple{\implies \pi r_1 ^2 - \pi r_2 ^2 = 462}$

$\purple{\implies}\pi(r_1 ^2 -r_2 ^2) = 462$

$\purple{\implies}r_1 ^2 - r_2 ^2 = \displaystyle \frac{462}{\pi}$

$\purple{\implies}(r_1 +r_2)(r_1 - r_2) = 462 \times \displaystyle \frac{7}{22}$

$\purple{\implies}(r_1 + r_2)(r_1 - r_2) = 147 \; \; \; -(i)$

$\boxed{\red{\bold{Given \ circumference \ relation:}}}$

$\red{\implies}\blue{ 2 \pi r_1 + 2 \pi r_2 = 132}$

$\red{\implies}2 \pi ( r_1 + r_2)= 132$

$\red{\implies}r_1 + r_2 = \displaystyle \frac{132}{2 \pi}$

$\red{\implies}r_1 + r_2 = \displaystyle \frac{66}{\pi}$

$\red{\implies}r_1 + r_2 = 66 \times \displaystyle \frac{7}{22}$

$\red{\implies}r_1 + r_2 = 21 \; \; \; -(ii)$

$\boxed{\red{\bold{Putting \ r_1 + r_2 = 21 \ in \ (i):}}}$

$\blue{\implies}21(r_1 - r_2) = 147$

$\blue{\implies}r_1 -r_2 = 7 \; \; \; -(iii)$

$\boxed{\red{\bold{Adding\ (ii) \ , \ (iii):}}}$

$\green{\implies r_1 + r_2 + r_1 - r_2 = 21 + 7}$

$\green{\implies}2r_1 = 28$

$\green{\boxed{\implies{\boxed{r_1 = 14}}}}$

$\green{\boxed{\implies{\boxed{r_2 = 21 - r_1 = 21 - 7 = 14}}}}$

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Answer 2

Answer:

hi bro

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