Comparing Areas ( 2 )

$\text \color{#3D99F6}{{Area of Blue is :}}$ $10\pi^2 - \text{ Area of White }$

$\text\color{#D61F06}{ {Area of Red Is }:}$ $(6\pi^2 + 4\pi^2 ) - \text { Area of White }$

$\text { Thus, \color{#D61F06}Red = \color{#3D99F6}Blue }$

Considering area of circle of radius 10

$A_1 = B + W_1 + W_2$

$π(10)^2 = B + W_1 + W_2$

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Considering area of circle of radius 6

$A_2 = R_2 + W_2$

$π(6)^2 = R_2 + W_2$

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Considering area of circle of radius 8

$A_3 = R_1 + W_1$

$π(8)^2 = R_1 + W_1$

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Now,

$6^2 + 8^2 = 10^2$

$π(6)^2 + π(8)^2 = π(10)^2$

$(R_2 + W_2) + (R_1 + W_1) = B + W_1 + W_2$

$R_1 + R_2 = B$

As $R_1 + R_2 = R$

Therefore, $\boxed{R = B}$