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(A) Let the radii of the two circles be $r_1$ and $r_2$ cm.Since the circles touch externally,the distance between their centres is $r_1 + r_2 = 14$.T

Vedclass · Accepted Answer

$3$

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(A) Let the radii of the two circles be $r_1$ and $r_2$ cm.Since the circles touch externally,the distance between their centres is $r_1 + r_2 = 14$.Thus,$r_2 = 14 - r_1$.The sum of their areas is given by $\pi r_1^2 + \pi r_2^2 = 130 \pi$.Dividing by $\pi$,we get $r_1^2 + r_2^2 = 130$.Substituting $r_2 = 14 - r_1$ into the equation:$r_1^2 + (14 - r_1)^2 = 130$$r_1^2 + 196 + r_1^2 - 28r_1 = 130$$2r_1^2 - 28r_1 + 66 = 0$Dividing by $2$,we get $r_1^2 - 14r_1 + 33 = 0$.Factoring the quadratic equation:$(r_1 - 11)(r_1 - 3) = 0$.So,$r_1 = 11$ or $r_1 = 3$.If $r_1 = 3$,then $r_2 = 14 - 3 = 11$. If $r_1 = 11$,then $r_2 = 14 - 11 = 3$.The radius of the smaller circle is $3 	ext{ cm}$.

Two circles touch each other externally. The sum of their areas is $130 \pi \text{ cm}^2$ and the distance between their centres is $14 \text{ cm}$. The radius of the smaller circle is . . . . . . $\text{cm}$.

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