Question

find the radius of the circle whose area is equal to the sum of the areas of three circle with radii 3 cm 4 cm and 12 CM

Answer 1

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Answer 2

$Given:-$

• $\: \: Radius \: \: circle:- \: \: 3 \: cm, \: \: 4 \: cm, \: \: and \: \: 12 \: cm.$

• $Area \: \: of \: \: bigger \: \: circle \: \: = \: \: sum \: \: area \: \: of \: \: circle \: \: having \: \: radiii \: \: 3 \: cm, \: \: 4 \: cm, \: \: and \: \: 12 \: cm.$

$To \: \: Find:-$

• $Radius \: \: of \: \: bigger \: \: circle.$

$Solutions:-$

• $Given \: \: radius \: \: of \: \: different \: \: circle \: \: = \: \: 3\: cm, \: \: 4 \: cm, \: \: and \: \: 12 \: cm$

$Area \: \: of \: \: circle \: \: = \: \: {π}{r}^{2}$

$Area \: \: of \: \: 1st \: \: circle \: \: having \: \: Radius \: \: 3 \: cm:-$

$⇢ Area \: \: of \: \: 1st \: \: circle \: \: \frac{22}{7} \: × \: {3}^{2}$

$⇢ Area \: \: of \: \: 1st \: \: circle \: \: \frac{22}{7} \: × \: {9}$

$⇢ Area \: \: of \: \: 1st \: \: circle \: \: \frac{198}{7} \: {cm}^{2}$

$Area \: \: of \: \: 2nd \: \: circle \: \: having \: \: Radius \: \: 4 \: cm:-$

$⇢ Area \: \: of \: \: 2nd \: \: circle \: \: \frac{22}{7} \: × \: {4}^{2}$

$⇢ Area \: \: of \: \: 1st \: \: circle \: \: \frac{22}{7} \: × \: {16}$

$⇢ Area \: \: of \: \: 1st \: \: circle \: \: \frac{352}{7} \: {cm}^{2}$

$Area \: \: of \: \: 3rd \: \: circle \: \: having \: \: Radius \: \: 12 \: cm:-$

$⇢ Area \: \: of \: \: 1st \: \: circle \: \: \frac{22}{7} \: × \: {12}^{2}$

$⇢ Area \: \: of \: \: 1st \: \: circle \: \: \frac{22}{7} \: × \: {144}$

$⇢ Area \: \: of \: \: 1st \: \: circle \: \: \frac{3168}{7} \: {cm}^{2}$

$Now,$

$\: \: \: \: \: it \: \: given \: \: that \: \: the \: \: area \: \: of \: \: bigger \: \: circle \: \: is \: \: equal \: \: to \: \: the \: \: sum \: \: of \: \: area \: \: of \: \: circle \: \: having \: \: Radius \: \: 3 \: cm, \: \: 4 \: cm, \: \: and \: \: 12 \: cm.$

$so,$

$\: \: \: \: \: Area \: \: bigger \: \: circle \: \: = \: \: area \: \: of \: \: 1st \: \: circle \: \: + \: \: area \: \: of \: \: 2nd \: \: circle \: \: + \: \: area \: \: of \: \: 3rd \: \: circle.$

$⇢ Area \: \: bigger \: \: circle \: \: = \: \: \frac{198}{7} \: + \: \frac{352}{7} \: + \: \frac{3168}{7}$

$⇢ Area \: \: bigger \: \: circle \: \: = \: \: \frac{{198} \: + \: {352} \: + \: {3168}}{7}$

$⇢ Area \: \: bigger \: \: circle \: \: = \: \: \frac{3718}{7} \: {cm}^{2}$

$⇢ Area \: \: bigger \: \: circle \: \: = \: \: {π}{r}^{2}$

$⇢ \frac{3718}{7} \: \: = \: \: {π}{r}^{2}$

$⇢ \frac{3718}{7} \: \: = \: \: \frac{22}{7} \: × \: {r}^{2}$

$⇢ \frac{3718}{7} \: × \: \: \frac{7}{22} \: \: = \: \: {r}^{2}$

$⇢ \frac{3718}{22} \: \: = \: \: {r}^{2}$

$⇢ {169} \: \: = \: \: {r}^{2}$

$⇢ {r} \: \: = \: \: \sqrt{169}$

$⇢ {r} \: \: = \: \: {13} \: cm.$

$So,$

$\: \: \: \: \: the \: \: radius \: \: of \: \: the \: \: circle \: \: is \: \: 13 \: cm, \: \: whose \: \: are \: \: is \: \: equal \: \: to \: \: the \: \: sum \: \: of \: \: the \: \: area \: \: of \: \: three \: \: circle \: \: with \: \: 3 \: cm, \: \: 4 \: cm, \: \: and \: \: 12 \: cm.$

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