Given Information :-
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A cone with dimensions :-
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Another cone with dimensions :-
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To Find :-
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Formula Used :-
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[tex] \qquad \diamond \: \underline{ \boxed{ \red{ \sf T.S.A._{Cone}= \pi r(r+l) }}} \: \star[/tex]
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Solution :-
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For the first cone,
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Since, we don't really have to find the exact values of the surface area, we will let pi remain as a sign itself, this will make the calculations easier.
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[tex] \sf \longrightarrow T.S.A. = \pi \times 3(3 + 7) \\ \\ \\ \sf \longrightarrow T.S.A. = \pi \times 3 \times 10 \: \: \: \\ \\ \\ \sf \longrightarrow T.S.A. =30 \pi \: {cm}^{2} \: \: \: \: \: \: \: \: \\ \\ [/tex]
Now, for the second cone,
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[tex] \sf \longrightarrow T.S.A. = \pi \times 5(5 + 9) \\ \\ \\ \sf \longrightarrow T.S.A. = \pi \times 5 \times 14 \: \: \: \: \\ \\ \\ \sf \longrightarrow T.S.A. =70 \pi \: {cm}^{2} \: \: \: \: \: \: \: \: \: \\ \\ [/tex]
Now, we just have to calculate the ratio of their surface areas, thus,
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[tex] \sf \longrightarrow Ratio = \dfrac{Surface ~area~of~first~cone}{Surface ~area~of~second~cone} \\ \\ \\ \sf \longrightarrow Ratio = \frac{30 \pi \: {cm}^{2} }{70 \pi \: {cm}^{2} } \: \: \: \: \: \: \: \: \: \qquad \qquad \qquad \\ \\ \\ \sf \longrightarrow Ratio = \frac{ 3 \cancel{0 \pi \: {cm}^{2}} }{ 7 \cancel{0 \pi \: {cm}^{2} } } \qquad \qquad \qquad \qquad \\ \\ \\\sf \longrightarrow Ratio = \frac{3}{7} = 3 : 7 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ [/tex]
Thus, the ratio between the surface areas of the cones is 3 : 7.
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[tex] \underline{ \rule{227pt}{2pt}} \\ \\ [/tex]
