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What is the ratio of the surface areas of two cones if the radius of one is 3 en and the slant height is 7 cm, and the other has a radius of 5 cm and a slan Height of 9 cm?​

Sagot :

Given Information :-

A cone with dimensions :-

  • Radius = 3 cm
  • Slant height ( l ) = 7 cm

Another cone with dimensions :-

  • Radius = 5 cm
  • Slant height = 9 cm

To Find :-

  • The ratio of their total surface area

Formula Used :-

[tex] \qquad \diamond \: \underline{ \boxed{ \red{ \sf T.S.A._{Cone}= \pi r(r+l) }}} \: \star[/tex]

Solution :-

For the first cone,

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.

[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,

[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,

[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.

[tex] \underline{ \rule{227pt}{2pt}} \\ \\ [/tex]

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