An inverted cone of vertical height …

According to the question, 
Height of an inverted cone = 12 cm
Radius of an inverted cone = 9 cm

{tex}\triangle \mathrm { ABE } \sim \triangle \mathrm { CDE }{/tex}
{tex}\therefore \frac { \mathrm { AB } } { \mathrm { CD } } = \frac { \mathrm { BE } } { \mathrm { DE } }\\ \Rightarrow \frac { 9 } { \mathrm { CD } } = \frac { 12 } { 4 } {/tex} 
{tex}\Rightarrow CD = 3\ cm{/tex}
Slant height of the cone = {tex}\sqrt { 12 ^ { 2 } + 9 ^ { 2 } } = \sqrt { 144 + 81 } = \sqrt { 225 }{/tex}{tex}= 15 \ cm{/tex}
CSA {tex}= \pi r l = \pi \times 9 \times 15 = 135 \pi{/tex} cm²
Slant height of the conical part containing water = {tex}\sqrt { 4 ^ { 2 } + 3 ^ { 2 } }{/tex} {tex}= 5 \ cm{/tex}
CSA of conical part containing water = {tex}\pi \times 3 \times 5{/tex}= 15{tex}\pi{/tex} cm²
Surface area not in contact with water = 135 {tex}\pi{/tex} cm² - 15 {tex}\pi{/tex} cm²
= 120{tex}\pi{/tex} cm²
= {tex}120 \times \frac { 22 } { 7 }{/tex} cm²
{tex}​​​​​​​= 377.14 {/tex}cm²